Question

Find a partial differential equation of all spheres of radius 10 and centres lying on the plane x+y=0.

Answer 1

Answer:

the equation for such spheres will be. (x − a). 2. + (y − b). 2 ... first-order PDE uτ − (y. 2 − 1)ux = 0, and the initial condition is given by u(x, y,0) = eye.

Step-by-step explanation:

Answer 2

Answer:

The required partial equation is -

$(\frac{x+y}{p+q})^{2}$ $(p^{2} +q^{2} +1)$ = 100 where, p = $\frac{dz}{dx}$ ; q = $\frac{dz}{dy}$

Step-by-step explanation:

The equation of the sphere with Centre (a,b,c) and the radius r is -

$(x-a)^{2} + (y-b)^{2} + (z-c)^{2} = r^{2}$ ..................(1)

Given, Centre lie on the plain

x + y =0, so a = -b;  

r = 10;

Therefore, equation (i) becomes-

$(x-a)^{2} + (y+a)^{2} + (z-c)^{2} = 10^{2}$  .............................(ii)

Differentiating Equation (ii), partially with respect to x, keeping y constant,

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

Dividing, whole equation by 2, and putting $\frac{dz}{dx}$ = p;

(x-a) + (z-c)(p) = 0                            .............................(iii)

(z-c)(p) = a - x

(z-c) = $\frac{a-x}{p}$                                       ..............................(iv)

Differentiating Equation (ii), partially with respect to y, keeping x constant,

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

Dividing, whole equation by 2, and putting $\frac{dz}{dy}$ = q;

(y+a) + (z-c)(q) = 0

From equation iv -

(y+a) +( $\frac{a-x}{p}$)q = 0                   ...........................(v)

$\frac{p(a+y)+(a-x)q}{p} = 0$

p(a+y) + (a - x)q = 0

pa + py + aq - xq = 0

a ( p + q) = xq - yp

⇒a = $\frac{xq - yp}{p + q}$

Putting Value of a in equation (iv)-

⇒z - c = $\frac{\frac{xq-yp}{p+q} -x}{p}$

⇒z - c = $\frac{xq - yp - xp - xq}{(p+q)p}$

⇒z - c = $\frac{-px-py}{p(p+q)}$

z - c = $\frac{-p(x+y)}{p(p+q)}$

z - c = $\frac{-(x+y)}{p+q}$              ..................(vi)

From equation (iii);

(x - a) = -p(z-c)

$(x - a )^{2}$ = $( -p(z-c))^{2}$         .......(1)

From equation (v)

(y + a) = - q (z-c)                      

$(y + a)^{2}$ = $(- q (z-c))^{2}$        ..........(2)

Using (1) and (2) in equation (ii);

$p^{2} (z-c)^{2}$ + $q^{2} (z-c)^{2}$ + $(z-c)^{2}$ = 100

$(z-c)^{2}$ $(p^{2} +q^{2} +1)$ = 100

Putting value of z - c from equation (vi)-

$(\frac{(x+y)}{p+q})^{2}$  $(p^{2} +q^{2} +1)$ = 100 which is the required partial equation, where

p = $\frac{dz}{dx}$ ; q = $\frac{dz}{dy}$

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