ࡱ> Rk#Fg&} u t  lbjbj [hhSeD2jj8~nE )))*III|||||||(p|JEIJJ|)*}XXXJ&)*|XJ|XXvlz*G2Lw*|}0~)xQTzzzIIXI ISIII||VIII~JJJJIIIIIIIIIj s: Section 3.1: Derivatives of Polynomials and Exponential Functions SOLs: APC.5: The student will apply formulas to find the derivative of algebraic, trigonometric, exponential, and logarithmic functions and their inverses. APC.6: The student will apply formulas to find the derivative of the sum of elementary functions. Objectives: Students will be able to: Find derivatives of polynomial functions Find derivatives of natural exponent functions Vocabulary: Natural number, e (approximately 2.71828)  is the number such that limit of (eh  1) / h = 1 as h ! 0 Derivative of a function  is the slope of the function at that point and is equal to the slope of the tangent line at that point. It is also a rate of change with respect to the variable the derivative is taken in respect to. Key Concept:  EMBED PowerPoint.Show.8  Linear operators: An operator L is linear if L(ku) = kL(u) and L(u + v) = L(u) + L(v) Note: differentiation is a linear operator! Constant function rule: If f(x) = k where k is a constant, then for any x, f(x) = 0 Identity function rule: If f(x) = x, then for any x, f(x) = 1 Constant multiple rule: If k is a constant and f(x) is a differentiable function, then for any x, (kf)(x) = k(f(x)) Sum & Difference rule: If f and g are differential functions, then (f+ g)(x) = f(x) + g(x) and (f g)(x) = f(x) g(x) Power rule: If f(x) = xn where n is a positive integer, then f(x) = nxn-1  EMBED PowerPoint.Show.8  Concept Summary: Differentiation operators can be distributed across constant multiples, addition and subtraction Homework: pg 191 192: 4, 7, 8, 11-13, 21-23, 27, 34, 45, 54 Read: Section 3.2 Section 3.2: The Product and Quotient Rules SOLs: APC.5: The student will apply formulas to find the derivative of algebraic, trigonometric, exponential, and logarithmic functions and their inverses. APC.6: The student will apply formulas to find the derivative of the sum, product, quotient, inverse and composite (chain rule) of elementary functions. Objectives: Students will be able to: Use product and quotient rules of differentiation Vocabulary: Function an independent variable (x or t) yields only one dependent variable value Key Concepts:  EMBED PowerPoint.Show.8  Product Rule: Let  EMBED Equation.3 , then  EMBED Equation.3  Proof:  EMBED Equation.3  Add & subtract:  EMBED Equation.3   EMBED Equation.3  Quotient Rule: Let  EMBED Equation.3 , then  EMBED Equation.3  Proof: Add & subtract:  EMBED Equation.3    EMBED PowerPoint.Show.8  Product & Quotient Rule Practice: 1.  EMBED Equation.3  2.  EMBED Equation.3  3.  EMBED Equation.3  4.  EMBED Equation.3  5.  EMBED Equation.3  6.  EMBED Equation.3  7.  EMBED Equation.3  8.  EMBED Equation.3  9.  EMBED Equation.3  10.  EMBED Equation.3  11.  EMBED Equation.3  12.  EMBED Equation.3  13. Find the equation of the tangent line to the curve  EMBED Equation.3  at x=4 Homework Problems: pg: 197-198: 3, 5-7, 9, 13, 14, 17, 31, 35 Read: Section 3.3 Section 3.3: Rates of Change in the Natural and Social Sciences SOLs: none Objectives: Students will be able to: Understand the mathematical modeling process of derivatives (rates of changes) in the real world Vocabulary: Mathematical Model an equation that models a process (usually in the real world) Key Concept:  EMBED Equation.3  is the rate of change of  EMBED Equation.3  with respect to  EMBED Equation.3 ; also the instantaneous rate of change.  EMBED Equation.3  is the average rate of change of  EMBED Equation.3  with respect to  EMBED Equation.3  over the interval [x1, x2] Applications: Particle motion, water flow, populations, etc Particle or Rectilinear Motion describes motion of an object along a line. s(t) - Position Function gives the position of an object at time t. The displacement over an interval [a, b] is s(b) s(a). Distance traveled must take the sign of the velocity into account, so distance traveled is |s(b) s(a)| - this means that you add up all the distances the particle travels left and right or up and down. Average velocity over [a, b] is  EMBED Equation.DSMT4 ; Instantaneous velocity at time t is defined as v(t) = s(t) Speed is |v(t)| Acceleration at time t is defined as a(t) = v(t) = s(t) 1. A particle is moving along an axis so that at time t its position is f(t) = t - 6t + 6 feet. What is the velocity at time t? What is the velocity at 3 seconds? Is the particle moving left or right at 3 seconds? 2. A stone is thrown upward from a 70 meter cliff so that its height above ground is f(t) = 70 + 3t - t. What is the velocity of the stone as it hits the ground? 3. A particle moves according to the position function, s(t) = t - 9t + 15t + 10, te"0 where t is in seconds and s(t) is in feet. Find the velocity at time t. When is the particle at rest? When is the particle moving to the right? Find the total distance traveled in the first 8 seconds. Draw a diagram to illustrate the particles motion. 4. Water is flowing out of a water tower in such a way that after t minutes there are 10,000 10t t gallons remaining. How fast is the water flowing after 2 minutes? 5. A space shuttle is 16t + t meters from its launch pad t seconds after liftoff. What is its velocity after 3 seconds? 6. The numbers of yellow perch in a heavily fished portion of Lake Michigan have been declining rapidly. Using the data below, estimate the rate of decline in 1996 by averaging the slopes of two secant lines. t , years199319941995199619971998P(t) , population (millions)4.24.03.73.63.33.1 Homework Problems: pg: 208-210: 8,9,10 Read: Section 3.4 Section 3.4: Derivatives of Trigonometric Functions SOLs: APC.5: The student will apply formulas to find the derivative of algebraic, trigonometric, exponential, and logarithmic functions and their inverses. APC.6: The student will apply formulas to find the derivative of the sum, product, and quotient of elementary functions. Objectives: Students will be able to: Use the differentiation rules of trigonometric functions Vocabulary: none new Key Concept:  EMBED PowerPoint.Show.8  Use the quotient rule to find these four trig derivatives. 3.  EMBED Equation.3  4.  EMBED Equation.3  5.  EMBED Equation.3  6.  EMBED Equation.3  Practice problems and more limits: 1. y = sin x  cos x 2.  EMBED Equation.3  3. y = sin (/4) 4. y = x sin x 5. y = x + 2x cos x 6. EMBED Equation.3  7.  EMBED Equation.3  8.  EMBED Equation.3  9.  EMBED Equation.3  10.  EMBED Equation.3  11.  EMBED Equation.3  12.  EMBED Equation.3  13.  EMBED Equation.3  14.  EMBED Equation.3  15.  EMBED Equation.3  A particle moves along a line so that at any time t>0 its position is given by x(t) = 2t + cos(2t). Find the velocity at time t. Find the speed (|v|) at t = . What are the values of t for which the particle is at rest? Homework  Problems: pg: 216  217: 1-3, 6, 9-11, 18, 29, 41 Read: Section 3.5 Section 3.5: The Chain Rule SOLs: APC.5: The student will apply formulas to find the derivative of algebraic, trigonometric, exponential, and logarithmic functions and their inverses. APC.6: The student will apply formulas to find the derivative of the sum, product, quotient, inverse and composite (chain rule) of elementary functions. Objectives: Students will be able to: Use the chain rule to find derivatives of complex functions Vocabulary: none new Key Concept:  EMBED PowerPoint.Show.8   EMBED PowerPoint.Show.8  Differentiating a Composite Function: If David can type twice as fast as Mary and Mary can type three times as fast as Joe, then David can type 23=6 times as fast as Joe. This is basically the Chain Rule. If y changes  EMBED Equation.3  times as fast as u and u changes  EMBED Equation.3  times x, then y changes  EMBED Equation.3 times as fast as x. Chain Rule: If f and g are both differentiable functions and h(x) = f(g(x)), then h(x) = f(g(x)) g(x) Examples: 1. If y = (2x - 4x + 1)60, find  EMBED Equation.3 . First decompose the function. Let y = u60, and u = 2x - 4x + 1 . Now find  EMBED Equation.3  and  EMBED Equation.3  , then multiply to find the answer. 2. Find  EMBED Equation.3  of  EMBED Equation.3  Practice: 1.  EMBED Equation.3  2.  EMBED Equation.3  3.  EMBED Equation.3  4.  EMBED Equation.3  5.  EMBED Equation.3  6.  EMBED Equation.3  7.  EMBED Equation.3  8.  EMBED Equation.3  9.  EMBED Equation.3  10.  