ࡱ>     nAy bjbj {{g&8&|sTFv %"%%%FFFSSSSSSS$VXnSuFCFFFS%%3-TNNNF%%SNFSNNVQQ%@ş GnQ~SCT0sTvQ,=YM=YQ=YQFFNFFFFFSSNFFFsTFFFF=YFFFFFFFFF : Derivatives Derivatives Video 1 Popper 09 Definition Geometric considerations Average rate of change The slope connection Shortcuts to the derivative: Power Rule Properties of the Derivative Scalar Multiplication Sums and Differences What does it mean when the derivative is zero? Increasing and decreasing Derivatives Video 2 Popper 10 Exponential Functions More shortcuts: Product Rule Quotient Rule Derivatives Video 3 Popper 11 Application Problems Appendices Increasing/Decreasing functions Definition of Derivative The Derivative of a function, f (x), is the  EMBED Equation.DSMT4 . It is denoted  EMBED Equation.DSMT4  And, while its important to know this, its also important to know what this means physically in the plane, so lets back up a bit and talk about rates of change. If I have a table of inputs and outputs, I can discuss the average rate of change of the outputs in the following way: The average rate of change from an initial to a final output is the difference in the outputs divided by the difference in the inputs doing the final first in each difference: AROC:  EMBED Equation.DSMT4  If y is your output and x is your input this is the slope formula:  EMBED Equation.DSMT4  Where final is defined to be Point 2 and initial is defined to be Point 1. Rates of change and derivatives are intimately connected to the formula for the slope of a line. Watch for how often this comes up in the next few pages. Heres a table of values: inputs (x)outputs (y)-39-24-1100112439416525 The average rate of change from (3 (initial) to (1 is  EMBED Equation.DSMT4  The average rate of change from (1 (initial) to 4 is  EMBED Equation.DSMT4  Now, Im using the function  EMBED Equation.DSMT4  as my output generator. Lets look at the geometry of these calculations: From ((3, 9) to ((1, 1), connect these points with a secant line, rather than the line of the graph of the function The average rate of change is the slope of the secant line. that joins these two points:  EMBED Equation.DSMT4   From ((3, 9) to ((1, 1), connect these points with a secant linethis secant line is really NOT the graphline but it provides two endpoints to the segment that are graph points and a systematic way to get from one to the other in steps From (1 to 4: The average rate of change from (1 to 4 is  EMBED Equation.DSMT4  Lets look at the line joining ((1, 1) and (4, 16)this is a secant line with slope 3. If we start at the initial point and step over 1 and up 3 successively where do we end up?  What is the slope of this line? Note that for each step to the right you go up a specific amount. The first point and second point are graph points, but the points in between on the secant line are not graph points unless your graph is actually a line. So Average Rate of Change is associated with the slope of the secant line joining two graph points. Let s look at one more example. Given  EMBED Equation.DSMT4 , what is the average rate of change from x = "3 to x = "1? Sketch this on the graph, then calculate the slope of the secant line joining these points.  Now let s put this in words: The initial graph point is The final graph point is For every unit step to the right we go up This approximates the graph, but NOTE, not exactly what IS the value at "2? And on the secant line, what is the approximate value? You can see that youll start and end on graph points but your motion doesnt follow the graph well. Popper 09, Question 1 Slopes, Difference Quotients, and derivatives are connected. So lets start with a given functions graph and put in  EMBED Equation.DSMT4 and  EMBED Equation.DSMT4 . Lets get the set-up from average rate of change through Difference Quotient through derivative using x and x + h as our xs. Note where h is on the x-axis.  