Math 222 Spring 2013 Exam 3 Review Problem Answers

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1 . (a) By the Chain ule, Math Spring 3 Exam 3 eview Problem Answers w s w x x s + w y y s (y xy)() + (xy x )( ) (( s + 4t) (s 3t)( s + 4t)) ((s 3t)( s + 4t) (s 3t) ) 8s 94st + 3t (b) By the Chain ule, w s w x x s + w y y s (t cos s) + x + y x + y (t) t cos s + t t sin s + st cos s + sin s + s Note that both parts (a) and (b) could be done by first writing w in terms of s and t and then computing w/ s directly.. We have to use the Chain ule for this one. We have dw dt w dx x dt + w dy y dt w x w (t + 3) + (5 t) y When t, we get x and y 6. Thus, when t, we have dw dt ( )(( ) + 3) + ()(5 ( )) t 6 When t, we get x 4 and y 4. Thus, when t, we have dw dt ()(() + 3) + ( 4)(5 ()) t (a) f yz e xy, xz e xy, ze xy

2 (b) This is asking for D u f(,, ), where u is the unit vector in the direction of, 3,. We have D u f(,, ) f(,, ) u e, e, e 3,, e (c) f(, 3, ) 3,, (d) f(,, ),, 5 4. (a) g y + z, x + 3z, x + 3y (b) This is asking for D u g(,, ), where u is the unit vector in the direction of 3, 4,. We have D u g(,, ) g(,, ) u, 4, 3 4,, (c) g(,, ), 4, (d) g(,, ), 4, 4 5. Let h(x, y) z.x.y. (a) Since u, is the unit vector pointing due south, we need to know the sign of D u h(5, 8), 3., 3.. Since D u h(5, 8) is positive, you will ascend if you go due south. (b) We need D u h(5, 8) where u,, which is D u h(5, 8), 3.,.. Since this is negative, you will descend if you head northwest. (c) The hill is the steepest in the direction of h(5, 8), 3. (in terms of cardinal directions, this is 7.65 south of west) 6. (a) A general form for the equation of the tangent plane at (,, ) is f x (, )(x ) + f y (, )(y ( )) (z ) Solving for f x (, ) and f y (, ), and plugging these in gives 8(x ) + 4(y + ) (z ) which is 8x + 4y z, or z 8x + 4y +. (b) The function f has horizontal tangent planes at any points where f x and f y are both. We have f x 6x + and f y y, so f x when x 3, and f y when y. Thus, f has one horizontal tangent place at the point ( 3,, f( 3, )) ( 3,, 3 ). 7. (a) A general form for the equation of the tangent plane at (3,, 3e ) is g x (3, )(x 3) + g y (3, )(y ) (z 3e ) Solving for g x (3, ) and g y (3, ), and plugging these in gives 9e (x 3) + 6e (y ) (z 3e ) which is 9e x + 6e y z 6e, or z e (9x + 6y 6).

3 (b) As in the previous problem, we look for points where both g x and g y are. We have g x (x + )e x +y and g y xye x +y. Since e x +y is never, the only way that g x can be is if x +. However, no value of x can make this equation true. So, g x is never, and hence there are no points where g has a horizontal tangent plane. 8. A general form for the linearization at (, ) is L(x, y) f x (, )(x ) + f y (, )(y + ) + f(, ) Note that f(, ). Since f x x sec (x y ) and f y y sec (x y ), we get f x (, ) and f y (, ). So, the linearization is L(x, y) (x ) + (y + ) + x + y We have tan(. (.96) ) L(.,.96). (the actual value of tan(. (.96) ) tan(.985) is about.99) 9. (a) We have f x x+4 and f y y 6, so the only critical point is (, 3). Furthermore, f xx, f yy, and f xy, so D f xx f yy f xy ] 4. At (, 3), we have D > and f xx >, so by the Second Partial Derivative Test there is a local minimum at (, 3) (b) We have g x x y + 9 and g y x + y 6, so g x when y x + 9. Plugging this into g y gives g y 3x +, so g y when x 4. Backsolving, we get y. Thus, there is one critical point at ( 4, ). At this point, g xx > and D g xx g yy g xy ] 3 >, so there is a local minimum at ( 4, ). (c) We have h x xe y/ and h y x e y/ + e y/ + yey/. Since e y/ is never, h x when x. Plugging this into h y gives h y e y/ + yey/ e y/ ( + y). So, h y when + y, i.e. when y. So, there is one critical point: (, ). Now, h xx e y/, h yy 4 x e y/ + ey/ + ey/ + 4 ey/, and h xy xe y/. At (, ), we have h xx (, ) e, h yy e, and h xy. Thus, D h xx h yy h xy ] is negative, so h has a saddle point at (, ). (d) We have k x 3x 6 and k y 4y 3 4y, so k x when x ± and k y when y,, or. This means that there are six critical points: (, ), (, ), (, ) (, ), (, ), (, ) Furthermore, k xx 6x, k yy y 4, and k xy. Hence, D 6x(y 4). The conclusions we reach are summarized in the following table Point D k xx Conclusion (, ) 4 6 saddle point (, ) 48 6 local minimum (, ) 48 6 local minimum (, ) 4 6 local maximum (, ) 48 6 saddle point (, ) 48 6 saddle point

