hw11sol - Math 53 Homework 11 Solutions 16.3 # 3: y (2 x 3...

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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....
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hw11sol - Math 53 Homework 11 Solutions 16.3 # 3: y (2 x 3...

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