Question

If x=√asin⁻¹t, y=√acos⁻¹t,prove dy/dx=-y/x | t | < 1,Find dy/dx :(Wherever y is defined as a function of x and dx/dt or dx/dθ≠0)

Answer 1

$\bf{x=\sqrt{asin^{-1}t}}$

differentiate with respect to t,

$\frac{dx}{dt}=\frac{d\{\sqrt{asin^{-1}t}\}}{dt}$

=$\frac{1}{2\sqrt{asin^{-1}t}}\frac{a}{\sqrt{1-x^2}}$

similarly, $\bf{y=\sqrt{acos^{-1}t}}$

differentiate with respect to t,

$\frac{dy}{dt}=-\frac{1}{2\sqrt{acos^{-1}t}}\frac{a}{\sqrt{1-x^2}}$

now, dy/dx = {dy/dt}/{dx/dt}

= $\frac{-\frac{1}{2\sqrt{acos^{-1}t}}\frac{a}{\sqrt{1-x^2}}}{\frac{1}{2\sqrt{asin^{-1}t}}\frac{a}{\sqrt{1-x^2}}}$

= $-\frac{\sqrt{asin^{-1}t}}{\sqrt{acos^{-1}t}}$

here, x = √(asin^-1t) and y = √(acos^-1t

so, $\frac{dy}{dx}=-\frac{x}{y}$

I think little mistake in question. I checked many times , but I got dy/dx = -x/y

Answer 2

In the attachment I have answered this problem.

The solution is simple and easy to understand.

See the attachment for detailed solution.