Question

form a partial differentiation equation from , X2 + y2 = ( z-c )2 tan2 alpha​

Answer 1

Answer:

partial differential equation is an equation involving a function of two ormore variables and some of its partial derivatives. Therefore a partial differentialequation contains one dependent variable and one independent variable. Here z will be taken as the dependent variable and x and y the independentvariable so that z = f ( x, y ) . We will use the following standard notations to denote the partial derivatives. ∂z ∂z ∂2z ∂2z ∂2z = p, = q, 2 = r , = s, 2 = t ∂x ∂y ∂x ∂x∂y ∂y The order of partial differential equation is that of the highest order derivativeoccurring in it.Formation of partial differential equation: There are two methods to form a partial differential equation. (i) By elimination of arbitrary constants. (ii) By elimination of arbitrary functions.ProblemsFormation of partial differential equation by elimination of arbitraryconstants:(1)Form the partial differential equation by eliminating the arbitrary constantsfrom z = ax + by + a 2 + b 2 .Solution: 1

2. Given z = ax + by + a 2 + b 2 ……………... (1) Here we have two arbitrary constants a & b. Differentiating equation (1) partially with respect to x and y respectively we get ∂z =a⇒ p=a ……………… (2) ∂x ∂z =b⇒q=a ………………. (3) ∂y Substitute (2) and (3) in (1) we get z = px + qy + p 2 + q 2 , which is the required partial differential equation.(2) Form the partial differential equation by eliminating the arbitrary constants x2 y2 z 2a, b, c from + + = 1. a2 b2 c2Solution: We note that the number of constants is more than the number of independent variable. Hence the order of the resulting equation will be more than 1. x2 y2 z 2 + + =1 .................. (1) a2 b2 c2 Differentiating (1) partially with respect to x and then with respect to y, we get 2x 2 y + p=0 .....................(2) a2 c2 2 y 2z + q=0 ......................(3) b2 c2 Differentiating (2) partially with respect to x, 1 1 2 + 2 ( zr + p 2 ) ……………..(4) a c ∂2z Where r = 2 . ∂x c 2 zp − 2 = ......................(5) a x From (2) and (4) , 2 . c − 2 = zr + p 2 ......................(6) a 2

3. From (5) and (6), we get 2 ∂2z  ∂z  ∂z xz 2 + x  = z , which is the required partial differential ∂x  ∂x  ∂x equation.(3) Find the differential equation of all spheres of the same radius c having theircenter on the yoz-plane. .Solution: The equation of a sphere having its centre at ( 0, a, b ) , that lies on the yoz -plane and having its radius equal to c is x 2 + ( y − a ) 2 + ( z − b) 2 = c 2 ……………. (1) If a and b are treated as arbitrary constants, (1) represents the family of spheres having the given property. Differentiating (1) partially with respect to x and then with respect to y, we have 2 x + 2( z − b ) p = 0 …………… (2) and 2( y − a ) + 2( z − b ) q = 0 …………….(3) x From (2), z −b = − …………….(4) p qx Using (4) in (3), y − b = ……………..(5) p Using (4) and (5) in (1), we get q2x2 x2 x + 2 + 2 = c2 2 p p . i.e. (1 + p 2 + q 2 ) x 2 = c 2 p 2 , which is the required partial differentialequation.

Answer 2

Concept:

A differential condition in science is a condition that interfaces at least one obscure capabilities and their subsidiaries. In applications, works frequently address actual amounts, subsidiaries show their paces of progress, and the differential condition lays out an association between the two. Any equation with one or more terms and one or more derivatives of the dependent variable with respect to the independent variable is referred to as a differential equation.

Given:

Here the equation that is given to us is $x^2 + y^2 = ( z-c )^2 tan^2\alpha$

Find:

We have to form a partial differentiation equation form the above equation.

Solution:

According to the question,

Differentiate w.r.t$'x'$ partially

$2ax + 2z(\delta z/ \delta x) = 0$

$a = (-zp)/x......(2)$

Differentiate w.r.t $'y'$, partially

$2by + 2z(\delta z/\delta y) = 0$

$by + zq = 0........{(\delta z/\delta y) = q}$

$b = (-zq)/y$ $.........(3)$

Now put the value of$'a'$and $'b'$ from equation $(2)$ and $(3)$ in equation $(1)$ we get

$-zpx - zqy + z2 = 1\\zpx + zqy = z^2 - 1\\z(px + qy) = z^2 - 1$

Hence we have form the differential equation from the above equation.

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