Question

find partial differential equation of family of all spheres whose centre lies on XY plane radius is 2​

Answer 1

Family of Spheres with the center in XY- plane and radius two, find their differential equation.

Explanation:

• Standard Equation of a sphere having the center C whose co-ordinates are given by (a, b, c) and radius denoted by 'r' is given by ,               $(x-a)^2+(y-b)^2+(z-c)^2=r^2$        
• given that the required family of spheres has their center in XY- plane we get the above equation as $(x-a)^2+(y-b)^2+z^2=4$  
• Partially differentiating the equation with respect to 'x' we get,                                $2(x-a)+2z\frac{dz}{dx}=0\\ x-a+z\frac{dz}{dx}=0$      ----------(a)
• Partially differentiating the equation with respect to 'y' we get,            $2(y-b)+2z\frac{dz}{dy}=0\\ y-b+z\frac{dz}{dy}=0$      ----------(b)      
• now equating in (a) and (b)  $p=\frac{dz}{dx} ,\ \ q=\frac{dz}{dy}$  we get,                                             $x-a=-zp\\y-b=-zq$    
• squaring on both sides and adding above equations we get,                           $(x-a)^2+(y-b)^2=z^2(p^2+q^2)$                                                                                                            
• from standard equation we get,                                                                                $4-z^2=z^2(p^2+q^2)\\z^2(1+p^2+q^2)=4$       ---->Required partial differential equation                                                                                                                                                      

Answer 2

Given,

sphere whose center lies on the x-y plane whose plane radius is 2.

To Find,

partial differential equation of the family of all spheres.

Solution,

the equation is $(x-a)^{2}+(y-b)^{2}+z^{2}=4$              (1)

differentiate equation (1) w.r.t 'x'

$2(x-a)+2z\frac{dz}{dx}=0$                                                    (2)

[ $\frac{dz}{dx}$ is the partial differential w.r.t 'x' and is denoted by p ]

now, squaring on both sides in equation (2) we get,

$(x-a)^{2}=z^{2}p^{2}$                                                            (3)

similarly partial differentiate equation (1) w.r.t 'y'

$2(y-b)+ 2z\frac{dz}{dy} =0$                                                      (4)

[ $\frac{dz}{dy}$ is the partial differential w.r.t 'y' and is denoted by q ]

now, squaring both sides in equation (4) we get,

$(y-b)^{2}=z^{2}q^{2}$                                                              (5)

now substituting the value of equation (3) and equation (5) in equation (1), we get

$z^{2}p^{2}+z^{2}q^{2} +z^{2}=4\\ \\ z^{2}(p^{2}+q^{2}+1)=4$

Hence the partial differential equation of the family of all spheres whose center lies on the XY plane is  $z^{2}(p^{2}+q^{2}+1)=4$ .