Another Differentiation Problem

**puggie** · June 2nd 2009, 11:22 PM

Find dy/dx for the following curve:

x = ( 1 - a sq ) to the power of 3/2
y = sin to the power of -1 ( a )

**pickslides** · June 2nd 2009, 11:32 PM

can you confirm the equations

Looks like,

$x= (1-\sqrt{a})^{\frac{3}{2}}$
$y= sin^{-1}(a)$

**craig** · June 2nd 2009, 11:33 PM

Originally Posted by puggie

Find dy/dx for the following curve:

x = ( 1 - a sq ) to the power of 3/2
y = sin to the power of -1 ( a )

As I said in your last post, this requires Parametric Differentiation.

Remember, $\frac{dy}{dA} \times \frac{dA}{dx} = \frac{dy}{dx}$

Is it differentiating the equations themselves you are struggling with?

**puggie** · June 3rd 2009, 06:29 PM

Hi pickslides

y as you have written is correct.

x should be a square and not square root of a!

The answer is -1/3a(1-a square)

Thanks everybody

**alexmahone** · June 3rd 2009, 08:22 PM

$x=(1-a^2)^{\frac{3}{2}}$
$y=sin^{-1}(a)$

Substitute $a=sin \theta$

$y=\theta$

$x=(1-sin^2\theta)^{\frac{3}{2}}$

$x=cos^3\theta$

$\frac{dx}{d\theta}=3cos^2\theta*-sin\theta$
$\frac{dx}{dy}=3cos^2\theta*-sin\theta$
$\frac{dx}{dy}=3(1-a^2)*-a$
$\frac{dx}{dy}=3a(a^2-1)$
$\frac{dy}{dx}=\frac{1}{3a(a^2-1)}$

**pickslides** · June 3rd 2009, 08:27 PM

$x= (1-a^2)^{\frac{3}{2}}$

from this find $\frac{dx}{da}$

$y= sin^{-1}(a)$

from this find $\frac{dy}{da}$

$then \frac{dy}{dx} = \frac{dy}{da} \times \frac{1}{\frac{dx}{da}}<br />$

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