This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 53 Homework 11 Solutions 16.3 # 3: y (2 x 3 y ) = 3 = x ( 3 x +4 y 8), and vector F is defined in the entire plane (open and simply connected), so vector F is conservative. So there is a function f such that f = vector F , i.e. f x = 2 x 3 y and f y = 3 x + 4 y 8. Integrating with respect to x , f x ( x,y ) = 2 x 3 y implies f ( x,y ) = x 2 3 xy + g ( y ) for some function g ( y ); differentiating both sides of this equation with respect to y gives f y = 3 x + g ( y ). Thus we should have 3 x + g ( y ) = 3 x +4 y 8, or g ( y ) = 4 y 8. Therefore g ( y ) = 2 y 2 8 y + c where c is a constant. Hence f ( x,y ) = x 2 3 xy + 2 y 2 8 y + c . 16.3 # 8: y ( xy cos xy + sin xy ) = x cos xy x 2 y sin xy + x cos xy = 2 x cos xy x 2 y sin xy , while x ( x 2 cos xy ) = 2 x cos xy x 2 y sin xy . So vector F is defined everywhere and satisfies P y = Q x , and hence it is conservative. We now look for f such that f = vector F , i.e. f x = xy cos xy + sin xy and f y = x 2 cos xy . Integrating with respect to y , f y = x 2 cos xy implies that f ( x,y ) = x sin xy + g ( x ) for some function g ( x ). Differentiating this equation with respect to x , we get f x = sin xy + x 2 cos xy + g ( x ); since we want f x = xy cos xy + sin xy , we deduce that g ( x ) = 0, i.e. g ( x ) = c is a constant. Thus f ( x,y ) = x sin xy + c . (The usual method, namely first integrating f x = P with respect to x to find f up to a function of y , and then differentiating with respect to y to find that function, would also work just fine. However in this example it is slightly easier to integrate Q with respect to y than to integrate P with respect to x .) 16.3 # 15: a) f x ( x,y,z ) = yz implies f ( x,y,z ) = xyz + g ( y,z ), and so f y ( x,y,z ) = xz + g y ( y,z ). But f y ( x,y,z ) = xz , so g y ( y,z ) = 0, and (integrating with respect to y ) g ( y,z ) = h ( z ). Thus f ( x,y,z ) = xyz + h ( z ), and f z ( x,y,z ) = xy + h ( z ). But f z ( x,y,z ) = xy +2 z , so h ( z ) = 2 z , and hence h ( z ) = z 2 + c . Thus (taking c = 0), one potential function is f ( x,y,z ) = xyz + z 2 . b) By the fundamental theorem, integraltext C vector F dvector r = f (4 , 6 , 3) f (1 , , 2) = 81 4 = 77. 16.3 # 19: Note that vector F = ( tan y,x sec 2 y ) is only defined when y negationslash = 2 + n . Therefore the path C from (1 , 0) to (2 ,/ 4) cant be arbitrary, it must lie in the region 2 < y < 2 . With this understood, we note that y (tan y ) = sec 2 y = x ( x sec 2 y ), and the region 2 < y < 2 is simply connected, so vector F is conservative....
View
Full
Document
 Fall '07
 Hutchings
 Math

Click to edit the document details