To determine

To find: The value of dzdt when (x,y,z)=(2,2,1).

Answer

The required value of dzdt when (x,y,z)=(2,2,1) is dzdt=−18_.

Explanation

Given:

The given values are dxdt=5 and dydt=4 and (x,y,z)=(2,2,1).

Formula used:

Chain rule: dydx=dydu⋅dudx

Calculation:

Since x2+y2+z2=9, where x, y, and z are function of the variable t.

Therefore, x, y and z changes when t changes.

Obtain dzdt when (x,y,z)=(2,2,1), dxdt=5 and dydt=4.

Differentiate the function x2+y2+z2=9 with respect to the variable t both sides.

ddt(x2+y2+z2)=ddt(9)ddt(x2)+ddt(y2)+ddt(z2)=02xdxdt+2ydydt+2zdzdt=02zdzdt=−[2xdxdt+2ydydt]

Therefore, dzdt=−12z[2xdxdt+2ydydt].

Substitute the value of x=2, y=2, z=1, dxdt=5 and dydt=4 in dzdt.

dzdt=−12×1[(2)(2)(5)+(2)(2)(4)]=−12[20+16]=−12[36]=−18

Therefore, dzdt=−18_ when (x,y,z)=(2,2,1).