Note that the average rate of change is a stair step like motion from initial to final. m = EMBED Equation.DSMT4  Do you see that this is  EMBED Equation.DSMT4 ? Now lets actually take the limit as h approaches zero: What does it mean that h approaches zero?  EMBED Equation.DSMT4  Now lets look at it at the end of the limiting process. The derivative is the SLOPE of the line tangent at x.  We call the derivative the instantaneous rate of change of y with respect to x. It is the slope of the tangent line at x. Amazingly, its a one-point slope. Now lets work with this a bit. Given  EMBED Equation.DSMT4 . Find  EMBED Equation.DSMT4  which is the instantaneous rate of change of x with respect to y at x = 2. First lets find  EMBED Equation.DSMT4 . Find the DQ Take the limit of the DQ Evaluate the solution at x = 2. Difference Quotient:  EMBED Equation.DSMT4  Evaluate it  Now lets put some words to it: The rate of change of x with respect to y at x = 2 is the expected change in y for an infinitesimal movement rightward from 2. If the derivative is 3, thats a small positive change. If its 110, thats a large positive change. In both of these cases, the graph is increasing. If the derivative is "0.3, that s a small negative change. If it s "57, that s a good sized negative change. In both of these cases, the graph is decreasing. Popper 09, Question 2 Shortcuts to getting the derivative: First let s do it the LONG way. Then well look at pictures and a shortcut. Given  EMBED Equation.DSMT4  Find its derivative. We will follow these steps. Find the DQ Take the limit of the DQ as  EMBED Equation.DSMT4  Finding the Difference Quotient:  EMBED Equation.DSMT4  Subtracting f(x) gives:  EMBED Equation.DSMT4  Dividing by h gives:  EMBED Equation.DSMT4  for the Difference Quotient Last:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Taking the limit as h approaches zero gives:  EMBED Equation.DSMT4  which is the derivative. Now lets look at what this tells us: In the plane: If x = 1, then f (1) = 1 and the slope of the line tangent to this graph at x = 1 is 3. You evaluate the derivative at the x value to get this number, and it will change as x changes. In the Big Picture: For an itsy bitsy step off of 1 to the right, the y value will go up 3 times the size of that step. That is to say, the instantaneous rate of change of y with respect to x is 3. This means that the measure of the change in the graph for places close to one is 3for every horizontal step horizontally you take 3 times that amount vertically. In fact, given this point (1, 1) and m = 3, I can come up with the FORMULA for this tangent line using the old point slope formula:  EMBED Equation.DSMT4  y ( 1 = 3 (x ( 1) i.e. y = 3x ( 2. This is nice to know and a standard question. Heres a picture of that line on the graph with the tangent line on it at x = 1. 3 is the instantaneous rate of change of x with respect to y for this function at x = 1. Note that the rate of change is NOT a constant. It changes depending on where you are on the graph.  Popper 09, Question 3 Now lets begin seeing the derivative as a function related to our original function that calculates instantaneous rates of change for the original function. It works like this: The original function takes x and computes the second coordinate for the graph point at x. The derivative takes x and computes the instantaneous rate of change of the function at x. Lets look at a table of these: xGraph point  EMBED Equation.DSMT4 Slope of tngt l  EMBED Equation.DSMT4  (2  (8 12  (1 (1 3 0  0 0 3 27 27 Lets talk about lines with slopes like the following: m < 0 m = 0 m > 0 m >> 0 For a tiny step horizontally the vertical motion is approximately multiple of the slope. Notice that the formula for the derivative above guarantees a positive slope over the course