Practice Midterm 2 Math 2153

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1 Practice Midterm 2 Math 23. Decide if the following statements are TRUE or FALSE and circle your answer. You do NOT need to justify your answers. (a) ( point) If both partial derivatives f x and f y exist at (a, b) then f is differentiable at (a, b). F (f x and f y must exist and be continuous on an open set containing the point (a, b) to ensure that f is differentiable at (a, b).) (b) ( point) If f has a local maximum at the point (a, b, c) then f. F (f can also have a local maximum at (a, b, c) when f does not exist.) (c) ( point) If f has a saddle point at (a, b) then f cannot have a local minimum at (a, b). T (The definition of saddle point precludes it from being a local maximum or mini- mum) (d) ( point) If f is differentiable at (a, b, c) then magnitude of the gradient vector f(a, b, c) is the maximal directional derivative D u (a, b, c) where u ranges over all unit vectors in R 3. T 2. Give examples of the following. Be as explicit as possible. You do NOT need to justify your answers. (a) (2 points) Give an example of a function f(x, y) continuous on R 2 such that there are infinitely many points (a, b) R 2 such that f has a local maximum at (a, b). f(x, y) (b) (2 points) Give an example of a function F (x, y, z) for which the graph of z sin(xy) is a level surface. F (x, y, z) z sin(xy) (c) (2 points) Give an example of a function f(x, y) with domain R 2 for which f x (, ) exists but f y (, ) does not exist. f(x, y) y (d) (2 points) Sketch level curves for a function f(x, y) with four local maximuma and no local minima. Make sure to include enough level curves to illustrate these properties.

2 3. Compute the following: ( ) ln(xy) (a) (2 points) y x y y ( ) ln(xy) x y y ye y ln x yx ( ) ln(xy) e y ln x ln(x)e y ln x ln(xy) e2y ln x xy ln(x)x y ln(xy) x 2y (b) (2 points) D u (3x 3 y y) where u Let f(x, y) 3x 3 y y. 2,. D u (3x 3 y y) f u 9x 2 y, 3x 3 2, 8 x 2 y (3x 3 ) (c) (2 points) Find the unit vector u u, u 2, u 3 pointing in the direction of maximum increase for the function f(x, y, z) xyz 2 at the point (,, ). f yz 2, xz 2, 2xyz f(,, ),, 2 f(,, ) u f(,, ),, 2,, 2 6, 6 2, 6 Page 2

3 (d) (2 points) π sin(x + y) dx dy. π sin(x + y) dx dy 2 sin y cos(x + y) π dy x cos(π + y) + cos y dy cos y + cos y dy 2 x 2 sin 2 2 sin (e) (2 points) xz y dz dx dy. xz y dz dx dy x 2 4y xz 2 2y z x dx dy 2y 2 dy x y dy ln y 2 y ln 2 ln ln 2 dx dy (f) (2 points) Compute z x in terms of x, y and z if z satisfies the implicit equation xy + yz + xz 7 (xy + yz + xz) x x () y + y z x + z + x z x y z x + x z x y z z (y + x) y z x Page 3

4 z x y z y+x 4. Change the order of integration for the following double integrals. You may have to express the new integral as a sum of double integrals. You do not need to evaluate the integrals. (a) (2 points) ln x x xy 2 dy dx. ln x x xy 2 dy dx x ln x ln 2 e y y+ ln 2 y+ xy 2 dy dx xy 2 dx dy + xy 2 dx dy ln 2 y+ e y xy 2 dx dy + ln 2 y+ xy 2 dx dy (b) (2 points) π/4 cos y sin y xy 2 dx dy. π/4 cos y sin y xy 2 dx dy 2/2 arcsin x xy 2 dy dx + 2/2 arccos x xy 2 dy dx. ( points) Give a triple integral which computes the volume of the bounded region in R 3 enclosed by the surfaces x 2 + y 2 z 2 9, z 4, z 4 Let D be the region in R 3 described above. Then Volume of D dv D 4 9+z 2 9+z 2 x z 2 9+z 2 x 2 dy dx dz 6. ( points) Compute the maximum and minimum values for the function f(x, y) x y 2 + xy on the region in the plane bounded by the ellipse y 2 + 9x 2 9. Page 4

5 First we find the critical points of f. f + y, 2y + x Thus + y and 2y + x. Hence y and ( 2)( ) + x. Thus the critical point of f is ( 2, ). This point is not in the region on which we are maximizing since ( ) 2 + 9( 2) 2 > 9. Thus there are no critical points in the interior of the ellipse. Now let g(x, y) y 2 + 9x 2 9. We wish to maximize f subject to the constraint g(x, y). Applying the method of Lagrange multipliers we compute Setting f λ g gives the two equations g 8x, 2y + y λ8x 2y + x λ2y Solving the second equation for λ we get λ 2y+x 2y. Substituting this into the first equation we get ( ) 2y + x + y 8x 2y 2y + 2y xy 8x 2 From the constraint equation we get y 2 9 9x 2 so 2y + 2y 2 36xy + 8x 2 2y + 2(9 9x 2 ) + 36xy 8x 2 2y xy 36x 2 (2 + 36x)y 36x 2 8 y 9(2x2 ) + 8x y 2 8(4x4 + 4x 2 ) + 36x + 324x 2 9 9x 2 8(4x4 + 4x 2 ) + 36x + 324x 2 You might have noticed by now that computing x here is not a reasonable task ( points) Use the method of Lagrange multipliers to find the maximum volume for a lidless can with surface area. The volume of a can with radius r and height h is V (r, h) πr 2 h. The surface area for a lidless can with radius r and height h is S(r, h) πr 2 + 2πrh. Page

6 We wish to maximize V subject to the constraint that S(r, h) V 2πrh, πr 2 S 2πr + 2πh, 2πr The vector equation V λ S gives the two equations 2πrh λ(2πr + 2πh) πr 2 λ2πr Solving for λ in the second equation we get λ r 2 Substituting λ r 2 into the first equation and cancelling π we get 2rh r (2r + 2h) 2 and hence 2rh r 2 + rh rh r 2 h r Substituting h r back into our constraint S(r, h) gives πr 2 + 2πr 2 3πr 2 ± 3π r Radii must be positive ( so only the positive root makes sense. We have h r so the maximum volume must occur at (r, h) 3π, 3π ). Thus the maximum volume is V ( ) ( ) 2 ( ), π 3π 3π 3π 3π 3 3π 8. ( points) Estimate the change in z xy 2 x 2 + y when (x, y) changes from (, 2) to (.,.9). Let f(x, y) xy 2 x 2 + y. Then f x y 2 2x f y 2xy + Page 6

7 so f x (, 2) 2 2 2() 2 f y (, 2) 2()(2) +. Thus we can estimate the change z in z using the differential dz z dz f x (, 2) dx + f y (, 2) dy f x (, 2) (. ) + f y (, 2) (.9 2) 2 (.) + (.).3 Page 7

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