Find the value K

**zorro** · May 3rd 2009, 10:53 PM

If $ siny = xsin(a+y) $ and $ \frac{dy}{dx} = K \frac{sin^2y}{x^2} $ , find K

**Danny** · May 4th 2009, 05:11 AM

Originally Posted by zorro

If $ siny = xsin(a+y) $ and $ \frac{dy}{dx} = K \frac{sin^2y}{x^2} $ , find K

I'm assuming that a is constant. If we expand your sine relationship and divide by x then $\frac{\sin y}{ x} = \sin a \cos y + \cos a \sin y$ then divide by $\sin y$ so

$\sin a \cot y + \cos a = \frac{1}{x}$ and upon differentiation gives $- \sin a \csc^2 y y' = - \frac{1}{x^2}$ which solving for y ' gives

$y' = \frac{1}{\sin a} \frac{\sin^2 y}{x^2}$

in which we can identify K.

**zorro** · December 20th 2009, 02:33 PM

Originally Posted by Danny

I'm assuming that a is constant. If we expand your sine relationship and divide by x then $\frac{\sin y}{ x} = \sin a \cos y + \cos a \sin y$ then divide by $\sin y$ so

$\sin a \cot y + \cos a = \frac{1}{x}$ and upon differentiation gives $- \sin a \csc^2 y y' = - \frac{1}{x^2}$ which solving for y ' gives

$y' = \frac{1}{\sin a} \frac{\sin^2 y}{x^2}$

in which we can identify K.

thanks mite
cheers

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