53 Worksheet Solutions 9_21

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online