मराठी

If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that [1+(dydx)2]32d2ydx2 is a constant independent of a and b. - Mathematics

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प्रश्न

If (x – a)2 + (y – b)2 = c2, for some c > 0, prove that `[1+ (dy/dx)^2]^(3/2)/((d^2y)/dx^2)` is a constant independent of a and b.

बेरीज

उत्तर

Here,  (x – a)2 + (y – b)2 = (Given)                    …(1)

On differentiating with respect to x,

`=> 2 (x - a) + 2(y - b)^2  dy/dx = 0`

`=> (x - a) + (y - b)  dy/dx = 0`                 ...(2)

Differentiating again with respect to x,

`1 + dy/dx * dy/dx + (y - b) (d^2 y)/dx^2` = 0

`1 + (dy/dx)^2 + (y - b) (d^2y)/dx^2` = 0

`=> (y - b) = - {(1 + (dy/dx)^2)/((d^2y)/dx^2)}`            ...(3)

Putting the value of (y – b) in (2),

`(x - a) = (1 + (dy/dx)^2)/((d^2 y)/dx^2) * dy/dx`

या `(x - a) = {(1 + (dy/dx)^2)/((d^2y)/dx^2)}(dy/dx)`            ...(4)

Putting the values ​​of (x - a) and (y - b) from (3) and (4) in (1),

`{1 + (dy/dx)^2}^2/((d^2y)/dx^2)^2 * (dy/dx)^2 + {(1 + (dy/dx)^2)/((d^2y)/dx^2)} = c^2`

On multiplying by `((d^2y)/dx^2)^2,`

`[1 + (dy/dx)^2]^2 (dy/dx)^2 + [1 + (dy/dx)^2]^2`

`= c^2 ((d^2y)/dx)^2`

`=> [1 + (dy/dx)^2]^2 [(dy/dx)^2 + 1] = c^2 ((d^2y)/dx^2)^2`

`=> {1 + (dy/dx)^2}^3 = c^2 ((d^2y)/dx^2)^2`

On taking the square root,

`therefore {1 + (dy/dx)^2}^(3//2)/((d^2y)/dx^2)` = c      ...(a constant independent of a and b.)

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पाठ 5: Continuity and Differentiability - Exercise 5.9 [पृष्ठ १९१]

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एनसीईआरटी Mathematics [English] Class 12
पाठ 5 Continuity and Differentiability
Exercise 5.9 | Q 15 | पृष्ठ १९१

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