53 Worksheet Solutions 9_21

53 Worksheet Solutions 9_21 - y to our original equation...

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Math 53: Multivariable Calculus Solutions for Worksheet 9/21/11 Exercise 0.1. Find both (first order) partial derivatives of f ( x,y ) = y x + x 2 ln( x + y ) . Explain how to interpet them as giving the slope of some curve in some cross section. Solution. We have f x ( x,y ) = y x ln( y ) + 2 x ln( x + y ) + x 2 x + y and f x ( x,y ) = y x - 1 + x 2 x + y . We may interpret f x ( x,y ) as the derivative of the curve defined by f ( x,y ) intersected with the plane y = y 0 (where y 0 is some constant). Similarly, f y ( x,y ) may be interpreted as the derivative of the curve defined by f ( x,y ) intersected with the plane y = x 0 . ± Exercise 0.2. Use implicit differentiation to find ∂z ∂x and ∂z ∂y for x - z = arctan( yz ) . Solution. Applying ∂x to both sides gives 1 - ∂z ∂x = y ∂z ∂x 1 + y 2 z 2 , which rearranges to 1 y 1+ y 2 z 2 + 1 = ∂z ∂x . Similarly, applying
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Unformatted text preview: y to our original equation gives-z y = z + y z y 1 + y 2 z 2 , which rearranges to z y =-z 1+ y 2 z 2 y 1+ y 2 z 2 + 1 =-z y + 1 + y 2 z 2 . 1 2 Exercise 0.3. Calculate the second order partials of f ( x,y ) = x sin( x + 2 y ), and check that f xy = f yx . Solution. We calculate f x = sin( x + 2 y ) + x cos( x + 2 y ) f y = 2 x cos( x + 2 y ) f xx = cos( x + 2 y ) + cos( x + 2 y )-x sin( x + 2 y ) f xy = 2 cos( x + 2 y ) + 2 x sin( x + 2 y ) f yx = 2 cos( x + 2 y ) + 2 x sin( x + 2 y ) f yy =-4 x sin( x + 2 y ) . As expected, we have f xy = f yx ....
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53 Worksheet Solutions 9_21 - y to our original equation...

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