EMBED Equation.3  Assume that f(x) and g(x) are differentiable functions about which we know information about a few discrete data points. The information we know is summarized in the table below: xf(x) f(x) g(x) g(x) -24-156-13-5170-6-38-5116232-151? Use your differentiation rules to determine each of the following. If p(x) = xf(x), find p(2) If q(x) = 3f(x)g(x), find q(-2) If r(x) = f(x) / (5g(x)) find r(0) If s(x) = f(g(x)), find s(1) If t(x) = (2 f(x)) / g(x) and t(2) = 4, find g(2) Homework Problems: pg 224 - 227: 3, 4, 7, 8, 11, 14, 15, 22, 29, 32, 43, 67 Read: Section 3.6 Section 3.6: Implicit Differentiation SOLs: APC.7: The student will find the derivative of an implicitly defined function. Objectives: Students will be able to: Use implicit differentiation to solve for dy/dx in given equations Use inverse trig rules to find the derivatives of inverse trig functions Vocabulary: Implicit Differentiation differentiating both sides of an equation with respect to one variable and then solving for the other variable prime (derivative with respect to the first variable) Orthogonal curves are orthogonal if their tangent lines are perpendicular at each point of intersection Orthogonal trajectories are families of curves that are orthogonal to every curve in the other family (lots of applications in physics (example: lines of force and lines of constant potential in electricity) Key Concept:  EMBED PowerPoint.Show.8  Up to now, we have worked explicitly, solving an equation for one variable in terms of another. For example, if you were asked to find  EMBED Equation.3  for 2x + y = 4, you would solve for y and get  EMBED Equation.3  and then take the derivative. Sometimes it is inconvenient or difficult to solve for y. In this case, we use implicit differentiation. You assume y could be solved in terms of x and treat it as a function in terms of x. Thus, you must apply the chain rule because you are assuming y is defined in terms of x. Differentiating with respect to x:  EMBED Equation.3  variables agree  EMBED Equation.3  variables disagree  EMBED Equation.3   variables disagree variables agree  EMBED Equation.3   variables disagree Consider the problem, find  EMBED Equation.3 for  EMBED Equation.3 . Treat y as a quantity in terms of x so   EMBED Equation.3   Different Same   EMBED Equation.3  Now solve for  EMBED Equation.3 .  EMBED Equation.3  Guidelines for Implicit Differentiation: Differentiate both sides of the equation with respect to x. Collect all terms involving  EMBED Equation.3  on one side of the equation and move all other terms to the other side. Factor  EMBED Equation.3  out of the terms on the one side. Solve for  EMBED Equation.3 by dividing both sides of the equation by the factored term. Practice: Find  EMBED Equation.3 : 1. y + 7y = x 2. 4xy 3y = x 1 3. x + 5y = x + 9 4. Find Dty if t + ty 10y4 = 0 5. Find the equation of the tangent line to the curve y xy + cos(xy) = 2 at x = 0. 6. Find  EMBED Equation.3  at (2,1) if 2xy 4y = 4. 7. Find the equation of the normal line (line perpendicular to the tangent line) to the curve 8(x + y) = 100(x y) at the point (3,1).  EMBED PowerPoint.Show.8  If y = arcsin x, find cos y If y = arcsin x, find  EMBED Equation.3   EMBED Equation.3  Now, take the derivative implicitly:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Find each of the following derivatives:  1.  2.  3.  4.  5.  6. Homework Problems: pg 233-235: 1, 6, 7, 11, 17, 25, 41, 47 Read: Section 3.7 Section 3.7: Higher Derivatives SOLs: APC.8: The student will find the higher order derivatives of algebraic, trigonometric, exponential, and logarithmic functions. Objectives: Students will be able to: Find second and higher order derivatives using all previously learned rules for differentiation Vocabulary: Higher order Derivative taking the derivative of a function a second or more times Key Concept:  EMBED PowerPoint.Show.8  There are many uses and notations for higher order derivatives: First derivative:  EMBED Equation.3  Second derivative:  EMBED Equation.3  Third derivative:  EMBED Equation.3  Fourth derivative:  EMBED Equation.3  Nth derivative:  EMBED Equation.3  Practice: 1. Find  EMBED Equation.3  for y = 5x + 4x + 6x + 3 2. Find  EMBED Equation.3  3. Find  EMBED Equation.3  where  EMBED Equation.3  4. Find a formula for f n(x) where f(x) = x-2 Homework Problems: pg 240 - 242: 5, 9, 17, 18, 25, 29, 49, 57 Read: Read 3.8 Chapter 3.8: Derivatives of Logarithmic Functions SOLs: APC.9: The student will use logarithmic differentiation as a technique to differentiate nonlogarithmic functions. Objectives: Students will be able to: Know derivatives of regular and natural logarithmic functions Take derivatives using logarithmic differentiation Vocabulary: None new Key Concept: Review of Logarithms  EMBED PowerPoint.Show.8   EMBED PowerPoint.Show.8  Find  EMBED Equation.3  using implicit differentiation. In this case  EMBED Equation.3   EMBED Equation.3    EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Find  EMBED Equation.3  using the rule above Find f(x) for each of the following: 1. f(x) = ln(2x) 2. f(x) = ln("x) 3. f(x) = ln(x  x  2) 4. f(x) = ln(cos x) 5. f(x) = xln(x) 6. f(x) = log2(x + 1) 7.  EMBED Equation.3   EMBED PowerPoint.Show.8  Logarithmic Differentiation: The derivative of a function y = f(x) may be found using logarithmic differentiation 1. Take the natural log of both sides of the equation 2. Simplify using the log rules 3. Differentiate implicitly  Find for each of the following:  1. 2. y = 6x  3. We can also use logarithmic differentiation to find the derivatives of such functions as y = xx. Find this derivative Know these values of  EMBED Equation.3 :  EMBED Equation.3   EMBED Equation.3  What is the  EMBED Equation.3 ? Homework Problems: pg 76: Day 1 (Review of Laws of Logs) 34 41, 49 Pg 249: Day 2: 7, 9, 11, 21, 24, 35, 40 Read: Section 3.10 Section 3.9: Hyperbolic Functions SOLs: None Objectives: Students will be able to: Vocabulary: Key Concept: Inverse Functions: NOT COVERED IN OUR COURSE Homework Problems: none Read: Section 3.10 Section 3.10: Related Rates SOLs: APC.12: The student will apply the derivative to solve problems, including related rates of change problems. Objectives: Students will be able to: Use knowledge of derivatives to solve related rate problems Vocabulary: Related rate problems Problems where variables vary according to time and their change with respect to time can be modeled with an equation using derivatives. Key Concept:  EMBED PowerPoint.Show.8   EMBED PowerPoint.Show.8  Homework Problems: pg 260 - 262: 7, 8, 11, 14, 19, 23, 26, 31 Read: read section 3.11 Examples: A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up at a rate of 8 feet per second, how fast is the distance from the observer to the balloon increasing when the balloon is 50 feet high? How fast is the angle of elevation increasing? Water is pouring into a conical cistern at the rate of 8 cubic feet per minute. If the height of the cistern is 12 feet and the radius of its circular opening is 6 feet, how fast is the water level rising when the water is 4 feet deep? A particle P is moving along the graph of y = " x - 4 , x e" 2, so that the x coordinate of P is increasing at the rate of 5 units per second. How fast is the y coordinate of P increasing when x = 3? Air leaks out of a balloon at a rate of 3 cubic feet per minute. How fast is the surface area shrinking when the radius is 10 feet? (Note: SA = 4r & V = 4/3 r) Group Problems A cube of ice is melting uniformly so that the sides of cube are being reduced by 0.01 in/min. Find the rate of change of the volume when the cube has a side of 2 in. A stone is thrown into a pond creating a circle with an expanding radius. How fast is the distributed area expanding when the radius of the circle is 10 feet and is expanding at 1 foot per second? Sand is pouring into a conical pile at the rate of 25 cubic inches per minute. The radius is always twice the height. Find the rate at which the radius of the base is increasing when the pile is 18 inches high. (Note: V = (1/3) rh) A television camera 2000 feet from the launch pad at ground level is filming the lift-off of the space shuttle. The shuttle is rising at a rate of 1000 feet per second. Find the rate of change of the angle of elevation of the camera when the shuttle is 5000 feet from the ground. A stone is dropped into a lake causing circular waves where the radius is increasing at a constant rate of 5 meters per second. At what rate is the circumference changing when the radius is 4 meters? (Note: C = 2r) A boat is pulled toward a pier by means of a cable. If the boat is 12 feet below the level of the pier and the cable is being pulled in at a rate of 4 feet per second, how fast is the boat moving toward the pier when 13 feet of cable is out? Section 3.11: Linear Approximations and Differentials SOLs: APC.12: The student will apply the derivative to solve problems, including tangent and normal lines to a curve, curve sketching, velocity, acceleration, related rates of change, Newton's method, differentials and linear approximations, and optimization problems. Objectives: Students will be able to: Use linear approximation and differentials to estimate functional values and changes Vocabulary: Linear approximation linear approximation to a function at a point, also known as a tangent line approximation Linearization of f at a  the line L(x) = f(a) + f (a)(x  a) Differential  dy or dx approximates "y or "x (actual changes in y and x) Relative error  differential of variable divided by variable Percentage errors relative error converted to percentage Key Concept:  EMBED PowerPoint.Show.8   EMBED PowerPoint.Show.8  Let y = f(x) be differentiable at x and suppose that dx, the differential of the independent variable x, denotes an arbitrary increment of x. The corresponding differential dy of the dependent variable y is defined to be dy = f(x)dx. Example: Find dy if a.  