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Use Theorem 12.34 to find the indicated derivatives in Exercises 31鈥36. Be sure to simplify your answers.

ddtwhen=x2+y2+z2,x=t,y=t2,z=t3

Short Answer

Expert verified

The value of ddt=121+4t3+3t5t+t4+t6-12.

Step by step solution

01

Step 1. Given Information. 

=x2+y2+z2x=ty=t2z=t3

02

Step 2. Calculation.

By Theorem 12.34, we have

ddt=xdxdt+ydydt+zdzdt------(1)

So first we find

x,dxdt,y,dydt,z,dzdt

So we have

x=12x2x2+y2+z2=xx2+y2+z2y=12y2x2+y2+z2=yx2+y2+z2z=12z2x2+y2+z2=zx2+y2+z2dxdt=12t12dydt=2tdzdt=3t2

03

Step 3. Calculation.

Use these above values in (1) we get,

ddt=xdxdt+ydydt+zdzdtddt=x2t12x2+y2+z212+2ytx2+y2+z212+3t2zx2+y2+z212

So from here, putting the value of x,y,zin terms of twe get

ddt=x2t12x2+y2+z212+2ytx2+y2+z212+3t2zx2+y2+z212ddt=1t+t4+t612+2t3+3t5ddt=121+4t3+3t5t+t4+t6-12

04

Step 4. Conclusion.

The value ofddt=121+4t3+3t5t+t4+t6-12.

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