MA261-A Calculus III 2006 Fall Midterm 2 Solutions 11/8/2006 8:00AM ~9:15AM

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1 MA6-A Calculus III 6 Fall Midterm Solutions /8/6 8:AM ~9:5AM. Find the it xy cos y (x;y)(;) 3x + y, if it exists, or show that the it does not exist. Assume that x. The it becomes (;y)(;) y cos y 3 + y (;y)(;) Assume that x y. The it becomes (x;x)(;) xx cos x 3x + x y. (;y)(;) x cos x cos x (x;x)(;) x (x;x)(;). We got two di erent its by approaching along two di erent paths. Thus, the it does not exist.. Find the it (x;y)(;) x + y ln x + y. Note that if (r; ) are polar coordinates of the point (x; y) with r, note that r + as (x; y) (; ). The polar coordinates are x r cos y r sin. This implies that x + y r. Thus, the it becomes x + y ln x + y (x;y)(;) r + r ln r r + r ln r ln r l Hospital r + r r + r r 3 r + r r 3 r. r + 3. Find an equation of the tangent plane to the given parametric surface to the point (u; v) (; ). r (u; v) uvi + uj + vk.

2 There are two rst partial derivatives. r u (u; v) vi + j. r v (u; v) ui + k. At the point (u; v) (; ), we have r (; ) () () i + () j + () k i + j + k, r u (; ) () i + j i + j, r v (; ) () i + k i + k. The normal vector at the point (u; v) (; ) is i j k r u r v (; ) i j + So, the equation of the tangent plane is or, (x ) (y ) (z ), x y z +.. Find dy where sin x + cos y sin x cos y. k i j k. Let We have Thus, by equation 6, we have 5. Find df dr df d where F (x; y) sin x + cos y sin x cos y. cos x cos x cos y sin y + sin x sin cos x cos x cos y sin y + sin x sin y. F (x; y) x + x + y where x r cos y r sin.

3 By the Chain Rule, we have dr (x + ) (cos ) + (y) (sin ) df d ( (r cos ) + ) (cos ) + ( (r sin )) (sin ) r @ (x + ) ( r sin ) + (y) (r cos ) ( (r cos ) + ) ( r sin ) + ( (r sin )) (r cos ) r sin. 6. Find the directional derivative of the function f (x; y) e y ln x at the point (; ; ) in the direction of the vector v h5; i. First, we need an unit vector of v. The unit vector u v h5; i jvj p(5) 5 ; +( ) 3 3. Second, the gradient of f is rf (x; ; e y x ; ey ln x. Thus, the directional derivative at the point (; ; ) is 5 u f (; ) rf (; ) 3 ; 5 h; i 3 3 ; 3 7. Find the critical points of the function First, we have rf (x; y) f (x; y) x x y y. 5 ; (x ) y y ; x x ( y). By assuming rf (x; y), we have (x ) (y y ) (x x) ( y). From the rst equation, we have x, or y y. This tells us that x, or, y or. Similarly, From the second equation, we have x x, or y. This tells us that x or, or, y. There are several combinations: (a) If x, then y must be from the second equation. (b) If y, then x can be or from the second equation. (c) If y, then x can be or from the second equation. 3

4 Therefore, we have ve critical points (; ), (; ), (; ), (; ) (; ). 8. Find the local maximum minimum values saddle point(s) of the function f (x; y) x x y y. From problem 7, we have ve critical points (; ), (; ), (; ), (; ) (; ). To classify them, we need the determinant (x; y) f xx f yy f xy () y y x x ( ) ((x ) ( y)) xy ( y) (x ) 6 (x ) ( y) f xx (x; y) () (y y ) y ( y). (a) For the critical point (; ), (; ) 6 > f xx (; ) >. Thus, it is a local minimum. (b) For the critical point (; ), (; ) 6 <. Thus, it is a saddle point. (c) For the critical point (; ), (; ) 6 <. Thus, it is a saddle point. (d) For the critical point (; ), (; ) 6 <. Thus, it is a saddle point. (e) For the critical point (; ), (; ) 6 <. Thus, it is a saddle point. 9. Find the critical points of the function First, we calculate f (x; y) x + 3y x 5. rf (x; ; hx ; 6yi. Assume that rf. This implies that (x; y) (; ).. Find the extreme values of subject to the constraint f (x; y) x + 3y x 5 x + y 6. Let g (x; y) x + y 6. Thus, g becomes a constraint. If rg (x; y), then (x; y) (; ; ) which does not satisfy the constraint g. Assume that rf rg. With the constraint, we have 8 < : x x 6y y x + y 6 If y 6, then by the second equation, we have 3. This implies that x y p. In this case, we have two critical points ; p 3 ; p 3. The values are both f ; p 3 f ; p 3 7. If y, then x. In this case, we have two critical points (; ) ( ; ). The values of them are f (; ) f ( ; ) 3..

5 Hence, the maximum is 7 at the points ; p 3 ; p 3 the minimum is at the point (; ).. Find the extreme values of f (x; y) x + 3y x 5 5 on the region described by x + y 6. Let g (x; y) x + y 6. First, we calculate rf (x; ; hx ; 6yi rg hx; yi. First, we nd the extreme values for the interior of x + y 6. Consider the region x + y < 6. Assume that rf. This implies that (x; y) (; ). Since (; ) satis es x + y < 6, (; ) is a critical point. The value is f (; ) 7. Next, we consider the boundary x + y 6. Thus, g becomes a constraint. If rg (x; y), then (x; y) (; ; ) which does not satisfy the constraint g. Assume that rf rg. With the constraint, we have 8 < : x x 6y y x + y 6 If y 6, then by the second equation, we have 3. This implies that x y p. In this case, we have two critical points ; p 3 ; p 3. The values are both f ; p 3 f ; p 3 7. If y, then x. In this case, we have two critical points (; ) ( ; ). The values of them are f (; ) f ( ; ) 3. Hence, the maximum is 7 at the points ; p 3 ; p 3 the minimum is 7 at the point (; ).. Evaluate the double integral ZZ (x + y) da, where is bounded by y p x y x..

6 6 Set p x x 3. We can solve for x or.thus, ZZ (x + y) da Z p x (x + y) dy x " xy + y y p # x yx " x p x + (p x) x 3 x + 5 x 5 x + 5 () 5 () + x 3 x x x 5 x () x () 5 " Z p # x (x + y) dy x # x x + (x ) 5 () 5 () + () () Evaluate the integral: Z y e x dy. According to the region, by interchanging the order, we have Z y e x dy Z x e x dy e x h (y)j yx y e x [(x) xe x e x x x i ()] e e e.

7 . Evaluate RR x + y 7 da where is the half disk x + y with x. Using polar coordinates, let x r cos, y r sin where r The reason is for right-half disk. Therefore, RR x + y 7 da Z Z Z Z (r cos ) + (r sin ) 7 jrj ddr r 8 ddr Z r 8 dr r r Z j d Find the volume of the solid that lies under the plane x y +z above the square R [ ; ] [; ]. The volume is ZZ ( x + y) da R Z h ( x + y) dy y xy + y i y y Z ( x + y) dy () x () + () () x () + () ( x) x + x x x () + () ( ) + ( ).