Question

The curve y2 = ux³ + v passes through a point P(2,3) and dy/dx = 4 at P. The values of u and v are​

Answer 1

Answer:

u = 2, v = - 7

Step-by-step explanation:

at P(2, 3), 3^2 = u * (2)^3 + v, or, 9 = 8u + v (1)

Now, differentiating given equation, 2y*(dy/dx) = 3*x^2*u

or, u = (2*3*4)/(3*2^2) = 2

From (1) we get, 9 = 8*2 + v

or, v = -7

Answer 2

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm :\longmapsto\: {y}^{2} = u {x}^{3} + v - - - (1)$

Since, it is given that curve (1) passes through P(2, 3).

$\rm :\implies\: {3}^{2} = u {(2)}^{3} + v$

$\rm :\longmapsto\:9 = 8u + v - - - (2)$

Now,

$\rm :\longmapsto\: {y}^{2} = u {x}^{3} + v$

Differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} {y}^{2} = \dfrac{d}{dx}(u {x}^{3} + v)$

$\rm :\longmapsto\:2y \dfrac{dy}{dx} =u \dfrac{d}{dx}{x}^{3} + \dfrac{d}{dx}v$

$\rm :\longmapsto\:2y\dfrac{dy}{dx} = 3u {x}^{2} + 0$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{3u {x}^{2} }{2y}$

Now it is given that,

$\rm :\longmapsto\:\dfrac{dy}{dx}_{(P)} = 4$

$\rm :\longmapsto\:\dfrac{dy}{dx}_{(2,3)} = 4$

$\rm :\longmapsto\:\dfrac{3u {(2)}^{2} }{2 \times 3} = 4$

$\rm :\implies\:\boxed{\bf \:u = 2}$

On substituting the value of u in equation (1), we get

$\rm :\longmapsto\:9 = 8 \times 2 + v$

$\rm :\longmapsto\:9 = 16 + v$

$\rm :\longmapsto\:9 - 16 = v$

$\bf\implies \: \: \boxed{\bf \: v = - \: 7}$

Additional Information :-

$\boxed{\bf \: \dfrac{d}{dx}k = 0}$

$\boxed{\bf \: \dfrac{d}{dx}x = 1}$

$\boxed{\bf \: \dfrac{d}{dx} {x}^{2} = 2x}$

$\boxed{\bf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }$

$\boxed{\bf \: \dfrac{d}{dx} \sqrt{x} = \dfrac{1}{2 \sqrt{x}}}$

$\boxed{\bf \: \dfrac{d}{dx}kf(x) = k\dfrac{d}{dx}f(x)}$

$\boxed{\bf \: \dfrac{d}{dx}u.v = u\dfrac{d}{dx}v + v\dfrac{d}{dx}u}$