Finding the Orthogonal Trajectory of x^p + Cy^p = 1

mbaron
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I am working on this problem, and have a simple question.

Determine the orthogonal trajectory of
x^p + Cy^p = 1
where p = constant.

I start out by taking the derivative with respect to x. My question is this. does
Cy^p become Cpy^{p-1} or C_1y^{p-1} ?

Thanks,
Morgan
 
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C1?? There isn't any "C1" in your original formula!

The derivative of yp with respect to y is pyp-1. The derivative of Cyp with respect to y is Cpyp-1. By the chail law, the derivative of Cyp is Cpy^{p-1}\frac{dy}{dx}. Solve the resulting equation for \frac{dy}{dx} to find the slope of the tangent line to the original trajectory at each point.
 
If p is a constant and C is a constant isn't
C_1
just another constant? Isn't
C_1y^{p-1}\frac{dy} {dx}
the same as what you have?

Thanks for pointing out the chain rule, I missed that.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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