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# Answer to Question #243733 in Chemical Engineering for Lok

Question #243733

The differential equation of (x-a)^{2}+(y-b)^{2}+z^{2}=2019 where a, b are arbitrary constants is.

A) z^2(p^2+q^2)=1. B)z^2(p^2+q^2)=2019.

C) z^2(p^2+q^2+1)=1.

D) z^2(p^2+q^2+1)=2019

1
2021-10-01T04:40:02-0400
"z=(x-a)^2+(y-b)^2"

Differentiating partially with respect to "x" and "y," we get

"\\dfrac{\\partial z}{\\partial x}=2(x-a)""\\dfrac{\\partial z}{\\partial y}=2(y-b)"

Squaring and adding these equations, we have

"(\\dfrac{\\partial z}{\\partial x})^2+(\\dfrac{\\partial z}{\\partial y})^2=(2(x-a))^2+(2(y-b))^2""(\\dfrac{\\partial z}{\\partial x})^2+(\\dfrac{\\partial z}{\\partial y})^2=4((x-a)^2+(y-b)^2)"

Since "(x-a)^2+(y-b)^2=z," we get

"(\\dfrac{\\partial z}{\\partial x})^2+(\\dfrac{\\partial z}{\\partial y})^2=4z"

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