of the function. This is important and it means that the graph is increasing on its domain. And now, a tiny summary: Average rate of change: slope of the secant line Instantaneous rate of change: slope of the tangent line For derivatives: If the slope is negative, the graph is decreasing. If the slope is positive, the graph is increasing. If the slope is zero, there is a turn around point at that x. Shortcut #1 The Power Rule! Given:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Example 1 again  EMBED Equation.DSMT4  Lets track n and n 1 in this Example 2  EMBED Equation.3  Example 2A:  EMBED Equation.DSMT4  Example 3 f(x) = 5 This will be a big deal in the next module on Integration! Popper 09, Question 4 To review:  EMBED Equation.DSMT4  Domain End Behavior Range y-intercept x-intercept DERIVATIVE! Properties of derivatives: Scalar Multiplication: Given a multiple of a function, the derivative has the same multiplier. Example 4:  EMBED Equation.3  numbers like 5 are called scalars. Dont forget + c  EMBED Equation.3  Popper 09, Question 5 Sums and Differences: The derivative of a sum or a difference is the sum or difference of the derivatives of the summands (terms). Example 5:  EMBED Equation.3  Example 6:  EMBED Equation.3  Popper 09, Question 6 Example 7: Find the equation of the tangent line at x = 3 for  EMBED Equation.3 .  EMBED Equation.3   EMBED Equation.3 = m f(3) = 27 + 9 (5 = 31. This is the point to use: (3, 31) in the point slope equation:  EMBED Equation.3  gives y ( 31 = 33(x ( 3) The tangent line is y = 33x ( 68. Sketch it on the graph provided.  Now lets talk about the typical polynomial question: Given this polynomial, find: Domain, range, intercepts, derivative, sketch, equation of tangent line at a given point. Example 8: Given:  EMBED Equation.3  Domain: all Real Numbers Vertex: CTS: y =  EMBED Equation.3 . (3, (4) x-intercepts: (1, 0) and (5, 0) y-intercept: (0, 5) Derivative:  EMBED Equation.3  Equation of tangent line at 2 and at 3 and at 5.  EMBED Equation.3  f(2) = (3 y + 3 = (2(x (2)y = (2x + 1 m < 0  EMBED Equation.3  f(3) = (4 y + 4 = 0(x ( 3).y = ( 4 m = 0  EMBED Equation.3  f(5) = 0 y + 0 = 4(x ( 5).y = 4x ( 20 m > 0 Sketch them on the graph:  Popper 09, Question 7 What does it mean when the derivative is zero? It means that the instantaneous rate of change is going from a negative number to a positive number (or vice versa) and that youve found a turn-around point. What does it mean when the derivative is positive? The graph is increasing. What does it mean when the derivative is negative? The graph is decreasing. Increasing and decreasing functions: A function is increasing if  EMBED Equation.DSMT4 . A function is decreasing if  EMBED Equation.DSMT4 . Lets look at  EMBED Equation.DSMT4   EMBED Equation.DSMT4   Where is this graph increasing or decreasing? Where is the turn around point? Lets put on some tangent lines in a few places Think of replacing the actual curve with a short section of a linedoes the line have positive or negative slope if the graph is increasing? decreasing? Lets look at some slopes of the lines tangent to our parabola: (x, f(x))  EMBED Equation.DSMT4  ((1, 5) (6 negative (0, 0) (4 x-intercept, negative (1, (3) (2 negative (2, (4) 0 VERTEX - zero (3, (3) 2 positive (4, 0) 4 x-intercept, positive (5, 5) 6 positive The graph is decreasing to the left of the vertex and the slopes of the tangent lines are negative. The graph is increasing to the right of the vertex and the slopes of the tangent lines are positive. At the turn around point, the vertex, the slope of the tangent line is zero. Popper 09, Question 8 How do you find turn around points? Set the derivative = 0 and solve for x. Then use that x in the original formula to get the y value of the graph point. Example 9:  EMBED Equation.DSMT4  Domain: All Real numbers Range: All Real numbers x-intercepts: (3, 3, 13 y-intercepts: (117 End behavior:  EMBED Equation.DSMT4  Its continuous everywhere because its a polynomial. Derivative:  EMBED Equation.DSMT4  Where is the graph increasing and decreasing? We have TA1 and TA2  Where are the TAPs? Set the derivative equal to zero and solve for x!  