4 . Since f x, and f y, f has no critical points. So, the absolute extrema of f occur on the boundary of S. The boundary has two pieces: y 4 x (for x ) and y (for x ). Along y 4 x, we have f(x, 4 x ) x (4 x ) x + x 4, which makes f a function of one variable. So, we need to find the extreme values of f(x) x + x 4 on, ]. We have f (x) x +, so f(x) has a critical point at x. Evaluating f(x) at this critical point and at the endpoints of, ], we have f( ) 4, f( ) 5, and f() 4. Along y, we have f(x, ) x. On the interval, ], the maximum value of x is 4 and the minimum value is 4. Comparing the extreme values of f along y 4 x with the extreme values of f along y, we see that the maximum value of f on S is 4 while the minimum value is (a) x 3xy + y da x 3xy + y dy dx (b) xy x + y da 48 ] y4 x y 3 xy + 3 y3 dx 4x 4x dx 4 3 x3 x x ] x x 3 4 xy x + y dy dx u x + y du ydy when y, u x when y 4, u x x +6 x x u du dx ] x ux +6 3 u3/ ux dx y x(x + 6) 3/ x(x ) 3/ dx x(x + 6) 3/ x 4 dx 5 (x + 6) 5/ 5 x5 ] x3 x

5 (c) π/ x sin y dx dy π/ π/ ] x 3 x3 sin y dy x 8 sin y dy 3 ] yπ/ 83 cos y y 8 3 (d) Integrating with respect to x first will require integration by parts, so we switch the order of integration: xe xy+ dx dy (e) The region can be described as either xe xy+ dy dx u xy + du x dy when y, u when y, u x + x+ e u du dx ] ux+ e u dx u e x+ e dx e x+ ex e e ] x x, y x x or y, x y so we can evaluate either or x y ln(x + ) dy dx () ln(x + ) dx dy () Integral () can be done, but we will need to use integration by parts to handle ln(x + ) dx. So, we will compute integral () (as we ll see, integration by parts

6 will also be required for (), but it will be a little easier). We have x ] yx ln(x + ) dy dx y ln(x + ) dx x ln(x + ) dx u x + y du x dx when x, u when x, u ln u du From here, we use integration by parts (with w instead of u, since we used u above): (f) The region is w ln u dw u du ln u du dv du v u ln u du u ln u ] u u ln ] ( ln ) ln u du ] u u ] ] which is horizontally simple and can be described as y, y x 8 y. So, we have 8 y ] x8 y y da y dx dy xy dy y xy 8 y 3 dy ] y 8y y4 8 y

7 (g) We can t integrate is ln y dy, so we need to switch the order of integration. The region and can be described as x ln, e x y. We want to change the way we describe the region so that y is bounded by constants. The curve y e x is the same as x ln y, so the region is Thus, ln y, x ln y e x ln y dy dx ln y x ln y ] y dx dy ln y ] xln y (h) We can t integrate e x dx, so we need to switch the order of integration. The region is y dy y x dy and can be described as y, 3y x 3. We want to change the way we describe the region so that x is bounded by constants. The curve x 3y is the same as y 3 x, so the region is x 3, y 3 x

8 Thus, 3 3y e x dx dy x 3 3 e x dy dx ye x] y 3 x dx y 3 xex dx u x du x dx when x, u when x 3, u 9 9 e u du e u ] u9 u 6 (e9 ) (i) The region is given in terms of polar coordinates, so we should integrate using polar coordinates. Using the substitutions x r cos θ and da r dr dθ, we have π/ sin θ 3x da 3(r cos θ)(r dr dθ) π/4 sin θ π/ sin θ π/4 sin θ π/ π/4 π/ π/4 π/ π/4 3r cos θ dr dθ) ] r sin θ cos θ r 3 dθ) rsin θ cos θ(8 sin 3 θ sin 3 θ) dθ 7 cos θ sin 3 θ dθ u sin θ du cos θ dθ when θ π/4, u / when θ π/, u 7u 3 du / 7 6 ] u 4 u4 u /

9 (j) The region is best described using polar coordinates: r, π θ 3π 4 Performing the substitutions r x + y and da r dr dθ, we get 3π/4 cos(x + y ) da cos(r )r dr dθ π/ 3π/4 π/ u r r cos(r ) dr dθ du r dr when r, u when r, u 4 3π/4 4 cos(u) du dθ π/ 3π/4 ] u4 sin u π/ u 3π/4 sin 4 dθ π/ ] θ3π/4 (sin 4)θ θπ/ π 8 sin 4 (k) The region is best described using polar coordinates: r 5, θ π Performing the substitutions r x + y and da r dr dθ, we get π 5 5 x y da (5 r )r dr dθ π 5 π π dθ 5r r 3 dr dθ ] r5 5 r 4 r4 dθ r 44 4 dθ 44 4 θ ] θπ θ 44π