EMBED Equation.3  b.  EMBED Equation.3  c.  EMBED Equation.3  The chief use for differentials is in approximations. Ex. 1 Use differentials to approximate the increase in the area of a soap bubble when its radius increases from 3 inches to 3.025 inches. Ex. 2 It is known that y = 3sin (2t) + 4cos t. If t is measured as 1.13 0.05, calculate y and give an estimate for the error. Comparing "y and dy. "y is the actual change in y. dy is the approximate change in y. Ex. 3 Let y = x. Find dy when x = 1 and dx = 0.05. Compare this value to "y when x = 1 and "x = 0.05. We also know we can use the tangent line to approximate a curve. The tangent line y = f(a) + f(a)(x a) is called the linear approximation of f at a. Example: Find the linear approximation to f(x) = 5x + 6x at x = 2. Then approximate f(1.98) for the function. Homework Problems: pg 267 - 269: 5, 15, 17, 21, 22, 31, 32, 41 Read: Review Chapter 3 Chapter 3: Review SOLs: None Objectives: Students will be able to: Know material presented in Chapter 3 Vocabulary: None new Key Concept: Non-Calculator 1. Find the derivative of  EMBED Equation.DSMT4  A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  2. If  EMBED Equation.DSMT4 , find  EMBED Equation.DSMT4 . A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  3. Find the derivative of  EMBED Equation.DSMT4  A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  4. The equation of the tangent line to the curve  EMBED Equation.DSMT4  at the point where the curve crosses the y-axis is A.  EMBED Equation.DSMT4  B.  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  5. The slope of the curve  EMBED Equation.DSMT4  at the point where  EMBED Equation.DSMT4  is A. 2 B. C. D. E. 2 In problems 6 8, the motion of a particle on a straight line is given by  EMBED Equation.DSMT4  6. The distance  EMBED Equation.DSMT4  is increasing for A.  EMBED Equation.DSMT4  B. all  EMBED Equation.DSMT4  C.  EMBED Equation.DSMT4  D.  EMBED Equation.DSMT4  E.  EMBED Equation.DSMT4  7. 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Differentiable functions f and g have the values shown in the table below: xffgg2531-2 If  EMBED Equation.DSMT4 , find  EMBED Equation.DSMT4  A. 20 B. 7 C. 6 D. 1 E. 13 15. The side of a cube is measured to be 3 inches. If the measurement is correct to within 0.01 inches, use differentials to estimate the propagated error in the volume of the cube. A. ( 0.000001 in3 B. ( 0.06 in3 C. ( 0.027 in3 D. ( 0.27 in3 E. ( 0.006 in3 Free Response: 1984 AB 5: The volume of a cone is increasing at a rate of  EMBED Equation.DSMT4  cubic inches per second. At the instant when the radius  EMBED Equation.DSMT4  of the cone is 3 inches, its volume is  EMBED Equation.DSMT4  cubic inches and the radius is increasing at inch/second. At the instant when the radius of the cone is 3 inches, what is the rate of change of the area of its base? At the instant when the radius of the cone is 3 inches, what is the rate of change of its height? 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PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TdyUrKKKAJAymLTcf(x % ( Rp@"Arial @|V Q|Hm|0 eV V ! PD|p|0PU0z08hdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TxUrKKKAJAmL\)] = c V  % ( Rp@"Arial P@|V Q|Hm|0 fV V ! PD|p|0PU0z080idv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TdUrKKKAJAmLT---- % ( Rp@"Arial P@|V Q|Hm|0 gV V ! PD|p|0PU0z08Hjdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33T`U-rKKKAJAmLTf(x % ( Rp@"Arial P@|V Q|Hm|0 hV V ! PD|p|0PU0z08jdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TT.U6rKKKAJA.mLP)  % ( Rp@"Arial P@|V Q|Hm|0 iV V ! PD|p|0PU0z08kdv ! 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PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33Td6KKKAJALT----  % ( Rp@"Arial P@|V Q|Hm| OV V ! PD|p|0PU0z08`kdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33T`>\KKKAJA>LTf(x  % ( Rp@"Arial P@|V Q|Hm| PV V ! PD|p|0PU0z08kdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33Tl]KKKAJA]LX) +  % ( Rp@"Arial P@|V Q|Hm| QV V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TdKKKAJALT---- % ( Rp@"Arial P@|V Q|Hm| RV V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33T`KKKAJALTg(x % ( Rp@"Arial P@|V Q|Hm| SV V ! PD|p|0PU0z08H dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TTKKKAJALP)  % ( Rp@"Arial P@|V Q|Hm| TV V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33T|KKKAJAL\Sum Rule % ( Rp@"Arial0 `<fi>0#0- *i0PU0z08gT0I0i0  - @{dvdv  @"Arialw0 H0 dv%    33TXIeKKKAJAILPdx % ( Rp@"Arial *i@|V Q|Hm|0 V V I0i0  - 0PU0z08h  @"Arialw0 0 dv !PD|p|m|Z|m|dv%    33TX8KKKAJALPdx % ( Rp@"Ariali0  -@|V Q|Hm|0 V V   0PU0z080idv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TXKKKAJALPdx % ( Rp@"Arial0 efg0 {@{,TiK0W0fg{{0PU0z08g -0x @{!0l`pm50lq(@{(dv%    33TTI%KKKAJAI ,Ld d d % ( Rp@"Arial0 <f >0#0- * 0PU0z08gT0I0i0  - @{dvdv  @"Arialw0 H0 dv%    33TdIh9KKKAJAI4LT---- % ( Rp@"Arial * @|V Q|Hm|0 ːV V I0i0  - 0PU0z08g  @"Arialw0 0 dv !PD|p|m|Z|m|dv%    33TTqx9KKKAJAq4LP[ % ( Rp@"Ariali0  -@|V Q|Hm|0 ̐V V   0PU0z08hdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33T`y9KKKAJAy4LTf(x % ( Rp@"Arial @|V Q|Hm|0 ͐V V ! PD|p|0PU0z08hdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TX9KKKAJA4LP)  % ( Rp@"Arial P@|V Q|Hm|0 ΐV V ! PD|p|0PU0z080idv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TT9KKKAJA4LP- % ( Rp@"Arial P@|V Q|Hm|0 ϐV V ! PD|p|0PU0z08Hjdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33T`9KKKAJA4LTg(x % ( Rp@"Arial P@|V Q|Hm|0 АV V ! PD|p|0PU0z08jdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33Tp9KKKAJA4LX)] =  % ( Rp@"Arial P@|V Q|Hm|0 ѐV V ! PD|p|0PU0z08`kdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33Td09KKKAJA4LT----  % ( Rp@"Arial P@|V Q|Hm|0 ҐV V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33T`8U9KKKAJA84LTf(x % ( Rp@"Arial P@|V Q|Hm|0 ӐV V ! PD|p|0PU0z08( dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TXVe9KKKAJAV4LP)  % ( Rp@"Arial P@|V Q|Hm|0 ԐV V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TTfm9KKKAJAf4LP- % ( Rp@"Arial P@|V Q|Hm|0 ՐV V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33Td|9KKKAJA|4LT----  % ( Rp@"Arial P@|V Q|Hm|0 ֐V V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33T`9KKKAJA4LTg(x % ( Rp@"Arial P@|V Q|Hm|0 אV V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TT9KKKAJA4LP)  % ( Rp@"Arial P@|V Q|Hm|0 ؐV V ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TW9KKKAJA4LlDifference Rule  % ( Rp@"Arial00 <fi>0#0- *i0PU0z08gT0I0i0  - @{dvdv  @"Arialw H dv%    33TXI/eLKKKAJAIGLPdx % ( Rp@"Arial *i@|V Q|Hm| V V I0i0  - 0PU0z08h  @"Arialw  dv !PD|p|m|Z|m|dv%    33TX/1LKKKAJAGLPdx % ( Rp@"Ariali0  -@|V Q|Hm| V V   0PU0z080idv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TX~/LKKKAJA~GLPdx % ( Rp@"Arial00 fg0 p{@{TiK0W0fgZ{p{0PU0z08g 0x @{!0l`pm50lq(@{(dv%    33TTIkWKKKAJAILPd % ( Rp@"Arial00 <fj>0#0- *jxɥ0PU0z08gT0I0i0  - @{dvdv  @"Arialw H dv%    33TdIhKKKAJAILT---- % ( Rp@"Arial *j@|V Q|Hm| V V I0i0  - 0PU0z08g  @"Arialw  dv !PD|p|m|Z|m|dv%    33TXqKKKAJAqLP(e % ( Rp@"Ariali0  -@|V Q|Hm| V V   0PU0z08hdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TTKKKAJALPx  % ( Rp@"Arial @|V Q|Hm| V V ! PD|p|0PU0z08hdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TlKKKAJALX) = e % ( Rp@"Arial P@|V Q|Hm| V V ! PD|p|0PU0z080idv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TTKKKAJALPx  % ( Rp@"Arial P@|V Q|Hm| V V ! PD|p|0PU0z08Hjdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    33TnKKKAJALlNatural Exponent  % ( Rp@"Arial0 <fg>0#0- *g0PU0z08gT0I0i0  - @{dvdv  @"Arialw0 H0 dv%    33TXIeKKKAJAILPdx % ( F4(EMF+*@$??!b Ld)??" 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\P|p|0PU0z08Udvd! \P|p|m|Z|m|dvd!\P|p|m|Z|m|dv%    33TdKKKAJALT=  % ( Rp@"Arial \L|xQ|Hm| cxxd! \P|p|0PU0z08HUdvd! \P|p|m|Z|m|dvd!\P|p|m|Z|m|dv%    33TdKKKAJALT----  % ( Rp@"Arial \L|xQ|Hm| dxxd! \P|p|0PU0z08Udvd! \P|p|m|Z|m|dvd!\P|p|m|Z|m|dv%    33TTKKKAJALP"  % ( Rp@"Arial \L|xQ|Hm| exxd! \P|p|0PU0z08|Udvd! \P|p|m|Z|m|dvd!\P|p|m|Z|m|dv%    33Td )KKKAJA LT----  % ( Rp@"Arial0= m䏥<0  >$hTiK0W0䏥<> >0PU0z08ܐdSi0 S$!0 Еm50Е'>($( dv%    33TX=YKKKAJA=LPdx % ( Rp@"ArialW0䏥<L|xQ|Hm|= nxx0 S$0PU0z08m50Е ($( dvd!\P|p|m|Z|m|dv%    33TXkKKKAJAkLPdx % ( Rp@"Arial SL|xQ|Hm|= oxxm50Е ($0PU0z08dvd! \P|p|m|Z|m|dvd!\P|p|m|Z|m|dv%    33TXKKKAJALPdu % ( Rp@"Arial L|xQ|Hm|= pxxd! \P|p|0PU0z08dvd! \P|p|m|Z|m|dvd!\P|p|m|Z|m|dv%    33TX*KKKAJALPdx % ( Rp@"Arial|/0NP$P\/00PU0z08P04OЕm50Е($(dv%    TKKKAJA@LIn words: the derivative of a composite function is equal to th     % ( Rp@"ArialPL|xQ|Hm| xx04O0PU0z08ܐm50Е ($(dvd!\P|p|m|Z|m|dv%    TKKKAJALle derivation of   % ( Rp@"Arial0= z䏥ܐ0 t>$@TiK0W0䏥ܐȬ>t>0PU0z08PĴSA0S$!0Еm50Е ($詥( dv%    T#KKKAJA@Lthe outer function times the derivative of the inner function.      % ( Rp@"Arial0= 䏥h0 >$JTiK0W0䏥hܬ>>0PU0z08PشSK0S$!0Еm50Е\ ($< dv%    T5RKKKAJAM@LThe notation on the right is Leibniz notation and is often refer     % ( Rp@"ArialW0䏥hL|xQ|Hm|= xx0S$0PU0z08ܐm50Е ($< dvd!