EMBED Equation.DSMT4  = 0 [at the TAPs the slope of the tangent line = 0] x = 1/3 and x = ( 9 What are the graph points?  EMBED Equation.DSMT4  f(1/3) =  EMBED Equation.DSMT4 =  EMBED Equation.DSMT4  f((9) =  EMBED Equation.DSMT4  Now lets talk about where the graph is increasing or decreasing with the appropriate xs in the intervals. And lets talk about the range, too!  So now the typical polynomial question looks like this: Given this polynomial: Domain, range, intercepts, derivative, sketch, equation of tangent line at a given point, location of turn-around points, where is it increasing/decreasing, where is it continuous ? Popper 09, Question 9 Lets look at another polynomial Example 10:  EMBED Equation.DSMT4  Well call the x-intercepts x1, x2, and x3; theyre actually irrational numbers.  Lets mark the graph for increasing/decreasing and find where the turn-around points are. Note that the RANGE depends of the y associated with the turnaround point on the rightmostits [f(TA3) ,  EMBED Equation.DSMT4 ). If I knew the x value at the turnaround point, I could calculate the associated y value and tell you exactly what the range is Take the derivative of f(x) and set it equal to zero:  EMBED Equation.DSMT4  Note that theres a common factor of twelve, divide it out:  EMBED Equation.DSMT4  Use the Rational Root theorem and synthetic division to find that the factors of the derivative are 12(x + 3)(x ( 2)(x + 1) So the turn around points are at x = (3, x = 2, and x = (1. Evaluate the original function to get the actual graph point ys for each x. and f (2) will give you what you need to state the range Popper 09, Question 10 End Video 1 Now not everything is a polynomial. For example, we have Exponential Functions. Lets look at two of these: the exponential function base e and the exponential function base 10. Example 11  EMBED Equation.DSMT4  Domain: all Real Numbers Range:  EMBED Equation.DSMT4  x-intercepts: none y-intercept: (0, 1) The actual value of e is 2.7183. and its irrational. I generally characterize it as 3-ish The graph is leftward asymptotic to y = 0, aka the x axis. On the right it goes off rather quickly to positive infinity. Lets talk turnaround points: NONE This graph is increasing everywhere!  Lets talk slope of the tangent lines. The Difference Quotient:  EMBED Equation.DSMT4  Since  EMBED Equation.3 isnt affected by the limit, Ill work with the factor that has an h in it: Now lets take the limit as h approaches zero:  EMBED Equation.DSMT4  Notice that there is not any nice disappearing h action. We actually MUST do a table! h = .01h = .001h = .00011.0050171.00050011.0000500002h = (.01h = (.001h = (.0001.99501.99950.999950002 So the  EMBED Equation.DSMT4 . This function is its own derivative! That is amazing and very NOT polynomialish! This astonishing fact is true and makes it one of the most interesting functions in the world. Now lets talk about slopes and formulas for the tangent lines. At x = 1  EMBED Equation.DSMT4  So the formula for the tangent line using the point (1, e) and the slope, m = e gives us: y ( e = e(x ( 1) y = ex. At x = 2  EMBED Equation.DSMT4   EMBED Equation.DSMT4  note that  EMBED Equation.DSMT4  at 2 the instantaneous rate of change is 9ish times 2. Lets sketch these onto the graph:  Example 12 So now lets look at  EMBED Equation.DSMT4 . The domain is all Real numbers and the range is  EMBED Equation.DSMT4 . The graph is leftward asymptotic to zero and rightward up quickly to positive infinity. There are no x-intercepts and the y-intercept is (0, 1). The graph is everywhere increasing and has no turnaround points.  