Study Guide/Practice Exam 2 Solution. This study guide/practice exam is longer and harder than the actual exam. Problem A: Power Series. x 2i /i!

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1 Study Guide/Practice Exam 2 Solution This study guide/practice exam is longer and harder than the actual exam Problem A: Power Series (1) Find a series representation of f(x) = e x2 Explain why the series you give converges to e x2 for all values of x (Hint: use the estimate for the errors of the MacLaurin approximations to e x ) Solution: Let g(x) = e x and let P (x) = 1 i=0 i! xi We know that P (x) = g(x) for all x R If x R, then x 2 R Hence, P (x 2 ) = g(x 2 ) = e x2 That is, e x2 = P (x 2 ) = x 2i /i! i=0 (2) Find a series representation of x 0 et2 dt (with x > 0) Explain why the series you give converges to x 0 et2 dt for all values of x > 0 Solution: By a theorem discussed in class, integrating (or differentiating) does not change the radius of convergence of a power series Consequently: x By the same theorem, this equals 0 e t2 dt = x 0 t 2i /i! dt i=0 x i=0 0 t 2i /i! dt = i=0 x 2i+1 (2i + 1)i! This last series is the one we are looking for (3) Find the MacLaurin series for: x 0 ( t t 0 ) e s2 ds dt Solution: The reasons are the same as in the previous problem Here is the work: ( x t ) t 0 0 es2 ds dt = x t t 2i+1 0 i=0 dt = (2i+1)i! x t 2i+2 0 i=0 dt = (2i+1)i! i=0 x 2i+3 (2i+3)(2i+1)i! 1

2 (4) Consider the differential equation and initial values: Find a MacLaurin series for y y = y y(0) = 1 y (0) = 0 Solution: We compute the derivatives of y: y = y y = y y (4) = y = y y (5) = y y (6) = y = y Plugging in 0 we get: The MacLaurin series is therefore: y (0) = y(0) = 1 y (0) = y (0) = 0 y (4) (0) = y(0) = 1 y (5) (0) = y (0) = 0 y (6) (0) = y(0) = 1 P (x) = 1 + x2 2 + x4 4! + x6 6! + In summation form we have: x 2k P (x) = (2k)! (5) Let P (x) = k=1 k=0 (2k + 1)(2k 1)(2k 3) x 5k+1 (2k)(2k 2)(2k 4) 4 2 (a) Determine the radius of convergence of P (x) (b) Find P (x) and determine its radius of convergence Partial solution: for (a) Use ratio test For (b) take the term-by-term derivative and use the theorem which says that differentiating a power series doesn t change its radius of convergence Problem B: Graphing functions f : R 2 R (1) Let f(x, y) = x 2 y Draw at least 3 x-slices, at least 3 y-slices, and at least 3 level sets Describe with words or pictures the graph in 3-dimensions of this equation 2

3 (2) For x R 2, define f(x) = x Carefully describe, with words and pictures the 3- dimensional graph of this equation (Hint: this is the analogue in 3-dimensions of the function f(x) = x in 2-dimensions) Solution It is a cone with a point at the origin and opening along the z-axis (3) Below is the temperature plot of a function f(x, y) Points colored red have high function values and points colored blue have low function values Sketch a contour line (ie a z-slice) You do not need to label it with the z value Solution: Here is one possible answer: Problem C: Vector Operations (1) Suppose a, b are vectors in R 2 Explain why a b = a b cos θ Partial Solution: Use the law of cosines (2) Calculate the cross product Solution: (3) Find the equation of the plane in R 3 which contains the point (7, 19, 21) and is perpendicular to the vector (1, 4, 3) 3

