Answer to Question #308159 in Differential Equations for vyom

Question

form a partial differential equation for (x− a)² + (y−b)² + z² = 1

Answer 1

(x - a)² + (y – b)² + z² = c²   ........(1)

Differentiating (1) w.r.t ‘x’

2(x−a)+0+2zdz/dx=0

2(x-a) =-2zdz/dx

(x-a)=- zdz/dx .......(2)

Differentiating (1) with respect to y

0+2(y-b)+2zdz/dy =0

2(y-b)=-2zdz/dy

(y-b) = -zdz/dy.....(3)

Inserting 2 and 3 in the equation 1

We have, (-zdz/dx)² + (-zdz/dy)² +z²=c²

Let dz/dx be p and dz/dy be q

We have,

(-zp)²+(-zq)²+z² =c²

z² (p² + q² + 1) = c²