These graphs look a lot alike, BUT if you look closely youll see that the first graph has the graph point (1, 3ish) and this one has (1, 10). It gets large faster. Now for the Difference Quotient:  EMBED Equation.DSMT4 . Note that  EMBED Equation.3  isnt affected by the limit so lets look at  EMBED Equation.DSMT4  and lets note that  EMBED Equation.DSMT4  h = .01h = .001h = .0001h = .00001h = .0000012.329292.3052382.302850h = (.01h = (.001h = ( .0001h = ( .00001h = ( .0000012.2762772.2999362.30232 So, we find that the multiplier is a value called ln(10) (pronounced: natural log of 10)  EMBED Equation.DSMT4  What is the instantaneous rate of change for  EMBED Equation.DSMT4 at x = 2? Well the derivative of the graph is approximately EMBED Equation.3 . For one tiny step sideways at 2 the y value goes up 230. Thats pretty steep, folks. This is a pattern for the exponential functions:  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Only  EMBED Equation.DSMT4  escapes that multiplier because  EMBED Equation.DSMT4 ! So the derivative of an exponential is a specific, dependable multiple of itself! Popper 10, Question 1 And there are functions that come from multiplying and dividing other functions: Now lets look at the effect of multiplying unlike functions with one another and well look at the instantaneous rate of change at various points along these graphs. Example 13  EMBED Equation.DSMT4  The domain is  EMBED Equation.DSMT4  Why is that?  EMBED Equation.DSMT4  rules! Heres the graph. Note that the range is from the turn-around point to positive infinity.  We have two ways to get the derivative: One way is to multiply the functions and use the Sum rule. The other way is to use the Product Rule. Lets do it both ways. Multiply and use the Sum Rule  EMBED Equation.DSMT4  The derivative is, then,  EMBED Equation.3  Heres some room for the steps: Now lets solve  EMBED Equation.3   EMBED Equation.DSMT4  multiply both sides by the square root of x  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Popper 10, Question 2 More shortcuts: Product Rule! If f(x) is a string of factors, you may use the Product Rule. For the simplest case, suppose f(x) = h(x)(p(x)).  EMBED Equation.3  Example 13, cont.:  EMBED Equation.DSMT4  here:  EMBED Equation.3  We know:  EMBED Equation.3  and p(x) = 1 So the derivative is:  EMBED Equation.3  Mnemonic: If f(x) = First function times Second Function = F(S)  EMBED Equation.3  Example 14:  EMBED Equation.DSMT4   Domain: all Real numbers Leftward asymptotic to x = 0. x-intercepts: 0 and 4 Where are the turn-around points? f(x) = E(Q) the derivative is E(Q) + E(Q) recall that the derivative of  EMBED Equation.DSMT4 is  EMBED Equation.DSMT4 .  EMBED Equation.DSMT4  Now the first factor is NEVER zero so Using the quadratic formula on the second factor we find that  EMBED Equation.DSMT4  Which is pretty reasonable if you look at the graphmost everybody can see that its around (1 and 3 that there are turn arounds.  Lets check for the y values:  EMBED Equation.DSMT4  The turn around on the right is (3.23, 63)ish.  EMBED Equation.DSMT4  The turn around on the left is ((1.23, (1.8)ish. Write down the increasing and decreasing information in interval notation for this graph: Increasing: Decreasing: Popper 10, Question 3 Now for the Quotient Rule: recall that quotient is what you get when you divideso were talking about rational functions here. The Quotient Rule for  EMBED Equation.DSMT4  QR says:  EMBED Equation.DSMT4 . Please know this by heart. Example 15  EMBED Equation.DSMT4  Domain: all Reals except 2 Range: all Reals except 1 x-intercept ((6, 0) y-intercept (0, (3) VA x = 2 discontinuous here, limits to infinity vertically HA y = 1 limits to  EMBED Equation.DSMT4 here, horizontally as  EMBED Equation.DSMT4   Where is the graph increasing and decreasing? So, lets talk tangent lines to the curve? Are there tangent lines to the curve at x = 5 or at x = (1. You bet and, more, each tangent line has a slopethe derivative at that number for x. So, using the quotient rule, we can calculate the derivative: The top polynomial is (x + 6) and its derivative is 1. The bottom polynomial is (x ( 2) and its derivative is 1.  