4 Solution: The solutions to the equation (1, 4, 3) (x, y, z) = 0 consist of all vectors perpindicular to the vector (1, 4, 3) since the dot product of two vectors is zero if and only if the vectors are perpendicular Every plane parallel to this plane is also perpindicular to (1, 4, 3) and so the plane asked for in the problem has equation (1, 4, 3) (x 7, y + 19, z 21) = 0 or equivalently (x 7) + 4(y + 19) 3(z 21) = 0 (4) Find the equation of the plane in R 3 which contains the points (1, 4, 1), (0, 3, 2), and ( 1, 1, 1) Solution: First we find the equation of the plane which contains the points (0, 0, 0), ( 1, 1, 3), and ( 2, 3, 0) (These points were obtained from the given points by translating them by (1, 4, 1)) A normal vector to the plane which contains these points can be obtained by using the cross product: ( 1, 1, 3) ( 2, 3, 0) This gives the normal vector (9, 6, 1) Thus the equation of the plane which contains these three points is 9x 6y + z = 0 The plane which contains the three given points is obtained by translating this plane by (1, 4, 1) We obtain the equation: 9(x 1) 6(y 4) + (z + 1) (5) Find the area of the parallelogram defined by the vectors (6, 2) and ( 1, 8) in R 2 Solution: We put these vectors into the xy plane in R 3 to obtain the vectors (6, 2, 0) and ( 1, 8, 0) The magnitude of the cross product is the area of the parallelogram defined by these vectors That is, the desired area is (6, 2, 0) ( 1, 8, 0) = (0, 0, 46) = 46 (6) Find the projection of the vector (4, 3, 1) onto the vector ( 2, 6, 1) Solution: Let u = (4, 3, 1) and v = ( 2, 6, 1) We have v = 41 and u v = 9 The projection is then: (7) Find the projection of the vector (4, 3, 1) onto the plane 2x + 6y z = 0 Solution: Notice that the vector v in the previous solution is the normal vector to the plane so we just need to consider: /41 54/41 1 9/41 4

5 (8) Find the projection of the point (5, 2, 10) onto the plane x + y + z = 1 (Note: You can t use the projection formulae as stated since the plane doesn t go through the origin) Solution: Notice that the given plane P contains the point (1, 0, 0) If we translate everything back by this point we can use the techniques of the previous problem Let Q be the plane given by: (x + 1) + y + z = 1 Then a point (x, y, z) is in the plane P If and only the point (x 1, y, z) is in the plane Q Let p = (5, 2, 10) Translate it to the point q = (5, 2, 10) (1, 0, 0) = (4, 2, 10) Project q onto Q The projection of q onto the vector N = (1, 1, 1) is proj N (q) = 16 (1, 1, 1) = (16/3, 16/3, 16/3) 3 The projection of q onto Q is: (4, 2, 10) (16/3, 16/3, 16/3) = ( 4/3, 10/3, 14/3) Translate back to find the projection of p onto P : ( 1/3, 10/3, 14/3) (9) A canoe is in the middle of a portion of the Kennebec river which faces north-south The current is moving at 3 mph southward In the absence of the current, the wind would blow the canoe at 1 mph northwest How fast and in what direction must the canoeists paddle to stay in the same position? Solution: Choose a coordinate system so that the canoe is at (0, 0) and so that north is in the direction (1, 0) Then the current c velocity vector is (0, 3) The wind velocity vector w has magnitude 1 and is at an angle of 135 from the positive x axis Thus w = ( 2/2, 2/2) The canoeists must paddle with a velocity of (w + c) = ( 2/2, 3 2/2) The magnitude of this vector is approx 24 mph Problem D: Limits and Continuity (1) Suppose that f : R 2 R Carefully state a formal definition of x a f(x) = L Solution: For every ɛ > 0 there exists a δ > 0 so that whenever x a < δ (and x a) we have f(x) L < ɛ (2) Let f(x, y) = Show that x 0 f(x) does not exist Solution: Notice that x2 y x 3 + y 3 (x,0) (0,0) f(x, y) = x 0 0/x3 = 0 5

6 On the other hand, f(x, y) = (x,x) (0,0) x 0 x3 /2x 3 = 1/2 Since these its are different, the it x 0 f(x) does not exist (3) Show that the function f(x, y) = does not have a it as (x, y) (0, 0) Solution: Consider Also, x2 + y 3x 2 2y x 2 f(x, y) = (x,0) (0,0) x 0 3x = 1/3 2 y f(x, y) = (0,y) (0,0) y 0 2y = 2 These its are different, thus the requested it does not exist (4) Use the it definition of differentiable to show that { 3x if y = x f(x, y) = 0 if y x is not differentiable at (0, 0) Solution: The y-slice when y = 0 is the straight line z = 0 The x-slice when x = 0 is the straight line z = 0 Both of these lines have slope 0 Thus, f x (0, 0) = 0 and f y (0, 0) = 0 Also, f(0, 0) = 0 Thus, the linear approximation to f at (0, 0) is L(x, y) = 0 The relative error of the approximation is f(x, y) L(x, y) (x, y) (0, 0) = f(x, y) x, y If y x, the relative error is equal to 0 which has a it of 0 as (x, y) (0, 0) along any path with y x If y = x, the relative error is equal to 3x x2 + y 2 = 3x = 3 2x 2 2 If also x > 0, the the relative error is 3 2 Thus, as (x, y) (0, 0) along the line y = x with x > 0, the relative error its to 3 2 Since this is different from the previous it, the it of the relative error as (x, y) (0, 0) does not exist Thus, f is not differentiable at (0, 0) 6 x x