\P|p|m|Z|m|dv%    T5bRKKKAJAM L`red to as  % ( Rp@"Arial0 䏥0  >$&TiK0W0䏥> >0PU0z08P(S'0S$!0Еm50Е̜ ($ dv%    TTT)oKKKAJAkLPu % ( Rp@"ArialW0䏥L|xQ|Hm| xx0S$0PU0z08ܐm50Е ($ dvd!\P|p|m|Z|m|dv%    T0SpKKKAJA0kLtsubstitution: y =   % ( Rp@"ArialSL|xQ|Hm| xxm50Е ($0PU0z08hdvd! 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FEMF+@ nMSPresentationPowerPoint.Show.89q %_ Chris HeadleeChris HeadleeOh+'0 X`p  Slide 1Chris HeadleeoChris Headleeo11iMicrosoft PowerPointP@Current User GSummaryInformation((PowerPoint Document(   DocumentSummaryInformation8)@N@`.n՜.+,0    KOn-screen Show -s : Arial WingdingsDefault DesignSlide 1  Fonts UsedDesign Template( `/ 0DArialngstt-0B 0"DWingdingstt-0B 0@ .  @n?" dd@  @@`` (   0AA @3ʚ;ʚ;g4\d\d B 0p2p? <4dddd8pC 0t- 80___PPT10   ` 33` Sf3f` 33g` f` www3PP` ZXdbmo` \ғ3y`Ӣ` 3f3ff` 3f3FKf` hk]wwwfܹ` ff>>\`Y{ff` R>&- {p_/̴>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>> f(    6z  `}  T Click to edit Master title style! !  0}  `  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0 ^ `  >*  0x ^   @*  0@ ^ `  @*H  0޽h ? 3380___PPT10.K tv Default Design jb (   E </N LDifferentiation Chain Rule  6T>S["  If f and g are both differentiable and F = f%g is the composite function defined by F(x) = f(g(x)) d dy dy du ---- [F(x)] = F (x) = f (g(x)) " g (x) or ---- = ---- " ---- dx dx du dx In words: the derivative of a composite function is equal to the derivation of the outer function times the derivative of the inner function. The notation on the right is Leibniz notation and is often referred to as u substitution: y = f(u) and u = g(x) Combine with the Power Rule: d du d ---- [un] = n" un-1 ---- or ----- [g(x)]n = n[g(x)]n-1 " g (x) dx dx dx Example: y = (x6  2x + 1)8 dy dy du du let u = (x6  2x + 1) then y = u8 ------ = ----- " ---- = 8 u7 " ----- dx du dx dx = 8 (x6  2x + 1)7 (6x5  2)c6A#AF,7 * $ $o   K&P $,$-$, $,  $$$U(((( (( ((((((,, 0044m88<<     $$((,, 004488<<$+&X  9L  Achv  H  0޽h ? 3380___PPT10.K tvr0w ey  Slide Titles dO)"Microsoft PowerPoint PresentationMSPresentationPowerPoint.Show.89q %_t! PChris HeadleeChris HeadleeOh+'0 X_1252052894dO)֝2۝2Ole EPRINT CompObj{      !"#$%&'(+=>-./0123456789:;<?@BACEDHIJKLMNOPQRSTUVWXYZ[\]^_`abcdefghijklmnopqrstuvwxyz{|}~l1'% EMFw\KhCF, EMF+@F\PEMF+"@ @ $@ 0@?!@ @ F(GDICFXLEMF+*@$?? @$yD;D!b $$==% % V0.x>.x>% % $$AA" FEMF+@ FGDICF(GDIC1FGDICF(GDICUFGDICF(GDIC,FGDICRp@"Arial04&\0H>!>'0PU0z08P&8t&$8 0$&0$Е`0Е($詥(dv%    T8&KKKAJA 'LSpecific Applications of the Chain Rule     % ( F(GDIC:FGDICRp@"Arial0! g<䏥ܐ>0#0-S2ܐ0PU0z08PT0I0i0M-S$dvdvM@"Arialw VH Vdv%    TD@]KKKAJAX)LDifferentiation of exponential functions:  % ( &% [6[% ( Rp@"Arial0q 䏥0 >$RTiK0W0䏥>>0PU0z08PnUS0hnU$!0Еm50Е($8Pdv%    33TDsKKKAJADLxd  % ( Rp@"ArialW0䏥L|xQ|Hm|q xx0hnU$0PU0z08ܐm50Е ($8Pdv !\P|p|m|Z|m|dv%    33TXsKKKAJALPdu % ( Rp@"ArialhnUL|xQ|Hm|q xxm50Е ($0PU0z08hdv ! \P|p|m|Z|m|dv !\P|p|m|Z|m|dv%    33TsuKKKAJALxd  % ( Rp@"Arial0q <䏥`U>0#0-S2`U0PU0z08ܐT0I0i0M-S$dvdvM@"Arialw! 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Ӱxxm50Е ($0PU0z08 dv ! \P|p|m|Z|m|dv !\P|p|m|Z|m|dv%    TX!4>KKKAJA9LPdu % ( Rp@"Arial L|xQ|Hm|! ԰xx ! \P|p|0PU0z08$dv ! \P|p|m|Z|m|dv !\P|p|m|Z|m|dv%    TX!>KKKAJA9LPdu % ( Rp@"Arial0! <䏥$U>0#0-S2$U0PU0z08PT0I0i0M-S$dvdvM@"Arialw H dv%    T6,SKKKAJANLlet u = (2x + 1) then y =   % ( Rp@"ArialS2$U@|x Q|Hm| xxI0i0M-S0PU0z08ܐM @"Arialw  dv !PD|p|m|Z|m|dv%    TT-6:SKKKAJA-NLPe % ( Rp@"Ariali0M-@|x Q|Hm| xxM 0PU0z08hdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    TT;8DJKKKAJA;GLPu  % ( Rp@"Arial @|x Q|Hm| xx ! PD|p|0PU0z08dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    Tp6SKKKAJANLX------  % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08 dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    TX6SKKKAJANLP=  % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    Tl6SKKKAJANLX-----  % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    TT6 SKKKAJANLP"  % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08SUdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    Td62SKKKAJANLT----  % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08TUdv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    TXA6VSKKKAJAANLP=  % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08@Udv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    TTW6dSKKKAJAWNLPe % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08Udv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    TTe8nJKKKAJAeGLPu  % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08Udv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    TTv6~SKKKAJAvNLP"  % ( Rp@"Arial P@|x Q|Hm| xx ! PD|p|0PU0z08Udv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    Tl6SKKKAJANLX-----  % ( Rp@"Arial0! 䏥0 >$VTiK0W0䏥>>0PU0z08ܐ(oUW0 hnU$!0 Еm50Е0'>($ dv%    TXKhKKKAJAcLPdx % ( Rp@"ArialW0䏥L|xQ|Hm|! xx0 hnU$0PU0z08m50Е ($ dv !\P|p|m|Z|m|dv%    TXKhKKKAJAcLPdu % ( Rp@"Arial hnUL|xQ|Hm|! xxm50Е ($0PU0z08 dv ! \P|p|m|Z|m|dv !\P|p|m|Z|m|dv%    TXK6hKKKAJAcLPdx % ( Rp@"Arial L|xQ|Hm|! xx ! \P|p|0PU0z08$dv ! \P|p|m|Z|m|dv !\P|p|m|Z|m|dv%    TXKhKKKAJAcLPdx % ( Rp@"Arial0 <䏥<>0#0-S2<0PU0z08hT0I0i0M-S$dvdv<=@"Arialw! H! dv%    TX?|TKKKAJA?LP=  % ( Rp@"ArialS2<@|x Q|Hm|! !xxI0i0M-S0PU0z08<= @"Arialw! ! dv !PD|p|m|Z|m|dv%    TXU|iKKKAJAULPe  % ( Rp@"Ariali0M-@|x Q|Hm|! "xx<= 0PU0z08dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    Tpj~KKKAJAjLX2x + 1    % ( Rp@"Arial @|x Q|Hm|! #xx ! PD|p|0PU0z08dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    T|KKKAJALh(2) = 2 e  % ( Rp@"Arial P@|x Q|Hm|! $xx ! PD|p|0PU0z08$dv ! PD|p|m|Z|m|dv !PD|p|m|Z|m|dv%    Tp ~NKKKAJA LX2x + 1    % ( Rp@"Arial0 <䏥ܐ>0#0-S2ܐ0PU0z08PT0I0i0M-S$dvdvM@"Arialw! )H! )dv%    T%KKKAJALxGreater length Chains:   % ( &% 6&% ( Rp@"Arial0 <䏥h>0#0-S2h0PU0z08PT0I0i0M-S$dvdvM@"Arialwq SHq Sdv%    TpXKKKAJALXy = e  % ( Rp@"ArialS2h@|x Q|Hm|q TxxI0i0M-S0PU0z08ܐM @"Arialwq Sq Sdv !PD|p|m|Z|m|dv%    TYKKKAJAY Ldcos(3x + 1)     % ( Rp@"Arial0 䏥<0 , >$nTiK0W0䏥<$>, >0PU0z08h($( dv%    TXiKKKAJAiLPdy  % ( Rp@"ArialW0䏥<L|xQ|Hm| xx0 hnU$0PU0z08m50Е ($( dv !\P|p|m|Z|m|dv%    TXKKKAJALPdy  % ( Rp@"Arial hnUL|xQ|Hm| xxm50Е ($0PU0z08dv ! \P|p|m|Z|m|dv !\P|p|m|Z|m|dv%    TX KKKAJALPdu % ( Rp@"Arial L|xQ|Hm| xx ! \P|p|0PU0z08dv ! \P|p|m|Z|m|dv !\P|p|m|Z|m|dv%    TX7TKKKAJA7LPdv % ( Rp@"Arial0 <䏥DU>0#0-S2DU0PU0z08PT0I0i0M-S$dvdvM@"Arialwq Hq dv%    T\  )KKKAJA$-Llet u = cos(3x + 1) and v = 3x + 1 then y =   % ( Rp@"ArialS2DU@|x Q|Hm|q xxI0i0M-S0PU0z08ܐM @"Arialwq q dv !PD|p|m|Z|m|dv%    TT  )KKKAJA $LPe % ( Rp@"Ariali0M-@|x Q|Hm|q xxM 0PU0z08hdv ! 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Rp@"Arial0e |0 8եOTiK0W0|^[8ե0PU0z08d2P02!0䗥ěhm50䗥p h(૥< hdv%    TeKKKAJA}@LIf we tried solving for y first in the above circle we would get      % ( Rp@"ArialW0|L|kQ|Hm|e kk020PU0z08hm50䗥 h(૥< dv|!c\P|p|m|Z|m|dv%    TeKKKAJA}Lhthe following: % ( Rp@"Arial0e x[0 [LեTTiK0W0x[h[Lե0PU0z08d2U02!0䗥ěhm50䗥h2h(DShdv%    TdCKKKAJALTy =  % ( Rp@"ArialW0x[L|kQ|Hm|e !kk020PU0z08hm50䗥 h(DSdv|!c\P|p|m|Z|m|dv%    TTDQKKKAJADLP % ( Rp@"Arial2L|kQ|Hm|e "kkm50䗥 h(0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTYfKKKAJAYLP" % ( Rp@"Arial hL|kQ|Hm|e #kk|!c \P|p|0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TdgKKKAJAgLT100  % ( Rp@"Arial \L|kQ|Hm|e $kk|!c \P|p|0PU0z08 dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTKKKAJALP- % ( Rp@"Arial \L|kQ|Hm|e %kk|!c \P|p|0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTKKKAJALPx % ( Rp@"Arial \L|kQ|Hm|e &kk|!c \P|p|0PU0z088dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTKKKAJALP % ( Rp@"Arial \L|kQ|Hm|e 'kk|!c \P|p|0PU0z08Ėdv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    T WKKKAJA Lthis gives us two functions y =   % ( Rp@"Arial \L|kQ|Hm|e (kk|!c \P|p|0PU0z08Pdv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTXeKKKAJAXLP" % ( Rp@"Arial \L|kQ|Hm|e )kk|!c \P|p|0PU0z08,[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TdfKKKAJAfLT100  % ( Rp@"Arial \L|kQ|Hm|e *kk|!c \P|p|0PU0z08`-[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTKKKAJALP- % ( Rp@"Arial \L|kQ|Hm|e +kk|!c \P|p|0PU0z08-[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTKKKAJALPx % ( Rp@"Arial \L|kQ|Hm|e ,kk|!c \P|p|0PU0z08\)[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTKKKAJALP % ( Rp@"Arial \L|kQ|Hm|e -kk|!c \P|p|0PU0z08a[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    Tx'KKKAJAL\and y =  % ( Rp@"Arial \L|kQ|Hm|e .kk|!c \P|p|0PU0z08|[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TT/6KKKAJA/LP- % ( Rp@"Arial \L|kQ|Hm|e /kk|!c \P|p|0PU0z08T[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TT7DKKKAJA7LP" % ( Rp@"Arial \L|kQ|Hm|e 0kk|!c \P|p|0PU0z08[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TdEuKKKAJAELT100  % ( Rp@"Arial \L|kQ|Hm|e 1kk|!c \P|p|0PU0z08t[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTv}KKKAJAvLP- % ( Rp@"Arial \L|kQ|Hm|e 2kk|!c \P|p|0PU0z08[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTKKKAJALPx % ( Rp@"Arial \L|kQ|Hm|e 3kk|!c \P|p|0PU0z08r[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    TTKKKAJALP % ( Rp@"Arial0e <0 եETiK0W0J[ե0PU0z08d2F02!0䗥ěhm50䗥̝ h(P hdv%    