Math 2400, Midterm 2

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1 Math 24, Midterm 2 October 22, 218 PRINT your name: PRINT instructor s name: Mark your section/instructor: Section 1 Kevin Berg 8: 8:5 Section 2 Philip Kopel 8: 8:5 Section 3 Daniel Martin 8: 8:5 Section 4 Albert Bronstein 9: 9:5 Section 5 Mark Pullins 9: 9:5 Section 6 Xingzhou Yang 9: 9:5 Section 7 Martin Walter 1: 1:5 Section 8 Kevin Manley 1: 1:5 Section 9 Albert Bronstein 1: 1:5 Section 1 Martin Walter 1: 1:5 Section 11 Xingzhou Yang 2: 2:5 Section 12 Taylor Klotz 2: 2:5 Section 13 Xingzhou Yang 3: 3:5 Section 14 Braden Balentine 4: 4:5 Section 15 Caroline Matson 4: 4:5 Question Points Score Total: 1 Honor Code On my honor, as a University of Colorado at Boulder student, I have neither given nor received unauthorized assistance on this work. No calculators or cell phones or other electronic devices allowed at any time. Show all your reasoning and work for full credit, except where otherwise indicated. Use full mathematical or English sentences. You have 9 minutes and the exam is 1 points. You do not need to simplify numerical expressions. For example leave fractions like 1/7 or expressions like ln(3)/2 as is. When done, give your exam to your instructor, who will mark your name off on a photo roster. We hope you show us your best work!

1. (1 points) Match each 3D surface with

(d) (1), (B) (e) (3), (D) (1) (2) (3) (4)

1+x 2 +y 2, which is equivalent to 1 + x

= k, which is a hyperbola, or lines (y =

Then the contour line equation is x + y =

2 1. (1 points) Match each 3D surface with one of the contour plots, and one of the equations. (a) (4), (E) (b) (5), (A) (c) (2), (C) (d) (1), (B) (e) (3), (D) (1) (2) (3) (4) (5) x y (A) z = 1 + x 2 + y 2 Let z = k, e.g., k = 1 2. Then the contour line eqn is 1 2 = x y 1+x 2 +y 2, which is equivalent to 1 + x 2 + y 2 = 2(x y) (x 1) 2 + (y + 1) 2 = 1 (B) z = ln(1 + x 2 + y 2 ) Let z = k. Then the contour line equation is x 2 + y 2 = e k 1. (C) z = y 2 x 2 Let z = k. Then the contour line equation is y 2 x 2 = k, which is a hyperbola, or lines (y = ±x as k = ). (D) z = x + y Let z = k. Then the contour line equation is x + y = k, [see figures (e) and (3)] (E) z = cos( x 2 + y 2 ) Note: Let z = k. Then the contour line equation is x 2 +y 2 = (arccos k + 2nπ) 2, n Z, which is a circle. [see figures (a) and (4)] The gradient of f f can also be used to determine the corresponding contour map (see arrows). Page 1 of 1

3 2. Note: No partial credit for this problem. Consider function f(x, y) = e x + cos(xy) + y 7. Compute (a) (2 points) f x (x, y) = e x y sin(xy). Solution: f x = ( e x + cos(xy) + y 7) = e x y sin(xy) (b) (2 points) f y (x, y) = x sin(xy) + 7y 6. Solution: f y = ( e x + cos(xy) + y 7) = x sin(xy) + 7y 6 (c) (2 points) f(, 1) = 1, 7. Solution: f = f x, f y = e x y sin(xy), x sin(xy) + 7y 6 f(, 1) = e (1) sin (()(1)), () sin(()(1)) + 7(1) 6 = 1, 7 Page 2 of 1

4 3. Note: No partial credit for this problem. Consider the surface S: z = f(x, y) = x 3 xy + 3. (a) (5 points) The directional derivative of f at (, 1) in the direction of v = 3 5, 4 5 is 3 5. Solution: v is a unit vector since ( 3 5) 2 + ( 4 5) 2 = 1. f x = ( x 3 xy + 3 ) = 3x 2 y f y = ( x 3 xy + 3 ) = x f = f x, f y = 3x 2 y, x f(, 1) = 3() 2 (1), () = 1, ( ) ( ) 3 4 D v f(, 1) = f(, 1) v = ( 1) + = (b) (5 points) The equation for the tangent plane to the surface S at P (, 1, 3) is x + z = 3. Solution: The normal direction of the tangent plane to the surface S at P (, 1, 3) is n = f x (, 1), f y (, 1), 1 = ( 1), (), 1 = 1,, 1. So the equation of the tangent plane is (1)(x ) + ()(y 1) + (1)(z 3) = x + z = 3 Page 3 of 1

