9 Notes 7

9 Notes 7 - MATH 20C Lecture 16 Monday November 1 2010...

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Unformatted text preview: MATH 20C Lecture 16 - Monday, November 1, 2010 Implicit differentiation Example: x 2 + yz + z 3 = 8 . Viewing z = z ( x,y ) , compute ∂z ∂x and ∂z ∂y . Take ∂ ∂x of both sides of x 2 + yz + z 3 = 8 . Get 2 x + y ∂z ∂x +3 z 2 ∂z ∂x = 0 , hence ∂z ∂x =- 2 x y +3 z 2 =- 2 3 . In general, consider a surface F ( x,y,z ) = c. The we can view z = z ( x,y ) as a function of two independent variables x,y and compute ∂z ∂x and ∂z ∂y . To do so, we take the partial derivative with respect to x of both sides of the equation F ( x,y,z ) = c and get (by the chain rule) ∂F ∂x ∂x ∂x + ∂F ∂y ∂y ∂x + ∂F ∂z ∂z ∂x = 0 . But ∂x/∂x = 1 and, since x and y are independent, ∂y/∂x = 0 (changing x does not affect y ). Hence the equation above really says that F x + F z ∂z ∂x = 0 which implies ∂z ∂x =- F x F z . Similarly, ∂z ∂y =- F y F z . Changing gears, let’s see how we can recover f from its gradient. Say ∇ f = h 3 x 2 y,x 3 +2 z, 2 y + cos z i . We proceed by successive integration. We are given that f x = 3 x 2 y. Integrating with respect to x (view y,z as constants), we see that f = x 3 y + g ( y,z ) . Therefore f y = x 3 + ∂g ∂y . But we know from the gradient that f y = x 3 + 2 z , hence g y = 2 z. Integrate with respect to y and get g = 2 yz + h ( z ) , hence f = x 3 y + 2 yz + h ( z ) . Since f z = 2 y + cos z we get that dh dz = cos z, so h ( z ) = sin z + C. Substituting in the expression of f gives f = x 3 y + 2 yz + sin z + C....
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9 Notes 7 - MATH 20C Lecture 16 Monday November 1 2010...

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