   Chapter 14, Problem 36RE

Chapter
Section
Textbook Problem

If v = x2sin y + yexy, where x = s + 2t and y = st, use the Chain Rule to find ∂v/∂s and ∂v/∂t when s = 0 and t = 1.

To determine

To find: The values of vsandvt at s=0andt=1 using Chain Rule if v=x2siny+yexy , x=s+2t and y=st .

Explanation

Given:

The function v=x2siny+yexy .

Chain Rule:

“Suppose that z=f(x,y) is a differentiable function of x and y , where x=g(s,t) andy=h(s,t) are both differentiable functions of sandt . Then zs=zxxs+zyys and zt=zxxt+zyyt ”.

Calculation:

The function v=x2siny+yexy (1)

Substitute s=0andt=1 in x=s+2t and obtain x ,

x=s+2tx=0+2(1)x=2

Substitute s=0andt=1 in y=st andobtain y ,

y=sty=0(1)y=0

Thus, the value of xandy at s=0andt=1 are x=2andy=0 .

Take the partial derivative with respect to x of the equation (1),

vx=x(x2siny+yexy)=2xsiny+yexy(y)=2xsiny+y2exy

Thus, the partial derivate, vx=2xsiny+y2exy .

Substitute x=2andy=0 in vx=2xsiny+y2exy ,

vx=2xsiny+y2exy=2(2)sin(0)+(0)2e0=0+0=0

Thus, the value of vx=2xsiny+y2exy at x=2andy=0 is 0.

Take the partial derivative with respect to y of the equation (1),

vy=x2siny+yexy=x2(cosy)+yexy(x)+(1)exy=x2cosy+xyexy+exy

Thus, the partial derivate, vy=x2cosy+xyexy+exy .

Substitute x=2andy=0 in vy=x2cosy+xyexy+exy ,

vy=x2cosy+xyexy+exy=(2)2cos(0)+2(0)e0+e0=4+0+1=5

Thus, the value of vy=x2cosy+xyexy+exy at x=2andy=0 is 5

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