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Now, at x = 5, the graph point is ( 5, 11/3) and the slope of the tangent line is (8/9. The equation of the tangent line is y ( 11/3 = (8/9(x ( 5). At x = (1, the graph point is ((1, (5/3) and the slope of the tangent line is (8/9. Amazing but true. However, the equation of the tangent line is different: y + 5/3 = (8/9(x + 1). So they are parallel at these two points! Example 16  EMBED Equation.DSMT4  Domain: all Reals except 3 Range: all Reals except 2 x-intercept ((3, 0) y-intercept (0, (2) VA x = 3, discontinuous here, goes off to  EMBED Equation.DSMT4 vertically HA y = 2, limits to  EMBED Equation.DSMT4 here, horizontally as  EMBED Equation.DSMT4   Where is this graph increasing and decreasing? Lets look at getting the derivative:  EMBED Equation.DSMT4  The top polynomial is 2x + 6 and its derivative is 2. The bottom polynomial is x ( 3 and its derivative is 1.  EMBED Equation.DSMT4  EMBED Equation.DSMT4  EMBED Equation.DSMT4  So, for example at x = 3 there is NO VALUE for the derivativethe graph is undefined at that place so theres not a tangent to the not-graph. The graph is discontinuous there, too, remember? BUT, at x = 1, the slope of the tangent line is (12/4 = (3. The graph point is (1, (4). And the equation of the tangent line is y + 4 = (3(x (1).  Popper 10, Question 4 Summarizing: The derivative is a calculated quantity that tells you the slope of the tangent line to any point on the graph. The definition of a derivative is taking a limit as h approaches zero, but well use the shortcuts to find them. This is the instantaneous rate of change of the graph at a chosen point. For a polynomial, the domain is all Real numbers and the function is continuous everywhere. An example of the sum/difference rule:  EMBED Equation.DSMT4  factor by grouping! The end behavior is  EMBED Equation.DSMT4 . The x-intercept are at (4 and the y-intercept is at 16. The derivative:  EMBED Equation.DSMT4  EMBED Equation.DSMT4   Where are the turn around points? (2 and (2/3 Where is the graph increasing and decreasing? Increasing ( EMBED Equation.DSMT4  Decreasing ((2, (2/3) Popper 10, Question 5 An example of the product rule:  EMBED Equation.DSMT4  Why is the domain all Real numbers? How does this show that theres no vertical asymptote? You might be tempted to say theres a VA a x = 4 but since you KNOW the domain, you can tell. Leftward asymptotic to y = 0, though.   EMBED Equation.DSMT4   EMBED Equation.DSMT4 . Since the exponential function is NEVER zero, we can solve the linear part for zero giving us x  EMBED Equation.DSMT4  Where is the graph increasing and decreasing? Increasing:  EMBED Equation.DSMT4  Decreasing:  EMBED Equation.DSMT4   This means that the slopes of the tangent lines are positive to the left of 1.55, zero at 1.55, and negative to the right of it. Popper 10, Question 6 An example of quotient rule:  EMBED Equation.DSMT4  Domain: all Real numbers except zero Range: all Real numbers except zero no intercepts VA x = 0, discontinuous here HA y = 0  The top polynomial is 1; its derivative is zero. The bottom polynomial is x; its derivative is 1.  