5 Consider the integral y dy dx (a) (4 points) Evaluate the integral. Note: No partial credit for this problem. Solution: 1 2 y dy dx = 1 2 dx y dy = (1) y2 2 2 = 2 (b) (4 points) The integral above can be interpreted as the volume of a solid. Which of the following is the best representation of the solid? The answer is (2). (1) (4) (2) (5) (3) Page 4 of 1

6 Consider the integral y sin(x 2 ) dx dy. y 2 (a) (4 points) Sketch the region of integration. Solution: Based on the integral D as follows, 3 9 y 2 y sin(x 2 ) dx dy, we can get the region D = { (x, y) y 3, y 2 x 9 } y 4 3 (9, 3) 2 x = y x (b) (8 points) Evaluate the integral. Solution: The integration region can be expressed as type I and type II regions as follows, D = { (x, y) y 3, y 2 x 9 } = { (x, y) x 9, y x } To evalute the integral, we need to exchange the integration order: x 9 y sin(x 2 ) dx dy = y sin(x 2 ) dy dx = sin(x 2 ) y2 y= x y 2 2 dx x= ( ) 9 1 = 2 x sin(x2 ) dx Let u = x 2 x 9, du = 2x dx u = 4 sin u du = 1 4 cos u = 1 [1 cos(81)] 4 Page 5 of 1

7 6. (12 points) Consider the function f(x, y) = y 2 + x y 2 + y x + x (x, y) (, ) 1 (x, y) = (, ) Is f continuous at (x, y) = (, )? Justify your answer. Solution: f(, ) = 1. f is continuous at (, ) Take a path y = x (x ) lim f(x, y) = lim (x, y) (, ) x + ( along y = x So the function f is discontinuous. x) 2 + x ( x) 2 + ( x) 2 + x = lim lim f(x, y) = f(, ). (x,y) (,) x + 2x 3x = 2 3 f(, ) = 1 Note 1: If we take another path, for example x =, then the limit is y 2 lim f(x, y) = lim y (x, y) (, ) y = lim 1 = y 3. along x = This means the limit lim f(x, y) does not exist. (x,y) (,) Note 2: There are infinitely many paths through (, ). If we take a path y = kx, where k is a constant, we can check that the limit is always 1 as k varies. This does not mean that the limit of f is 1 as (x, y) (, ). Note 3: We may also take a path x = y 2, which passes through (, ). y 2 + y 2 lim f(x, y) = lim y (x, y) (, ) y 2 + y y + y = lim 2y 2 y 2y + y. along x = y 2 This limit doesn t exist since 2y lim y + 2y + y = lim 2y y + 2y + y = 2 3, lim 2y y 2y + y = lim 2y y 2y y = 2. This also shows that the limit of f does not exist as (x, y) (, ). Hence f is not continuous at (, ). Page 6 of 1

8 7. (12 points) Find all critical points of the function f(x, y) = 2x 3 6xy + y and classify each as a local maximum, local minimum, saddle point, or not enough information. Justify your answer. Solution: f x = ( 2x 3 6xy + y ) = 6x 2 6y f y = ( 2x 3 6xy + y ) = 2y 6x { 6x 2 6y = 1 Solving f =, for (x, y). 2y 6x = 2 From 2 we get y = 3x. Plug it into 1 and we get 6x 2 18x =, 6x(x 3) =, x = or x = 3. Plug them into y = 3x and we obtain two critical points: (, ) and (3, 9). To classify them, we use the second derivative test. f xx = (f x) = ( 6x 2 6y ) = 6 f yy = (f y) = (2y 6x) = 2 f xy = (f x) = ( 6x 2 6y ) = 6 f xx f xy D(x, y) = = f xxf yy (f xy ) 2 = (12x)(2) ( 6) 2 = 24x 36 f yx f yy At (, ), D(, ) = 24() 36 = 36 <, and so (, ) is a saddle point. At (3, 9), D(3, 9) = 24(3) 36 = = 36 >, f xx (3, 9) = 6 >, so (3, 9) is a local minimum point. Page 7 of 1

8. You are hiking up a mountain whose elevation is given roughly by You are at (1, 1, 5) in this model. f(x, y) = x 2 y 3 + xy 2 + 2.