T,KKKAJA%LDifferentiating the first (upper semi     % ( Rp@"ArialW0L|kQ|Hm|e =kk020PU0z08hm50䗥 h(P dv|!c\P|p|m|Z|m|dv%    TTKKKAJALP- % ( Rp@"Arial2L|kQ|Hm|e >kkm50䗥 h(0PU0z08|dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    T&KKKAJALcircle) of these would give us:   % ( Rp@"Arial0e PĖ0 |ե`TiK0W0Ė[|ե0PU0z08d\2a0 2!0 䗥ěhm50䗥>h(Ьh dv%    33TX5/KKKAJA*LPdy  % ( Rp@"ArialW0ĖL|kQ|Hm|e Qkk0 20PU0z08|hm50䗥 h(Ь dv|!c\P|p|m|Z|m|dv%    33TT/KKKAJA*LP- % ( Rp@"Arial 2L|kQ|Hm|e Rkkm50䗥 h(0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33T:/KKKAJA*L|2x  % ( Rp@"Arial hL|kQ|Hm|e Skk|!c \P|p|0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT;B/KKKAJA;*LP- % ( Rp@"Arial \L|kQ|Hm|e Tkk|!c \P|p|0PU0z08 dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TC/KKKAJAC*Ltx  % ( Rp@"Arial \L|kQ|Hm|e Ukk|!c \P|p|0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT/KKKAJA*LP- % ( Rp@"Arial \L|kQ|Hm|e Vkk|!c \P|p|0PU0z088dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT/KKKAJA*LPx % ( Rp@"Arial0e {<[>0#0-2@3[0PU0z08dT0I0i0 k-2hdvdv k@"Arialw H dv%    33Td'9DKKKAJA?LT---- % ( Rp@"Arial2@3[@|k Q|Hm| kkI0i0 k-20PU0z08 k @"Arialw  dv|!cPD|p|m|Z|m|dv%    33TXA'VDKKKAJAA?LP=  % ( Rp@"Ariali0 k-@|k Q|Hm| kk k 0PU0z08|dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TTW'kDKKKAJAW?LP % ( Rp@"Arial @|k Q|Hm| kk|!c PD|p|0PU0z08dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33Tls'DKKKAJAs?LX(100  % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TT'DKKKAJA?LP- % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08 dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TT'DKKKAJA?LPx % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TT'DKKKAJA?LP % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z088dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TT'DKKKAJA?LP) % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08Ėdv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TT);KKKAJA8LP- % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08Pdv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TT);KKKAJA8LP % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08,[dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TT'DKKKAJA?LP( % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08`-[dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TT'DKKKAJA?LP- % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08-[dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33Tx'LDKKKAJA?L\2x) =  % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08x[dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33TM'DKKKAJAM?Lp-----------------      % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08[dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33T`' DKKKAJA?LT=  % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08[dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33T 'DKKKAJA ?Lh--------------     % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08([dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33Td'DKKKAJA?LT=  % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08[dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33T|'DKKKAJA?L\--------   % ( Rp@"Arial P@|k Q|Hm| kk|!c PD|p|0PU0z08H[dv|!c PD|p|m|Z|m|dv|!cPD|p|m|Z|m|dv%    33Tl'8DKKKAJA?LX!!!!! % ( Rp@"Arial0e [0 [\եXTiK0W0[p[\ե0PU0z08d2Y02!0䗥ěhm50䗥X>h(>@hdv%    33TX<6YKKKAJATLPdx % ( Rp@"ArialW0[L|kQ|Hm|e kk020PU0z08hm50䗥 h(>@dv|!c\P|p|m|Z|m|dv%    33TTU<aYKKKAJAUTLP2  % ( Rp@"Arial2L|kQ|Hm|e kkm50䗥 h(0PU0z08|dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TTb<oYKKKAJAbTLP" % ( Rp@"Arial hL|kQ|Hm|e kk|!c \P|p|0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33Tdp<YKKKAJApTLT100  % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT<YKKKAJATLP- % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08 dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT<YKKKAJATLPx % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT<YKKKAJATLP % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08Ėdv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT< YKKKAJATLP" % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08Pdv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33Td!<PYKKKAJA!TLT100  % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08,[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TTQ<XYKKKAJAQTLP- % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08`-[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT`<mYKKKAJA`TLPx % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08-[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TTn<uYKKKAJAnTLP % ( Rp@"Arial \L|kQ|Hm|e kk|!c \P|p|0PU0z08x[dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    33TT<YKKKAJATLPy  % ( Rp@"Arial|/0LNN\/00PU0z08d04M䗥ěhm50䗥ph(<hdv%    T0vKKKAJA&Lsolving for the second one (lower semi   % ( Rp@"ArialNL|kQ|Hm|e kk04M0PU0z08hm50䗥 h(<dv|!c\P|p|m|Z|m|dv%    TTvKKKAJALP- % ( Rp@"Arial04L|kQ|Hm|e kkm50䗥 h(0PU0z08|dv|!c \P|p|m|Z|m|dv|!c\P|p|m|Z|m|dv%    T8vKKKAJA'Lcircle) again gives us the same answer    % ( Rp@"Arial0 70 ԥ+TiK0W0[ԥ0PU0z08dо2,02!0䗥ěhm50䗥h((hdv%    TPKKKAJA+Lbecause y is negative for all values in it.   % ( 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FEMF+@ CompObj{ObjInfocgPicturesCurrent UserfiG     "%&),-03458;>ABCFIJKNQRSTWZ]`bcdeghijklmpstwz{~%_Q HChris HeadleeChris HeadleeOh+'0 X`p  Slide 1Chris HeadleeoChris Headleeo12iMicrosoft PowerPointP@z"@N@@:1՜.+,0SummaryInformation((PowerPoint Document(hj*uDocumentSummaryInformation8_1119764524mFP2P2( `/ 0DArialngsttS0B 0"DWingdingsttS0B 0@ .  @n?" dd@  @@`` (  0AA @3ʚ;ʚ;g47d7d B 0p2p? <4dddd8pC 0tS 80___PPT10   ` 33` Sf3f` 33g` f` www3PP` ZXdbmo` \ғ3y`Ӣ` 3f3ff` 3f3FKf` hk]wwwfܹ` ff>>\`Y{ff` R>&- {p_/̴>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>> f(    6 S  `} S T Click to edit Master title style! !  0S  ` S RClick to edit Master text styles Second level Third level Fourth level Fifth level!     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H! dv%    TT3@KKKAJA3LP2 % ( Rp@"Arial1 *(`1 @|x Q|Hm|! xxI0i0 T-1 0PU0z08g T @"Arialw! ! dvc!PD|p|m|Z|m|dv%    TXATKKKAJAALPnd  % ( Rp@"Ariali0 T-@|x Q|Hm|! xx T 0PU0z084hdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    T\KKKAJA\ LhDerivative:   % ( Rp@"Arial @|x Q|Hm|! xxc! PD|p|0PU0z08hdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Td KKKAJALT----  % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08Lidvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    TT KKKAJA LP(l % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08idvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Td7KKKAJALTf (x % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08djdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Tl8eKKKAJA8LX)) =   % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08jdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    TlfKKKAJAfLXf  (x % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08|kdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    TdKKKAJALT) =  % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08ldvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    TxKKKAJAL\-------   % ( Rp@"Arial0 fi0 \{7TiK0W0fi0PU0z08g1 801 \{!0l|pЀm50l0#0-1 *(`1 0PU0z08gT0I0i0 T-1 \{Ѐdvdv T@"Arialw! H! dv%    TT3@:KKKAJA35LP30 % ( Rp@"Arial1 *(`1 @|x Q|Hm|! xxI0i0 T-1 0PU0z08g T @"Arialw! ! dvc!PD|p|m|Z|m|dv%    TXAQ1KKKAJAA.LPrd  % ( Rp@"Ariali0 T-@|x Q|Hm|! xx T 0PU0z084hdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    TY:KKKAJAY5 LhDerivative:   % ( Rp@"Arial @|x Q|Hm|! xxc! PD|p|0PU0z08hdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Td:KKKAJA5LT----  % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08Lidvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    TT:KKKAJA5LP( % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08idvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Tl;:KKKAJA5LXf  (x % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08djdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Tl<h:KKKAJA<5LX)) =  % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08jdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Tpi:KKKAJAi5LXf   (x % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08|kdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Td:KKKAJA5LT) =  % ( Rp@"Arial P@|x Q|Hm|! xxc! PD|p|0PU0z08ldvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    Tp:KKKAJA5LX------   % ( Rp@"Arial0 fi0 \{6TiK0W0fi0PU0z08g1 701 \{!0l|pЀm50l0#0-1 *h81 0PU0z08gT0I0i0 T-1 \{Ѐdvdv T@"Arialw H dv%    T3lKKKAJA3LThe classic example of the 2 % ( Rp@"Arial1 *h@|x Q|Hm| xxI0i0 T-1 0PU0z08g T @"Arialw  dvc!PD|p|m|Z|m|dv%    TXnKKKAJA}LPnd  % ( Rp@"Ariali0 T-@|x Q|Hm| xx T 0PU0z084hdvc! PD|p|m|Z|m|dvc!PD|p|m|Z|m|dv%    TDlnKKKAJA)Lderivative is with the position function:  % ( Rp@"Arial0 fh0 \{HTiK0W0fh0PU0z08gX1 I01 \{!0l|pЀm50lr Ѐ(\{pP Ѐdv%    T`3JKKKAJA3LTf(t % ( Rp@"ArialW0fhL|xQ|Hm| xx01 \{0PU0z08gЀm50l Ѐ(\{pP dvc!\P|p|m|Z|m|dv%    TKKKKAJAK@L) is the position of something (like a particle) with respect to     % ( Rp@"Arial1 L|xQ|Hm| xxm50l Ѐ(\{0PU0z084hdvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TdNKKKAJALTtime % ( Rp@"Arial0 fLi0 \{;TiK0W0fLi0PU0z08gl1 <01 \{!0l|pЀm50lr Ѐ(\{d Ѐdv%    TT3;KKKAJA3LPf % ( Rp@"ArialW0fLiL|xQ|Hm| xx01 \{0PU0z08gЀm50l Ѐ(\{d dvc!