7 is differen- (5) Use a theorem (say which one!) to explain why the function f(x, y) = x x 2 +y 2 tiable at the point ( 4, 6) Solution: The partial derivatives of f are: f x (x, y) = ((x 2 + y 2 ) 2x)/(x 2 + y 2 ) 2 f y (x, y) = 2yx/(x 2 + y 2 ) 2 These are continuous at every (x, y) (0, 0) Thus, if a R 2 and a (0, 0), then the partial derivatives of f are continous on the region {x R 2 : x a < a } By a theorem stated in class this is enough to guarantee that f is differentiable at a In particular, f is differentiable at a = ( 4, 6) Problem E: Partial and Directional Derivatives (1) Carefully describe the geometric meaning of the (first) partial derivatives Solution: The partial deriviative f is the slope of the tangent lines in a y-slice The x partial derivative f is the slope of the tangent line in the x-slice y (2) For each of the following calculate / x, / y and find the equation of the tangent plane to the graph of the function at (1, 1) (a) f(x, y) = xe x y 2 (b) f(x, y) = y cos(x) (c) f(x, y) = (x 2 + y 2 )e x2 +y 2 Solution: The tangent planes are: (a) f x (x, y) = e x y 2 + xe x y 2 (a) f y (x, y) = 2yxe x (b) f x (x, y) = y sin(x) (b) f y (x, y) = cos(x) (c) f x (x, y) = (x 2 + y 2 )2xe x2 +y 2 + 2xe x2 +y 2 (c) f y (x, y) = (x 2 + y 2 )2ye x2 +y 2 + 2ye x2 +y 2 (a) z = 2e(x 1) + 2e(y 1) + e (b) z = sin(1)(x 1) + cos(1)(y 1) + sin(1) (c) z = 6e 2 (x 1) + 6e 2 (y 1) + 2e 2 (3) Calculate all second partial derivatives of the following functions (a) f(x, y) = x 3 y 2 (b) f(x, y) = x 3 + y 2 7

8 Solution: (a) f xx (x, y) = 6xy 2 (a) f xy (x, y) = 6x 2 y (a) f yx (x, y) = 6yx 2 (a) f yy (x, y) = 2x 3 (b) f xx (x, y) = 6x (b) f xy (x, y) = 0 (b) f yx (x, y) = 0 (b) f yy (x, y) = 2 (4) Let (5) f(x, y) = { x 2 y 2 x 4 +y 4 if (x, y) (0, 0) 0 if (x, y) = (0, 0) Find f x (0, 0) (Hint: Use the geometric interpretation of partial derivatives) Solution: In the y = 0 slice we have the function: { 0 if (x, y) (0, 0) f(x, 0) = x 4 0 if (x, y) = (0, 0) We see that this is just the 0 function and the derivative of the 0 function is 0 So f x (0, 0) = 0 (6) Suppose that u is a unit vector and that f : R 2 R is differentiable at a Give a thorough explanation of why f u (a, b) = f(a) u Solution: Since f is differentiable, there is a linear function L such that f(x) L(x) x a x a = 0 Let γ(h) = a + hu Notice that as h 0, γ(h) a Also γ(h) a = hu Thus, It follows that The linear function L has equation: f(γ(h)) L(γ(h)) h 0 h f(γ(h)) L(γ(h)) h 0 h = 0 = 0 L(x) = f(a) (x a) + f(a) The definition of the directional derivative is: Thus, f(γ(h)) f(a) f u (a) = h 0 h f(γ(h)) L(γ(h)) + L(γ(h)) f(a) f u (a) = h 0 h 8

9 By our previous comments f(γ(h)) L(γ(h)) h 0 Using the equation for L and γ(h) (when h 0): L(γ(h)) f(a) = h Consequently, f(a) (γ(h) a) + f(a) f(a) h f u (a = f(a) u = f(a) u (7) Explain the meaning of the direction and magnitude of f(a) (assuming that f(a) 0 and that f is differentiable at a) Solution: The direction is the direction of greatest increase in f The magnitude is the amount of the greatest increase of f (Equivalently, the direction is the direction of maximum rate of change and the magnitude is the amount of maximum rate of change of f) (8) Explain why (if f is differentiable) the gradient f(a) is orthogonal to every contour line through a Solution: Let u be a unit vector based at a pointing along the contour line Since the value of f doesn t change along the contour line: f u (a) = 0 By our previous remarks concerning directional derivatives, this means: f(a) u = 0, but this means f(a) and u are orthogonal (9) Suppose that f : R 2 R is differentiable at a = (a 1, a 2 ) R 2 Suppose also that L(x, y) = A(x a 1 ) + B(y a 2 ) + f(a 1, a 2 ) is the linear function approximating f near a from the definition of differentiable Show that A = f x (a 1, a 2 ) Solution: By the definition of differentiable: f(x) L(x) x a x a = 0 Let γ(t) = a + t(1, 0) As t 0, γ(t) a Let x = (x, y) Thus, 0 = t 0 f(γ(t)) L(γ(t)) t a 1 Recall that Thus, so A = f x (a 1, a 2 ) = t 0 f(a 1 + t, a 2 ) A(t a 1 ) f(a 1, a 2 ) t a 1 f(a 1 + t, a 2 ) f(a 1, a2) f x (a 1, a 2 ) = t 0 t a 1 0 = f x (a 1, a 2 ) A, 9