EMBED Equation.DSMT4  EMBED Equation.DSMT4   EMBED Equation.DSMT4  Which is always negative and never zero. And its true, this graph is decreasing on its domain. Note that there is NO tangent line at x = 0 because the function is not defined there. Popper 10, Question 7 Popper 10, Questions 8 Popper 10, Question 9 Popper 10, Question 10 End of Video 2 Application problems! Always important! * The revenue in dollars from the sale of x car seats for infants is given by  EMBED Equation.DSMT4  A. Graph the revenue curve over its domain. B. Find the revenue for a production level of 1,000 car seats. C. Find the average rate of change in revenue from 500 to 1,000 car seats. Write a brief verbal interpretation of the results. D. Find the instantaneous rate of change at 1,000 car seats. Write a brief verbal interpretation of the results. For the graph:   EMBED Equation.DSMT4  AROC from 500 to 1000:  EMBED Equation.DSMT4  The slope of the secant line joining these two graph points in 22.5. This means that you are getting, on average, $22.5 per baby seat when production changes from 500 seats to 1000 seats IROC at 1000 The derivative is  EMBED Equation.DSMT4  The slope of the tangent line at x = 1,000 is 10. This is a fairly large positive number and means that the revenue from the 1001st car seat will be $10 for that unit. Note that the AVERAGE revenue is $22.5, but the actual income from the 1000th unit is $10. So revenue is slowing down as we approach the vertex. Note that at 1200 units, the slope of the tangent line is zero. And for 1400 units the slope of the tangent line is "10 which means revenue, while still positive, is decreasing with each unit sold. Right at the vertex is where we want to have a sale, or change our production methods or do something before revenue goes negative at 2401 units. Popper 11, Question 1 * The annual US production of talc, a soft mineral, in thousands of metric tons, is given by  EMBED Equation.DSMT4  where t is time in years and t = 0 corresponds to 1997. A. Graph this function from 1997 to 2007. B. Find the annual production in 2000. C. Find the average rate of change in production from 1998 to 2002. Write a brief verbal interpretation of the results. D. Find the instantaneous rate of change of production in 2002. Write a brief verbal description of the results.  To graph this up to 2007, not that its 10 years from 1997 to 2007. We will use the interval [0,10] as our domain.  EMBED Equation.DSMT4  Here are our starting and ending points. Where is the peak production?  EMBED Equation.DSMT4  .57 of a year is about the 30th week. So the peak production is 30 weeks into 1999. Production increased til then and has decreased since. AROC 1998 to 2002 1998 is zero, 2002 is 5  EMBED Equation.DSMT4  This is the slope of the secant line joining the two graph points. Note that the slope is still positive because we havent gotten enough into the decline! In fact, if a manager is not careful, she might think that things are ok because the average revenue per thousand metric ton is positive. IROC 2002  EMBED Equation.DSMT4  While the derivative is a large negative number. Note that REVENUE is still positive but it is decreasing with each additional thousand metric tons. The revenue for the next week will be 730 LESS than the preceeding week. And in which year will the revenue zero out?  EMBED Equation.DSMT4  Use the quadratic formula!  EMBED Equation.DSMT4  [ Theres a negative x as well, but well ignore that one thats over on the left of zero!] In the first 3 months of 2008! They need to do something before then! Popper 11, Question 2 Popper 11, Question 3 * The number of female newborn deaths per 100,000 births in France is given by  EMBED Equation.DSMT4  Where t is in years and t = 0 corresponds to 1980. A. Graph this function from 1980 to 2005. B. Find the number of newborn girls who died in 2001. C. Find the instantaneous rate of change of mortality in 1996. Write a brief verbal interpretation of the results. From 1980 to 2005 is 25 years so our domain is [0, 25].   