\P|p|m|Z|m|dv%    TT;AKKKAJA;LP  % ( Rp@"Arial1 L|xQ|Hm| xxm50l Ѐ(\{0PU0z084hdvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TXBQKKKAJABLP(t % ( Rp@"Arial ЀL|xQ|Hm| xxc! \P|p|0PU0z08hdvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TRKKKAJAR7L) is the velocity of the particle with respect to time        % ( Rp@"Arial0! fl0 |\{"TiK0W0fl|0PU0z08gИ1 #0 1 \{!0 l|pЀm50lt Ѐ(\{8 Ѐ dv%    TTdlKKKAJAdLPf % ( Rp@"ArialW0flL|xQ|Hm|! xx0 1 \{0PU0z084hЀm50l Ѐ(\{8  dvc!\P|p|m|Z|m|dv%    TTlrKKKAJAlLP  % ( Rp@"Arial 1 L|xQ|Hm|! xxm50l Ѐ(\{0PU0z08hdvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TXsKKKAJAsLP(t % ( Rp@"Arial ЀL|xQ|Hm|! xxc! \P|p|0PU0z08Lidvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TdKKKAJALT) =  % ( Rp@"Arial \L|xQ|Hm|! xxc! \P|p|0PU0z08idvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    T`KKKAJALTv(t % ( Rp@"Arial \L|xQ|Hm|! xxc! \P|p|0PU0z08djdvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TlKKKAJALX)) =  % ( Rp@"Arial \L|xQ|Hm|! xxc! \P|p|0PU0z08jdvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TRKKKAJA L`d(f(t))/dt  % ( Rp@"Arial0! 'fLi0 \{@TiK0W0fLi0PU0z08gl1 A01 \{!0l|pЀm50lr Ѐ(\{d Ѐdv%    TT3 ;)KKKAJA3$LPf % ( Rp@"ArialW0fLiL|xQ|Hm|! (xx01 \{0PU0z08gЀm50l Ѐ(\{d dvc!\P|p|m|Z|m|dv%    TX; H)KKKAJA;$LP   % ( Rp@"Arial1 L|xQ|Hm|! )xxm50l Ѐ(\{0PU0z084hdvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TXI X)KKKAJAI$LP(t % ( Rp@"Arial ЀL|xQ|Hm|! *xxc! \P|p|0PU0z08hdvc! \P|p|m|Z|m|dvc!\P|p|m|Z|m|dv%    TY )KKKAJAY$;L) is the acceleration of the particle with respect to time      % ( Rp@"Arial0 f1 0 1 \{1TiK0W0f1 0PU0z08g41 201 \{!0l|pЀ m50l@ Ѐ( \{, Ѐdv%    TTd*lGKKKAJAdBLPf % ( Rp@"ArialW0f1 L|xQ|Hm| xx01 \{0PU0z084hЀm50l Ѐ( \{, dv c!\P|p|m|Z|m|dv%    TXl*yGKKKAJAlBLP   % ( Rp@"Arial1 L|xQ|Hm| xxm50l Ѐ( \{0PU0z08hdvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    TXz*GKKKAJAzBLP(t % ( Rp@"Arial ЀL|xQ|Hm| xxc! \P|p|0PU0z08Lidvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    Td*GKKKAJABLT) =  % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z08idvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    T`*GKKKAJABLTa(t % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z08djdvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    Td*GKKKAJABLT) =  % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z08jdvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    T`*GKKKAJABLTd(v % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z08|kdvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    TT*GKKKAJABLP  % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z08ldvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    Tx*]GKKKAJABL\(t))/dt % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z08(`1 dvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    T`e*GKKKAJAeBLT= d % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z08`1 dvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    TT*GKKKAJABLP % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z081 dvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    T*GKKKAJAB L`(f(t))/dt   % ( Rp@"Arial \L|xQ|Hm| xxc! \P|p|0PU0z08 1 dvc! \P|p|m|Z|m|dv c!\P|p|m|Z|m|dv%    TT*GKKKAJABLP % ( Rp@"Arial>0#01 *0lr0PU0z08g0l(r>p0m\{\{820Cmadv%    Tl3fsKKKAJA3~LXThe 3 % ( Rp@"Arial@|x Q|Hm| (xx0l(r>p0PU0z08g\{820 Cmadv\{c!PD|p|m|Z|m|dv%    TXthzKKKAJAtwLPrd  % ( Rp@"Arial(r>@|x Q|Hm| )xx820 C0PU0z084hdvc! PD|p|m|Z|m|dv\{c!PD|p|m|Z|m|dv%    Tf<KKKAJA~=Lderivative of the position function is the rate of change of     % ( Rp@"Arial0! Xfh0 \{&TiK0W0fh0PU0z08gX1 '01 \{!0l|pЀm50lr Ѐ(\{HP Ѐdv%    T3KKKAJA3!Lacceleration and is known as the   % ( Rp@"ArialW0fhL|xQ|Hm|! Yxx01 \{0PU0z08gЀm50l Ѐ(\{HP dvc!\P|p|m|Z|m|dv%    TdKKKAJALTjerk  % ( Rp@"Arial0 6f4h0 \{0TiK0W0f4h0PU0z08gD1 101 \{!0l|pЀm50l(rЀ(\{H<Ѐdv%    TT3:KKKAJA3LP( % ( Rp@"ArialW0f4hL|xQ|Hm| 7xx01 \{0PU0z08gЀm50l Ѐ(\{H<dvc!\P|p|m|Z|m|dv%    Th;rKKKAJA;/Lnot to be confused with the Steve Martin movie)  % ( F4(EMF+*@$??!b Ld)??" FEMF+@ PicturesCurrent UserGSummaryInformation((PowerPoint Document(_Oh+'0 X`p  Slide 1Chris HeadleeoChris Headleeo2riMicrosoft PowerPointP@2m@bY@ ?՜.+,0    B( `/ 0DArialtt&0B 0"Drialtt&0B 0@ .  @n?" dd@  @@`` x   0AA@3ʚ;ʚ;g4YdYd B 0Vppp@ <4dddd8pC 0t& 80___PPT10   ` 33` Sf3f` 33g` f` www3PP` ZXdbmo` \ғ3y`Ӣ` 3f3ff` 3f3FKf` hk]wwwfܹ` ff>>\`Y{ff` R>&- {p_/̴>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>> f(    6b  `}  T Click to edit Master title style! !  0  `  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0  ^ `  >*  0 ^   @*  0ܤ ^ `  @*H  0޽h ? 3380___PPT10. M Default Design  , $  (    <,91Q + JHigher Order Derivatives  0 f If f is a differentiable function, then its derivative f is also a function and we can take the derivative of the derivative! d dy 2nd Derivative: ----(f (x)) = f  (x) = ------- dx dx d dy 3rd Derivative: ----(f  (x)) = f   (x) = ------ dx dx The classic example of the 2nd derivative is with the position function: f(t) is the position of something (like a particle) with respect to time f (t) is the velocity of the particle with respect to time f (t) = v(t)) = d(f(t))/dt f  (t) is the acceleration of the particle with respect to time f  (t) = a(t) = d(v (t))/dt = d(f(t))/dt The 3rd derivative of the position function is the rate of change of acceleration and is known as the jerk (not to be confused with the Steve Martin movie)jFF c  4c (+ "I "_ "y (_  3 L%P$MG? C |0H  0޽h ? 3380___PPT10. [r0_ e#DocumentSummaryInformation8_1216743435F22Ole CompObjfOn-screen Show -s_: ArialDefault DesignSlide 1  Fonts UsedDesign Template Slide Titles FMicrosoft Equation 3.0 DS Equation Equation.39qObjInfoEquation Native _1216743588F22Ole ,\ dydx,2fx(),ddxfx()[],D x fx()[] FMicrosoft Equation 3.0 DS Equation Equation.39qCompObjfObjInfoEquation Native _1216743592F22x3, d 2 ydx 2 ,2fx(),d 2 dx 2 fx()[],D x  2 fx()[] FMicrosoft Equation 3.0 DS EqOle CompObjfObjInfoEquation Native uation Equation.39q' d 3 ydx 3 ,2fx(),d 3 dx 3 fx()[],D x  3 fx()[]_1216743621F22Ole CompObjfObjInfo FMicrosoft Equation 3.0 DS Equation Equation.39q(; d 4 ydx 4 ,f iv x(),d 4 dx 4 fx()[],D x  4 fx()[]Equation Native ,_1216743652F22Ole CompObjf FMicrosoft Equation 3.0 DS Equation Equation.39q ( d n ydx n ,f n x(),d n dx n fx()[],D x  n ObjInfoEquation Native (_1216743754F22Ole fx()[] FMicrosoft Equation 3.0 DS Equation Equation.39q=L4 d 2 ydx 2CompObjfObjInfoEquation Native Y_1216743780F22Ole CompObjfObjInfoEquation Native  FMicrosoft Equation 3.0 DS Equation Equation.39qmo\ d 2 dx 2  x 2 +1  [] FMicrosoft Equation 3.0 DS Eq_1216743876F22Ole CompObjfObjInfouation Equation.39q2cts f iv x() FMicrosoft Equation 3.0 DS Equation Equation.39qEquation Native N_1216743877F22Ole CompObjfObjInfoEquation Native _1285355886mdO)2\2Ole rL9 fx()=x 10 90+x 5 60 dO)1Microsoft Office PowerPoint 97-2003 PresentationMSPresentationPowerPoint.Show.89q CompObjObjInfoPicturesCurrent UserG%_< Chris HeadleeChris HeadleeOh+'0 `h  PowerPoint PresentationChris Headlee2Microsoft PowerPoint@ʚA@@@Y,՜.+,0SummaryInformation(@PowerPoint Document(7`DocumentSummaryInformation8_1254000299OdO)\2 2 (d/ 00DTimes New Roman`TN5hTDArialNew Roman`TN5hT DCalibriw Roman`TN5hT"0DWingdingsRoman`TN5hT@0.  @n?" dd@  @@`` x4       0@Hʚ;ʚ;<4dddd5h: Ⱥ5hg4GdGdd d dȺppp@ i)___PPT12 %0___PPT10 ?  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\P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    Tl')CKKKAJA?LX+ 6x)  % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08tdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TT*'0CKKKAJA*?LP % ( Rp@Times New Roman0.+\p\0\Fh7,p0PU0z08+8t+8 0+0$8m508| (ȱd dv%    Tlm<XKKKAJAmTLXy = :  % ( Rp@Times New Roman,L|~Q|Hm| ~~8t+8 0+0PU0z08Dm508 (ȱd dv !#\P|p|m|Z|m|dv%    T<7XKKKAJATLt-------------------       % ( Rp@Times New Roman+8L|~Q|Hm| ~~m508 (0PU0z08Иdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    Tp<XKKKAJATLXfind y  % ( Rp@Times New Roman L|~Q|Hm| ~~ !# \P|p|0PU0z08\dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TT<XKKKAJATLP  % ( Rp@Times New Roman0 L0 TiK0W0LF0PU0z080 !08m5084 ( dv%    T`QmKKKAJAiLT(7x  % ( Rp@Times New RomanW0LL|~Q|Hm| ֨~~0 0PU0z08Dm508 ( dv !#\P|p|m|Z|m|dv%    TTQmKKKAJAiLP % ( Rp@Times New Roman L|~Q|Hm| ר~~m508 (0PU0z08Иdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TdQmKKKAJAiLT+ 3x  % ( Rp@Times New Roman L|~Q|Hm| ب~~ !# \P|p|0PU0z08\dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTQmKKKAJAiLP % ( Rp@Times New Roman \L|~Q|Hm| ٨~~ !# \P|p|0PU0z08虥dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTQmKKKAJAiLP) % ( Rp@Times New Roman \L|~Q|Hm| ڨ~~ !# \P|p|0PU0z08tdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTQmKKKAJAiLP % ( F(GDICB0FGDICRp@Times New Roman0 LD0 4%TiK0W0LDF40PU0z08 &0 !08m508h (P( dv%    T,HNdKKKAJA`%LSteps in Logarithmic Differentiation:        % ( Rp@Times New Roman>0#0-_H@h0-0T000PU0z08\ -_dvdv|տ0844p0:40.dv%    TXfKKKAJA~LP1.  % ( Rp@Times New Roman|-w~ !#`|-w 0w ~ !