10 (10) The temperature of a point (x, y) on a metal plate is given by T (x, y) = 60x/(1 + x 2 + y 2 ) If an ant is standing at (1, 2) in what direction should the ant walk (leaving from (1, 2) to get the coolest quickest? Solution: T (x, y) = T (1, 2) = ( 60(1+x 2 +y 2 ) 120x 2, (1+x 2 +y 2 ) 2 ( 240, ) ) 120xy (1+x 2 +y 2 ) 2 Since T (1, 2) is the direction of greatest increase in T, we need to use T (1, 2), the direction of greatest decrease in T Thus, the direction of greatest decrease in T is ( 20 3, 20 3 ) Problem F: Tangent Planes (1) Carefully explain why the function f(x, y) = x 2 + y 2 is differentiable at (0, 0) Solution: The function f(x, y) is differentiable at (0, 0) (by definition) if and only if both partial derivatives exist and the relative error goes to 0 as (x, y) (0, 0) It is clear that f x (0, 0) = 0 and f y (0, 0) = 0 both exist Then L(x, y) = 0 and so the it of relative error is f(x,y) L(x,y) (x,y) (0,0) = (x,y) (0,0) x 2 +y 2 (x,y) (0,0) = x 2 +y 2 (x,y) (0,0) x2 + y 2 = 0 Thus, f(x, y) is differentiable at (0, 0) Alternatively, we can appeal to the theorem which says that since the partial derivatives of f are continuous, the function is differentiable (2) Find the equation of the plane tangent to the graph of f(x, y) = x y 3 at the point (2, 1) Solution: L(x, y) = f(2, 1) (x 2, y 1) + f(2, 1) L(x, y) = (1, 3) (x 2, y 1) + 1 L(x, y) = (x 2) 3(y 1) + 1 (3) Find the equation of the linear approximation to the function f(x, y) = x/y near the point (1, 1) Solution: The local linearization is the same as the function whose graph is the tangent plane L(x, y) Thus, L(x, y) = (x 1) (y 1) + 1 (4) Let f(x, y) = 5(x 1)+3(y 2)+8 Use the it definition of the derivative to show that f is differentiable at (1, 2) (Hint: What linear function L will provide a good approximation to f?) Solution: Let L(x, y) = f(x, y) Notice that L(x, y) is a linear function with L(1, 2) = f(1, 2) Since L(x, y) = f(x, y): f(x, y) L(x, y) x (1,2) (x, y) (1, 2) = 0 = 0 x (1,2) 10

11 Thus, f is differentiable at (1, 2) Problem G: Second Derivatives For the following, let f : R 2 R be a differentiable function (1) Explain the geometric interpretation of 2 x 2 f(a) Solution: It measures the curvature of the graph of f in the y-slice y = a 2 (2) Explain the geometric interpretation of 2 y 2 f(a) Solution: It measures the curvature of the graph of f in the x-slice x = a 1 (3) Explain the geometric interpretation of 2 x y f(a) Solution: In each x-slice we calculate the slope of the graph of f at a in that slice We then measure the rate of change of that slope as x changes and y is fixed (4) Explain the geometric interpretation of 2 y x f(a) Solution: In each y-slice we calculate the slope of the graph of f at a in that slice We then measure the rate of change of that slope as y changes and x is fixed (5) Give an informal explanation why, if a is a (non-degenerate) maximum for f, then 2 f(a) < x 2 0 and 2 f(a) < 0 y 2 Solution: In each y-slice and in each x-slice the graph of f must be concave down (6) Give an informal explanation why, if a is a (non-degenerate) minimum for f, then 2 f(a) > x 2 0 and 2 f(a) > 0 y 2 Solution: In each y-slice and in each x-slice the graph of f must be concave up 11