Find the partial differential equations of all spheres whose centers lie on the z-axis
Answers
Answer:
Find the partial differential equation of all spheres Whose centres lie on the z axis
1.x2+y2+(z−c)2=r2...( ...
2.2x+2(z−c)p=0.
3.2y+2(z−c)q=0.
2y+2(z−c)q=0.⇒(z−c)p=−x...( ...
2y+2(z−c)q=0.⇒(z−c)p=−x...( ...(z−c)q=−y
Step-by-step explanation:
Given :
Sphere whose center lies on z-axis.
To find :
Partial differential equation of sphere.
Solution :
The equation of sphere whose center lie on z-axis with radius r is :
{x}^{2} + {y}^{2} + {(z - c)}^{2} = {r}^{2}x
2
+y
2
+(z−c)
2
=r
2
(equation 1)
to find the partial differential equation of sphere,
firstly we are differentiating with respect to x.
\frac{ \delta({x}^{2} + {y}^{2} + {(z - c)}^{2} )}{ \delta \: x} = \frac{ \delta {r}^{2} }{ \delta \: x} \: = 0
δx
δ(x
2
+y
2
+(z−c)
2
)
=
δx
δr
2
=0
2x \: + \: 2(z - c)p = 02x+2(z−c)p=0
(equation 2)
where
\frac{ \delta \: z}{ \delta \: x} = p
δx
δz
=p
then we are differentiating with respect to y.
2y \: + \: 2(z - c)q = 02y+2(z−c)q=0
(equation 3)
where
\frac{ \delta \: z}{ \delta \: y} = q
δy
δz
=q
by solving equation 2 and 3 :
\begin{gathered}(z - c)p = - x \\ and \\ (z - c)q = - y\end{gathered}
(z−c)p=−x
and
(z−c)q=−y
dividing these we get :
\frac{p}{q} = \frac{x}{y}
q
p
=
y
x
it means
py = qx
So the desired partial differential equation is :
py - qx = 0
hope it will help you...