#|6w;w 0w @G-nwNnw 80PU0z08 0PD@Times New Romanm5000\8~ؿdv%    TxfKKKAJA~2LTake natural log of both sides of an equation y =                 % ( Rp@Times New Romanw @G-nwL|~Q|Hm| ~~0PD0PU0z08虥m5000 8~ؿdv !#\P|p|m|Z|m|dv%    T`fKKKAJA~LTf(x   % ( Rp@Times New RomanL|~Q|Hm| ~~m5000 8~0PU0z08tdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTf#KKKAJA~LP) % ( Rp@Times New Roman>0#0-_H@虥Th0-0T000PU0z08\ -_dvdvdvdv@Times New Roman p |0 dv%    TXKKKAJALP2.  % ( Rp@Times New Roman|-w~ !#`|-w 0w ͨ~ !#|6w;w 0w 8G-nwNnw 80PU0z08 0PD@Times New Romanm5000\8ؿdv%    T.KKKAJALUse to laws of logs to simplify          % ( Rp@Times New Roman>0#0-_H@虥Th0-0T000PU0z08\ -_dvdvdvdv@Times New Roman p |0 dv%    TXKKKAJALP3.  % ( Rp@Times New Roman|-w~ !#`|-w 0w ~ !#|6w;w 0w @G-nwNnw 80PU0z08 0PD@Times New Romanm5000\8ؿ dv%    THKKKAJA*LDifferentiate implicitly with respect to x          % ( Rp@Times New Roman>0#0-_H@h0-0T000PU0z08\ -_dvdvdvdv@Times New Roman p |0 dv%    TXKKKAJALP4.  % ( Rp@Times New Roman|-w~ !#`|-w 0w ~ !#|6w;w 0w 8G-nwNnw 80PU0z08 0PD@Times New Romanm5000\8ؿdv%    T,kKKKAJA%LSolve the resulting equation for y (           % ( Rp@Times New Romanw 8G-nwL|~Q|Hm| ~~0PD0PU0z08虥m5000 8ؿdv !#\P|p|m|Z|m|dv%    TllKKKAJAlLXdy/dx   % ( Rp@Times New RomanL|~Q|Hm| ~~m5000 80PU0z08tdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP) % ( Rp@Times New Roman>0#0-_H@虥Th0-0T000PU0z08\ -_dvdvdvdv@Times New Roman p |0 dv%    TXKKKAJALP5.  % ( Rp@Times New Roman|-w~ !#`|-w 0w ~ !#|6w;w 0w @G-nwNnw 80PU0z08 0PD@Times New Romanm5000\8ؿdv%    T<KKKAJA(LSubstitute back in what y was in step 1.          % ( F(GDICbFGDICRp@Times New Roman0 L0 t5TiK0W0LFt0PU0z08D60 !08m508|(-dv%    TXKKKAJALPln % ( Rp@Times New RomanW0LL|~Q|Hm| ~~0 0PU0z08Иm508 (-dv !#\P|p|m|Z|m|dv%    TpKKKAJALXy = 2   % ( Rp@Times New Roman L|~Q|Hm| ~~m508 (0PU0z08\dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TXKKKAJALPln % ( Rp@Times New Roman L|~Q|Hm| ~~ !# \P|p|0PU0z08虥dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    Tp.KKKAJALXx + 3   % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08tdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TX/CKKKAJA/LPln % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKRKKKAJAKLP( % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TXSlKKKAJASLP5x  % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTmsKKKAJAmLP % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TpzKKKAJAzLX+ 6x)   % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08{dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP  % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08x|dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TXKKKAJALP3  % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TXKKKAJALPln % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP( % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TXKKKAJALP7x  % ( Rp@Times New Roman \L|~Q|Hm| ~~ !# \P|p|0PU0z08$>dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP % ( Rp@Times New Roman \L|~Q|Hm| !~~ !# \P|p|0PU0z08Ldv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    Td MKKKAJA LT+ 3x  % ( Rp@Times New Roman \L|~Q|Hm| "~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTNTKKKAJANLP % ( Rp@Times New Roman \L|~Q|Hm| #~~ !# \P|p|0PU0z08\dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTU\KKKAJAULP) % ( Rp@Times New Roman0 0Lt0 :TiK0W0LtF0PU0z08p;0 !08m508أ(xdv%    TdmKKKAJAmLT1  % ( Rp@Times New RomanW0LtL|~Q|Hm| 1~~0 0PU0z08Dm508 (xdv !#\P|p|m|Z|m|dv%    TXKKKAJALPdy  % ( Rp@Times New Roman L|~Q|Hm| 2~~m508 (0PU0z08Иdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TKKKAJA"L1 10x + 6 21x     % ( Rp@Times New Roman L|~Q|Hm| 3~~ !# \P|p|0PU0z08\dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP % ( Rp@Times New Roman \L|~Q|Hm| 4~~ !# \P|p|0PU0z08虥dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    Td;KKKAJALT+ 6x  % ( Rp@Times New Roman0 fL{0 9TiK0W0L{F0PU0z08t:0 !08m508  (-| dv%    TXm|KKKAJAmLP-- % ( Rp@Times New RomanW0L{L|~Q|Hm| g~~0 0PU0z08Иm508 (-| dv !#\P|p|m|Z|m|dv%    TlKKKAJALX-----   % ( Rp@Times New Roman L|~Q|Hm| h~~m508 (0PU0z08\dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TdKKKAJALT= 2   % ( Rp@Times New Roman L|~Q|Hm| i~~ !# \P|p|0PU0z08虥dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP"  % ( Rp@Times New Roman \L|~Q|Hm| j~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    T`KKKAJALT---  % ( Rp@Times New Roman \L|~Q|Hm| k~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    T`)KKKAJALT+ 3  % ( Rp@Times New Roman \L|~Q|Hm| l~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TT08KKKAJA0LP"  % ( Rp@Times New Roman \L|~Q|Hm| m~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    T?KKKAJA? Ld-----------    % ( Rp@Times New Roman \L|~Q|Hm| n~~ !# \P|p|0PU0z08Ldv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP  % ( Rp@Times New Roman \L|~Q|Hm| o~~ !# \P|p|0PU0z08$>dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TXKKKAJALP3  % ( Rp@Times New Roman \L|~Q|Hm| p~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP"  % ( Rp@Times New Roman \L|~Q|Hm| q~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    THKKKAJALh--------------     % ( Rp@Times New Roman0 lL{0 :TiK0W0L{F0PU0z08;0 !0 8 m508$((  dv%    TdmKKKAJAmLTy  % ( Rp@Times New RomanW0L{L|~Q|Hm| m~~0 0PU0z08Dm508 (  dv !#\P|p|m|Z|m|dv%    TXKKKAJALPdx  % ( Rp@Times New Roman L|~Q|Hm| n~~m508 ( 0PU0z08Иdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TAKKKAJA Ldx  % ( Rp@Times New Roman L|~Q|Hm| o~~ !# \P|p|0PU0z08\dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TXB[KKKAJABLP5x  % ( Rp@Times New Roman \L|~Q|Hm| p~~ !# \P|p|0PU0z08虥dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TT\bKKKAJA\LP % ( Rp@Times New Roman \L|~Q|Hm| q~~ !# \P|p|0PU0z08tdv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TiKKKAJAiLl+ 6x   % ( Rp@Times New Roman \L|~Q|Hm| r~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TXKKKAJALP7x  % ( Rp@Times New Roman \L|~Q|Hm| s~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TTKKKAJALP % ( Rp@Times New Roman \L|~Q|Hm| t~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    Td/KKKAJALT+ 3x  % ( Rp@Times New Roman \L|~Q|Hm| u~~ !# \P|p|0PU0z08dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TT06KKKAJA0LP % ( Rp@Times New Roman0 Lt0 =TiK0W0LtF0PU0z08Dp>0 !08m508أ(xdv%    TX(DKKKAJA@LPdy  % ( Rp@Times New RomanW0LtL|~Q|Hm| ~~0 0PU0z08Иm508 (xdv !#\P|p|m|Z|m|dv%    T( DKKKAJA@"L1 10x + 6 21x     % ( Rp@Times New Roman L|~Q|Hm| ~~m508 (0PU0z08\dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    TT (DKKKAJA @LP % ( Rp@Times New Roman L|~Q|Hm| ~~ !# \P|p|0PU0z08虥dv !# \P|p|m|Z|m|dv !#\P|p|m|Z|m|dv%    Td(GDKKKAJA@LT+ 6x  % ( Rp@Times New Roman0 L\0 :TiK0W0L\F0PU0z08Dt;0 !08m508  (-| dv%    Tl=YKKKAJAULX-----   % ( 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Some dimensions in the problem remain fixed as time passes. Label these as constants in the diagram. Other information defines the point in time at which you are to calculate the rate of change. Do not label these dimensions as constants as they vary with time. Identify what is given and what is wanted in terms of the established variables. Write a general equation relating the variables. Differentiate this equation with respect to t. 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If the rocket is rising vertically at 880 ft/s w     % ( Rp@"ArialJL|Q|Hm| 40m4I0PU0z08gЀm50l fЀ(\{H(fdv'!q\P|p|m|Z|m|dv%    Te}KKKAJA} L`hen it is  % ( Rp@"Arial|/0BD\{mD\/00PU0z08gHm0m4Cl|pЀm50lq,Ѐ(\{H(,Ѐdv%    T^}gKKKAJA^@L4000 ft above the launching pad, how fast must the camera elevat   % ( Rp@"ArialDL|Q|Hm| 0m4C0PU0z08gЀm50l ,Ѐ(\{H(,dv'!q\P|p|m|Z|m|dv%    Tdh}KKKAJAhLTion  % ( Rp@"Arial0 <fg0  \{=TiK0W0fgF 0PU0z08gH3 >03 \{!0l|pЀm50lq Ѐ(\{ ( Ѐdv%    T^+KKKAJA^=Langle at that instant to keep the camera aimed at the rocket?    % ( Rp@"Arial0 Efh0 Ķ \{DTiK0W0fhT Ķ 0PU0z08gp3 E03 \{!0l|pЀm50lr Ѐ(\{pP Ѐdv%    T2KKKAJA2LlSolution: Let  % ( Rp@"ArialW0fhL|Q|Hm| F03 \{0PU0z08gЀm50l Ѐ(\{pP dv'!q\P|p|m|Z|m|dv%    TTKKKAJALPt % ( Rp@"Arial3 L|Q|Hm| Gm50l Ѐ(\{0PU0z084hdv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TaKKKAJA3L= number of seconds elapsed from the time of launchl   % ( Rp@"Arial0 3fi0  \{LTiK0W0fid 0PU0z08g3 M03 \{!0l|pЀm50l'!q \P|p|0PU0z08hdv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TXKfKKKAJAbLPdt ( Rp@"Arial \L|Q|Hm| '!q \P|p|0PU0z08idv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    T`J5gKKKAJAbLT(at ( Rp@"Arial \L|Q|Hm| @'!q \P|p|0PU0z08djdv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    T`=KafKKKAJA=bLTh=  ( Rp@"Arial \L|Q|Hm| '!q \P|p|0PU0z08jdv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TlbJgKKKAJAbbLX4000) ( Rp@"Arial \L|Q|Hm| B'!q \P|p|0PU0z08ldv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TJ%gKKKAJAb Ldgiven that  ( Rp@"Arial \L|Q|Hm| '!q \P|p|0PU0z08d dv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    T`&KMfKKKAJA&bLTdh/ ( Rp@"Arial \L|Q|Hm| D'!q \P|p|0PU0z08 dv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TXKKdfKKKAJAKbLPdt ( Rp@"Arial \L|Q|Hm| '!q \P|p|0PU0z08 dv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TdjJgKKKAJAjbLT(at  ( Rp@"Arial \L|Q|Hm| F'!q \P|p|0PU0z085 dv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TTKfKKKAJAbLPh ( Rp@"Arial \L|Q|Hm| '!q \P|p|0PU0z08 dv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TJ`gKKKAJAbLp=4000) = 880 ft/s ( F(GDICk{F(GDICVFxlEMF+*@$??