If point x 2 2y
[PDF File]Score - SJSU
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3. (20 points) Show that the ellipsoid 3x2 +2y2 +z2 = 9 and the sphere x 2+y +z2 −8x−6y −8z +24 = 0 have a common tangent plane at the point (1,1,2). Solution: Let f(x,y,z) = 3x2 +2y2 +z2 = 9 and g(x,y,z) = x 2+y2 +z −8x−6y −8z +24. We have ∇f(x,y,z) = h6x,4y,2zi and ∇g(x,y,z) = h2x−8,2y −6,2z −8i,
[PDF File]Lecture 13
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2 x + 1 2 y −3 = −2. (8) Therefore the desired point P is (1,1,−2). The distance between P and the origin is √ 1 +1 +4 = √ 6. We should really check that the above point represents the absolute minimum. Firstly, the point (x,y) = (1,1) was found to be the only critical point of h(x,y) defined above. A look at the second-order ...
[PDF File]Math 111 Quiz #6 Solutions 1. Consider the equation = 1, which ...
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Consider the equation x2 +2y2 = 1, which describes an ellipse in the xy-plane. (a) Use implicit differentiation to calculate dy dx for this ellipse. x 2+2y = 1 ⇒ 2x+4y dy dx = 0, so dy dx = −2x 4y = − x 2y. (b) Find the point(s) where the tangent line to the given ellipse has slope 1. The points where the tangent line to the ellipse has ...
[PDF File]Homework # 3 - Department of Physics
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another equilibrium point at x= 1, but it is a maximum.) SOLUTION - First we expand U(x) in a Tylor series around x = 3. A Taylor series looks like this for a given function f(x): f(x) = f(x 0) + f0(x 0) 1! + f00(x 0) 2! + f000(x 0) 3! + (39) Thus for this potential energy we nd U(x= 3) = 0 (40) U0(x= 3) = (x 3)2 3 + 2x(x 3) 3 x=3 = 0 (41) U00 ...
[PDF File]Lagrange Multipliers - Illinois Institute of Technology
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Find the extreme values of f(x,y)=x 2+2y2 on the disk x2+y ≤1. •Solution: Compare the values of f at the critical points with values at the points on the boundary. Since f x =2x and f y =4y, the only critical point is (0,0). We compare the value of f at that point with the extreme values on the boundary from Example 2: •f(0,0)=0 •f(±1,0)=1
[PDF File]Equations with regular-singular points (Sect. 5.5). Equations with ...
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Equations with regular-singular points (Sect. 5.5). I Equations with regular-singular points. I Examples: Equations with regular-singular points. I Method to find solutions. I Example: Method to find solutions. Recall: The point x 0 ∈ R is a singular point of the equation P(x) y00 + Q(x) y0 + R(x) y = 0 iff holds that P(x 0) = 0. Equations with regular-singular points.
[PDF File]Unit #5 - Implicit Di erentiation, Related Rates
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y2(y2 4) = x2(x2 5) at the point (x;y) = (0; 2). (The devil’s curve) This point happens to be at the bottom of the loop on the yaxis, so the slope there is zero. Therefore, the tangent line equation is simply: y= 2 11.Use implicit di erentiation to nd the (x;y) points where the circle de ned by x2 + y2 2x 4y= 1 has horizontal and vertical ...
[PDF File]Ch 14.6 : Directional Derivatives and Gradient vector
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1 + x2 + 2y2 + 3z2 where T is measured in degrees Celsius, and x;y;z in meters. In which direction does the temperature increase the fastest at the ... point (1;2;3) to the ellipsoid x2 4 + y2 + z2 9 = 3. Signi cance of the Gradient Vector If we consider a function f of two variables and a point P(x 0;y 0) in its domain
[PDF File]Name: SOLUTIONS Date: 10/06/2016
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de ned by z= x 2+ y2 and x + 2y2 + z2 = 7 at the point ( 1;1;2). Hint: Think about the geometry of the gradient vectors. You don’t have to parametrize ... = h2x;2y; 1i: Also, the gradient of the level surface G(x;y;z) = x 2+ 2y2 + z = 7 is rG(x;y;z) = h2x;4y;2zi: The tangent vector at ( 1;1;2) of the curve of intersection between these two ...