9 8. You are hiking up a mountain whose elevation is given roughly by You are at (1, 1, 5) in this model. f(x, y) = x 2 y 3 + xy (a) (6 points) What is the steepest uphill direction from your position? Solution: The max rate of change of f occurs along the direction of the gradient. For this problem, the steepest uphill direction from your position (1, 1, 5) is f(1, 1). f x = ( x 2 y 3 + xy ) = 2x + y 2 f y = ( x 2 y 3 + xy ) = 3y 2 + 2xy f = f x, f y = 2x + y 2, 3y 2 + 2xy f(1, 1) = 3, 5 (b) (4 points) Suppose you wish to move from your current location while maintaining your current elevation. Find a vector parallel to the direction in which you should move. Solution: The gradient direction is orthogonal or perpendicular to the level curve or the contour line. So in order to maintain your current elevation from your current location, we find the direction v = v 1, v 2 such that v f(1, 1). v f(1, 1) v f(1, 1) = v 1, v 2 3, 5 = 3v 1 5v 2 = = v 1 = 5 3 v 2 5 v = 3 v 2, v 2 = v 2 5, 3 3 where v 2 can be any nonzero number. Page 8 of 1

10 9. (1 points) Use the method of Lagrange multipliers to find the extreme values of the function f(x, y) = 2x y 2 subject to the constraint x 2 + 3y 2 = 36. Solution: Let g(x, y) = x 2 + 3y 2. Then we solve { f = λ g g(x, y) = k { 2, 2y = λ 2x, 6y x 2 + 3y 2 = 36 2 = 2xλ 1 2y = 6yλ 2 x 2 + 3y 2 = 36 3 From 2, 2y 6yλ = 2y(1 + 3λ) = y = or λ = (a) y = : = x = 36 = x = ±6 = 1 λ = 1 if x = 6, λ = 1 if x = So we get two points on the boundary: (x, y, λ) = (6,, 1 ), ( 6,, 1 ). 6 6 (b) λ = 1 3 : 1 = x = 3 3 = ( 3) 2 + y 2 = 36 = y 2 = 9 = y = ±3 So we get two points subject to the constraint in this case: (x, y) = ( 3, 3), ( 3, 3). Evaluate the function values of f at the points obtained in the above steps. f(6, ) = 2(6) () 2 = 12, f( 6, ) = 2( 6) () 2 = 12 f( 3, 3) = 2( 3) (3) 2 = 15, f( 3, 3) = 2( 3) ( 3) 2 = 15 So after comparing these values, we get f attains its max value 12 at (6, ), and its min value 15 at ( 3, ±3). Page 9 of 1

11 1. Suppose that z = F (x, y) and that x = X(s, t) and y = Y (s, t), where F, X and Y all have continuous partial derivatives at all points. Suppose that the following facts are given: (1, 11) = b (c, d) = 5 (1, 11) = 11 (c, d) = 12 (c, d) = 1 (c, d) = 7 (c, d) = a (c, d) = X(a, b) = 1 Y (a, b) = 7 X(c, d) = 1 Y (c, d) = 11 Note: One can view z as a function of x and y, and one can view z as a function of s and t. (a) (5 points) Find (c, d) Solution: X(c, d) = 1, Y (c, d) = 11. By Chain Rule, (c, d) = F x(x(c, d), Y (c, d)) X (c, d) + F y(x(c, d), Y (c, d)) Y (c, d) = (1, 11) (c, d) + = (b)(1) + (11)(a) = 11a + 1b (1, 11) (c, d) (b) (5 points) Find (c, d) Solution: Note X(c, d) = 1, Y (c, d) = 11. Differentiate both sides of z = F (X(s, t), F (s, t)) with respect to t, and we get = F x(x(s, t), F (s, t)) X + F y(x(s, t), F (s, t)) Y (c, d) = F x(x(c, d), F (c, d)) (c, d)f y(x(c, d), F (c, d)) So we have (c, d) = (1, 11) (c, d) + = (b)(7) + (11) (1, 11) (c, d) (c, d) = (c, d) = 7b 11 (c, d) Page 1 of 1

Math 2400, Midterm 1

Math 2400, Midterm 1 Math 4, Midterm 1 February 1, 18 PRINT your name: PRINT instructor s name: Mark your section/instructor: Section 1 Kevin Berg 8: 8:5 Section Xingzhou Yang 8: 8:5 Section 3 Albert Bronstein 9: 9:5 Section