@4(`QCjC`QC9NDcA9ND@( !b $$=='%  % V(U F L%L%%  % $$AAF`TEMF+@@4^UU?A@$$==_888% % V(W F L%L%% % $$AA( " FEMF+@ FGDICF(GDICFGDICRp@"Arial0#\m,0 Ț m0PU0z08g8t\{8 0\{0$l|pЀm50lqЀ(\{ (Ѐdv%    TT KKKAJALPh  ( F(GDIC#Y|FGDICRp@Wingdings0#\m,0 Ț m0PU0z08g8t\{8 0\{0$l|pЀm50lrЀ(\{pPЀdv%    TT.`AtKKKAJA.qLP ( Rp@"Arial L|Q|Hm| 8t\{8 0\{0PU0z08gЀm50l Ѐ(\{pPdv'!q\P|p|m|Z|m|dv%    T`_tKKKAJA`q Lh3000  ( Rp@Wingdings\{8L|Q|Hm|l Bm50l Ѐ(\{0PU0z084hdv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TT`tKKKAJAqLP ( F(GDICL&dFtEMF+*@$??@<0\sAhD@>|ADD ADp0BeD@( !b $$=='%  % V,M$by$_%&5%%  % $$AAF\PEMF+@<0VUU?A@: $$==_888 % % V,A0ny$_%&5%% % $$AA( : " FEMF+@ FGDICF(GDICkF(GDICkFxlEMF+*@$??@4(PC(C9KCCUCC@33( !b $$=='33%  % V(m  kT k%  % $$AAF`TEMF+@@4^UU?A33@$$==_88833% % V(l  kT k% % $$AA( " FEMF+@ FGDICF(GDICFXLEMF+*@$?? @$339KCUC@APUA!b $$==% % V0 s FT FT s s% % $$AAFEMF+@<0VUU?A33@<09KCUC9KCjCUCjCUCUC@: $$==_888 33% % V,{ s FT FT s% % $$AA( : " FEMF+@ FGDICF(GDICFxlEMF+*@$??@4(VCxCVCrTC_CrTC@33!b $$==%  % V(Y Y C C%  % $$AAF`TEMF+@ @4^UU?A33@ $$==_88833% % V(Y Y C C% % $$AA( " FEMF+@ FGDICF(GDICFxlEMF+*@$??@ 4(xKCxCxKCrTCACrTC@ 33!b $$==%  % V(  C C%  % $$AAF`TEMF+@ @4^UU?A33@  $$==_88833% % V(  C C% % $$AA( " FEMF+@ FGDICFGDICF(GDIC,9M\FGDICRp@"Arial0#\m,0 Ț m0PU0z08g8t\{8 0\{0$l|pЀm50lqЀ(\{ (Ѐdv%    7QRp@"Arial\m,0 Ț m0PU0z08gww!q4 0w'!qpd !  /T w'!q(w,Xdv% %  TT7?AUKKKAJA7QLP  % % (  ( FGDICF(GDICvFGDICRp@"Arial0l Nfg0  \{TiK0W0fg  0PU0z08gH3 03 \{!0l|pЀm50lqЀ(\{ (Ѐdv%    TTuKKKAJALPh ( Rp@"Arial0 ifdj0 ̶ \{FTiK0W0fdjX ̶ 0PU0z08g3 G03 \{!0l|pЀm50lsЀ(\{聥Ѐdv%    Td9KKKAJALTtan  ( Rp@"ArialW0fdjL|Q|Hm| j03 \{0PU0z08gЀm50l Ѐ(\{聥dv'!q\P|p|m|Z|m|dv%    :Rp@"ArialdjL|Q|Hm| j03 \{0PU0z08gЀww!q4 0w'!qpd +  /T w'!q(w Xdv% %  TT:HKKKAJA:LP % % (  ( Rp@"Arial3 L|Q|Hm|l Qm50l Ѐ(\{0PU0z08hdv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    T`OkKKKAJAOLT=  ( Rp@"Arial ЀL|Q|Hm|l R'!q \P|p|0PU0z08Lidv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TlKKKAJAl L`----------   ( Rp@"Arial \L|Q|Hm| '!q \P|p|0PU0z08idv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    TfKKKAJAfL(relate variables involved)   ( Rp@"Arial0 fg0  \{TiK0W0fg  0PU0z08gH3 03 \{!0l|pЀm50lqЀ(\{ (Ѐdv%    Td~KKKAJA~LT3000 ( Rp@"Arial0l _fLi0 L \{&TiK0W0fLi L 0PU0z08g3 '03 \{!0l|pЀm50lrЀ(\{dЀdv%    TTiyKKKAJAiLPd ( Rp@"ArialW0fLiL|Q|Hm|l `03 \{0PU0z084hЀm50l Ѐ(\{ddv'!q\P|p|m|Z|m|dv%    xRp@"ArialLiL|Q|Hm|l `03 \{0PU0z084hЀww!q4 0w'!qpd +  /T w'!q(wXdv% %  TTxKKKAJAxLP % % (  ( Rp@"Arial3 L|Q|Hm| m50l Ѐ(\{0PU0z08hdv'!q \P|p|m|Z|m|dv'!q\P|p|m|Z|m|dv%    T<KKKAJA Ld1 dh ( Rp@"Arial0. 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FEMF+@ CompObj~{ObjInfo[_PicturesCurrent User^aG %_*& :Chris HeadleeChris HeadleeOh+'0 X`p  Slide 1Chris HeadleeoChris Headleeo2riMicrosoft PowerPointP@~@У@"oSummaryInformation((PowerPoint Document(`bN&DocumentSummaryInformation8_1254463737YgdO)p*2p $2( `/ 0DArialngstt&0B 0"DWingdingstt&0B 0@ .  @n?" dd@  @@`` (  0AA@3ʚ;ʚ;g4\d\d B 0Lppp@ <4dddd8pC 0t& 80___PPT10   ` 33` Sf3f` 33g` f` www3PP` ZXdbmo` \ғ3y`Ӣ` 3f3ff` 3f3FKf` hk]wwwfܹ` ff>>\`Y{ff` R>&- {p_/̴>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>> f(    6b  `}  T Click to edit Master title style! !  0  `  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  0  ^ `  >*  0 ^   @*  0ܤ ^ `  @*H  0޽h ? 3380___PPT10.p9Y Default Design  (    <  GRelated Rates Example  0 pld\___PPT9>6 HtProblem: A camera is mounted at a point 3000 ft from the base of a rocket launching pad. If the rocket is rising vertically at 880 ft/s when it is 4000 ft above the launching pad, how fast must the camera elevation angle at that instant to keep the camera aimed at the rocket? Solution: Let t = number of seconds elapsed from the time of launch  = camera elevation angle in radians after t seconds h = height of the rocket in feet after t seconds We must find d/dt (at h= 4000) given that dh/dt (at h=4000) = 880 ft/s:;P'L"+"" ""&"""""""" """""XtgGF e>L  >eL`b B 0x p    B܊ *   3h   Bm  L j* 3000 $f   6Ȝ,e^ T r v # \> x ZR   s *r <Z   s *<Zb  s *>svZb B s *s  B|'    L"  <p) 2Z h tan  = ---------- (relate variables involved) 3000 d 1 dh sec  ------ = -------- ------ (differentiating with respect to t) dt 3000 dt d 1 dh ----- = -------- ------ cos  (solving for d /dt) dt 3000 dt d 1 3 66 ----- = -------- (880) (-----) = ------ H" 0.11 rad/s H" 6.05 deg/s dt 3000 5 625 FqFF d      S            6       "    M$ $$ $.$ $$ $.fS;i.  <3 O2 6 adj 3 cos  = ------ = --- hyp 5&FF)"b  <T:9a = at h = 4000  H  0޽h ? 3380___PPT10.`Yr0s e&՜.+,0    KOn-screen Show -sN&: Arial WingdingsDefault DesignSlide 1  Fonts UsedDesign Template Slide TitlesOle EPRINTdfbCompObj{ObjInfoei dO)"Microsoft PowerPoint PresentationMSPresentationPowerPoint.Show.89q %_~( @Chris HeadleeChris HeadleeOh+'0 X`p                            ! 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Find the maximum error in volume.@     % ( F(GDIC,DCFGDICRp@"Arial0 <؊0 Z>8 TiK0W0<؊~Q>Z>0PU0z08 @0@8!0(m50(l(8tddv%    T|EJgKKKAJAEbL\V = 4/3  % ( Rp@"ArialW0<؊L|0Q|Hm| 880@80PU0z084m50( (8tddv!d\P|p|m|Z|m|dv%    bRp@"Arial؊L|0Q|Hm| 880@80PU0z084ww!d4 0w!d B  /T w!d(wppdv% %  TTJgKKKAJAbLP % % (  ( Rp@"Arial@L|0Q|Hm| 88m50( (80PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TTJgKKKAJAbLPr  ( Rp@"Arial L|0Q|Hm| 88!d \P|p|0PU0z08Ldv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TTJgKKKAJAbLP ( Rp@"Arial0 <|0 Z>8 TiK0W0<|~Q>Z>0PU0z08H@0@8!0(m50((8좥dv%    TX7hVKKKAJA7LPdV ( Rp@"ArialW0<|L|0Q|Hm| 880@80PU0z084m50( (8좥dv!d\P|p|m|Z|m|dv%    Td^hKKKAJA^LT= 4  ( Rp@"Arial@L|0Q|Hm| C88m50( (80PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    Rp@"Arial@L|0Q|Hm| C88m50( (80PU0z08dvww!d4 0w!d B  /T w!d(wpdv% %  TThKKKAJALP % % (  ( Rp@"Arial L|0Q|Hm| 88!d \P|p|0PU0z08Ldv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TThKKKAJALPr  ( Rp@"Arial \L|0Q|Hm| 88!d \P|p|0PU0z08؊dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TThKKKAJALP ( Rp@"Arial \L|0Q|Hm|= n88!d \P|p|0PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TXhKKKAJALPdr  ( Rp@"Arial0 <0 Z>8TiK0W0<Q>Z>0PU0z084@0@8!0(m50($(8Ģdv%    TX7VKKKAJA7LPdV ( Rp@"ArialW0<L|0Q|Hm| 880@80PU0z084m50( (8Ģdv!d\P|p|m|Z|m|dv%    Td^KKKAJA^LT= 4  ( Rp@"Arial@L|0Q|Hm| 88m50( (80PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    Rp@"Arial@L|0Q|Hm| 88m50( (80PU0z08dvww!d4 0w!d B  /T w!d(wpdv% %  TTKKKAJALP % % (  ( Rp@"Arial L|0Q|Hm| 88!d \P|p|0PU0z08Ldv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TdKKKAJALT(10) ( Rp@"Arial \L|0Q|Hm| 88!d \P|p|0PU0z08؊dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TTKKKAJALP ( Rp@"Arial \L|0Q|Hm| G88!d \P|p|0PU0z08ddv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TxKKKAJAL\(0.05)  ( Rp@"Arial0.6\`0dQ>0\>7`0PU0z0868t688 0860$(m50( (8LP dv%    TaKKKAJAa Ld= 62.832 cm ( Rp@"Arial>7L|0Q|Hm|= 888t688 0860PU0z084m50( (8LP dv!d\P|p|m|Z|m|dv%    TTKKKAJALP ( Rp@"Arial688L|0Q|Hm| 88m50( (80PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TIKKKAJAIL(maximum error in volume)    ( Rp@"Arial0 d<0 Z>8TiK0W0<Q>Z>0PU0z084p@0 @8!0 (m50(k>(8< dv%    TX>]KKKAJA>LPdV ( Rp@"ArialW0<L|0Q|Hm| e880 @80PU0z08Lm50( (8< dv!d\P|p|m|Z|m|dv%    TXKKKAJALP4  ( Rp@"Arial @L|0Q|Hm|= 88m50( (80PU0z08؊dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    Rp@"Arial@L|0Q|Hm|= 88m50( (80PU0z08؊dvww!d4 0w!d B  /T w!d(w@pdv% %  TTKKKAJALP % % (  ( Rp@"Arial L|0Q|Hm| g88!d \P|p|0PU0z08ddv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    T`KKKAJALTr  ( Rp@"Arial \L|0Q|Hm| h88!d \P|p|0PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TXKKKAJALPdr  ( Rp@"Arial \L|0Q|Hm| 88!d \P|p|0PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    TX/GKKKAJA/LPdr  ( Rp@"Arial0 <0 p[>8CTiK0W0<Q>p[>0PU0z084@D0@8!0(m50($(8Ģdv%    Tl7 _)KKKAJA7$LX-----  ( Rp@"ArialW0<L|0Q|Hm| 880@80PU0z084m50( (8Ģdv!d\P|p|m|Z|m|dv%    TXg |)KKKAJAg$LP=  ( Rp@"Arial@L|0Q|Hm| l88m50( (80PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    T} )KKKAJA}$Lh--------------     ( Rp@"Arial L|0Q|Hm| 88!d \P|p|0PU0z08Ldv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    Td ")KKKAJA$LT= 3  ( Rp@"Arial \L|0Q|Hm| n88!d \P|p|0PU0z08؊dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    Tl# K)KKKAJA#$LX-----  ( Rp@"Arial \L|0Q|Hm| 88!d \P|p|0PU0z08ddv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    T )KKKAJA$LlRelative Error     ( Rp@"Arial0 <؊0 Z>8 TiK0W0<؊Q>Z>0PU0z08 @!0@8!0(m50(l(8tddv%    TE!>KKKAJAE9 L`V  ( Rp@"ArialW0<؊L|0Q|Hm| 880@80PU0z084m50( (8tddv!d\P|p|m|Z|m|dv%    Td!>KKKAJA9LT4/3  ( Rp@"Arial@L|0Q|Hm| r88m50( (80PU0z08dv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    9Rp@"Arial@L|0Q|Hm| r88m50( (80PU0z08dvww!d4 0w!d B  /T w!d(wpdv% %  TT!>KKKAJA9LP % % (  ( Rp@"Arial L|0Q|Hm| 88!d \P|p|0PU0z08Ldv!d \P|p|m|Z|m|dv!d\P|p|m|Z|m|dv%    T!@>KKKAJA9Llr r   ( F4(EMF+*@$??!b Ld)??" 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