[PDF File]Unit #22 - Unconstrained Optimization Local Extrema - Queen's U
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x = 0 : 0 = 6x 4 + 2y y = 2 + 3x If f y = 0 : 0 = 2x 10y + 48 y = 1 5 (x+ 24) For both derivatives to be zero simultaneously, we must have 2 + 3x = 1 5 (x+ 24) 10 + 15x = x+ 24 x = 1 Subbing back into an earlier equation, y = 5 Thus, there is a single critical point at (x;y) = (1;5). To determine the type of critical point, we need the second ...
[PDF File]1998 AP Calculus AB Scoring Guidelines
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6. Consider the curve de ned by 2y3 + 6x2y 12x2 + 6y= 1. (a) Show that dy dx = 4x 2xy x2 + y2 + 1. (b) Write an equation of each horizontal tangent line to the curve. (c) The line through the origin with slope 1 is tangent to the curve at point P. Find the x{and y{coordinates of point P. (a) 6y2 dy dx + 6x2 dy dx + 12xy 24x+ 6 dy dx = 0 dy dx ...
[PDF File]Section 4.2 selected answers - Clark University
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section to see that (0;0) is the only critical point and it’s a minimum. Since x2 + y2 + 1 is a paraboloid, and ln is an increasing function, the ... z = 2(x + 2y2 + 1)sinz Set these all to 0 and solve to nd the critical points. Note that (x2 + 2y2 + 1)sinz= 0 implies sinz= 0 since x2 + 2y2 + 1 can’t be 0. Therefore
[PDF File]Unit #23 - Lagrange Multipliers Lagrange Multipliers - Queen's U
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4. f(x;y) = x2 + 2y2;x2 + y2 4 Note that we are dealing with an inequality for the constraint. We can consider any point in or on the boundary of a circle with radius 2. To look on the ... critical point we see: f(0;2) and f(0; 2) are global maxes with values of f= 8 f(0;0) is the global min on the region, with f= 0.
[PDF File]Math 222 Final Exam Solutions Spring 2008 - Cornell University
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that are closest and farthest from the point (2,1,−2). Solution. We want to find max/min for the function f(x,y,z) = (x−2)2+(y ... (−1)− (x−y)2y] (x2 +y2)2 = x2 + y2 − 2x2 − 2xy + x2 + y2 + 2xy − 2y2 (x2 +y2)2 = 0 (6 pts) (b) Compute the line integral of F around the circle x2 +y2 = a2. Solution. If we could apply Green’s ...
[PDF File]Midterm 2 Solutions ) that represents the curve of intersection of the
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Midterm 2 Solutions 1. (a) (6 pts.) Find a vector-valued function ~r(t) that represents the curve of intersection of the surfaces z= x 2+ 4y2 and y= x: Solution: We have ydetermined by xand zdetermined by xand y, so we can just use ~r(t) =
[PDF File]Problem 1. Find the point on the plane 4x+ 3y z = 10 nearest to (2 0 1)
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equation is rf= rg, so 2xy= 2x and x2 = 2y. In addition to these two equations, we have the third equation x 2+ y 1 = 0. Now, if xis not 0, the rst equation just says y= , and the second then gives x2 = 2y2. Plugging this into the third equation, 2y2+y 2 1 = 0, so y = 1=3, and we have y= 1= p 3. Then x2 = 2=3, so x= 2= p 3. It could be that x ...
[PDF File]Find and classify critical points - West Virginia University
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Find and classify critical points Useful facts: The discriminant ∆ = f xxf yy − f xy 2 at a critical point P(x 0,y 0) plays the following role: 1. If ∆(x 0,y 0) > 0 and f xx(x 0,y 0) > 0, then f has a local minimum at (x 0,y 0). 2. If ∆(x 0,y 0) > 0 and f xx(x 0,y 0) < 0, then f has a local maximum at (x 0,y 0). 3. If ∆(x
[PDF File]Math 2280 - Assignment 11 - University of Utah
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(x2 − 1)y′′ +4xy′ +2y = 0. State the recurrence relation and the guaranteed radius of conver-gence. Solution - A power series solution y(x) and its derivatives will have the forms: y(x) = X∞ n=0 c nx n; y′(x) = X∞ n=1 nc nx n−1; y′′(x) = X∞ n=2 n(n −1)c nxn−2. If we plug these into the ODE we get: X∞ n=0 n(n − 1)c ...
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