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

[mbaron]

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

 

[HallsofIvy]

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

The derivative of y^p with respect to y is py^p-1. The derivative of Cy^p with respect to y is Cpy^p-1. By the chail law, the derivative of Cy^p 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.

 

[mbaron]

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.