EMBED Equation.DSMT4  So France halved the mortality rate for female newborns in 25 years. In 2001 approximately 11 baby girls per 100,000 births died. 11/100,000 is .00011 which is .01%.  EMBED Equation.DSMT4  The instantaneous rate of change in mortality per 100,000 births in 1996 is  EMBED Equation.DSMT4   EMBED Equation.DSMT4  We would really talk about 49% of a baby girl, wed note that the rate is decreasing. Popper 11, Question 4 * The total sales of a company in millions of dollars t months from this month is given by  EMBED Equation.DSMT4  Find the sales a year from now and find the derivative at that time and interpret these results. Use the results to predict the sales in 13 months. A year from now is t = 12 months.  EMBED Equation.DSMT4 million The prediction will use the derivative.  EMBED Equation.DSMT4  0.144(1,000,000) = 144,000 Well earn about $144,000 dollars more than this month. Lets check this:  EMBED Equation.DSMT4 million 7.60 7.46 = .14 million which is 144,000 dollars. Popper 11, Question 5 * If $100 is invest in an account that earns 6% compounded annually, then the amount in the account after t years is given by  EMBED Equation.DSMT4  EMBED Equation.DSMT4  Find the amount after 5 years and the instaneous rate of change at that time.  EMBED Equation.DSMT4  use your  EMBED Equation.DSMT4 key to do this. mine is a second fcn  EMBED Equation.DSMT4  This says that the rate of change of the amount with respect to a year will be about $7.80. Lets check this:  EMBED Equation.DSMT4  So, not quite perfect, but only 23 cents off! You can use derivatives to predict what will happen in the next unitthe cost of the next unit to produce, the revenue from the next unit, the income from the next unit. This can be very handy. Popper 11, Question 6 * Suppose a person learns y items in x hours as given by  EMBED Equation.DSMT4  Graph the learning curve and give the rate of learning at 1 hour and at 6 hours.   EMBED Equation.DSMT4  So starting out the student knows none of the items but after 9 hours the student knows 150 items.  EMBED Equation.DSMT4  This says that the rate of learning items at 1 hour is 50 items and at 6 hours its still positive but its slowed down to about 20 items in the 6th hour. 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hs(>UVjhs(>Uhs(>hujhuUjrhuhuEHU"UVWmnopqrstuvwxyz{|}~܃݃gd<\؄ل <=uvgd<΄τЄvЅ ņƆdžȆ2濲ʮʦʢzobjh5h5EHUjkS h5UVjhEhEEHUjkS hEUVjh5Uh5hh5h5h-05hs(>j h<h<EHUjhS h<UVh-0jh<h<EHUjgS h<UVh<jh<Ujh<h<EHU& Ɇ ^`gd5Ɇʆl :;i ^`gd5gd5 ^`gd52345ABYZ[\ 67895KMS\b~۸ۧۜߧ{{shj$pS hEUVjhEUhEhE6hhEhE5j h5h5EHUjbmS h5UVhEjh5h5EHUjmS h5UVjh5h5EHUj|lS h5UVh5jh5Uj$h5h5EHUj7lS h5UV%ijk./012345KLMNOPQRSTUVWXYZ[\]gd5]^_`abcef/01 gdE ^`gdEgd5+,-./EGPfhjkv|<ȽȥȐȈȄȀ~vk^vj/hhEHUj/sS hUVjhUUhhhh.5h.h.H*j+h.h.EHUj=qS h.UVj (h.h.EHUjpS h.UVh.jh.Uh5jh.UmHnHuhEjhEUjd$hEhEEHU#NOPfghijklmnopqrstuvwxyz{|}~ gdE~<=>Z[\3456789:;<Suppose that in a given gourmet food shop people are willing to buy x pounds of chocolate candy per day at $p per quarter pound, as given by the following equation:  EMBED Equation.DSMT4  What is the instantaneous rate of change of demand with respect to price when the price is 5 dollars per quarter pound?  EMBED Equation.DSMT4  When the price is $5  EMBED Equation.DSMT4  The rate of change is negative so as price goes up demand goes down. The demand is decreasing at a rate of 7 quarter pounds per dollar. Popper 11, Question 8 * The total sales in thousands of CDs for a compact disk are given by  EMBED Equation.DSMT4  Find the derivative using the quotient rule. Find the sales and the rate of change of the sales at t = 10 Use the results to estimate the number of sales at t = 11  EMBED Equation.DSMT4   EMBED Equation.DSMT4  This says that after 10 months sales are 60,000 and the rate of change is 4,000 at 10 months. The estimated sales at 11 months will be 64,000. 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