Answer to Question #223828 in Chemical Engineering for Lokika

Question #223828

The differential equation of (x-a)^{2}+(y-b)^{2}+z^{2}=21


1
Expert's answer
2021-08-16T02:20:56-0400
"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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