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.