Partial differentiation of cos (in vector calculus)

[Eight]

Homework Statement

So using standard spherical polar co-ordinates, my notes define a sphere as

r(s,t) = aCos(s)Sin(t) i + aSin(s)Sin(t) j + aCos(t) k

and the normal to the surface is given by the cross product of the two partial differentials:

\partialr/\partials X \partialr/dt

So my issue is what is in my notes where:

\partialr/\partial s= -aSin(s)Sin(t) i + aCos(s)Sin(t) j + aCot(t) k

It is the final part that I don't understand. Why is the partial differential of Cos(t) with respect to s Cot(t)? I would have calculated it as zero.

Homework Equations

\partial/\partials (aCos(t)) = aCot(t) ?

The Attempt at a Solution

All I can think is that it's using a trig identity? As I understand it the k component should only depend upon t, not just from the above equations but in spherical polar co-ordinates in general, at least how I'm picturing it.



Thanks in advance for any help.

 

[Fredrik]

I agree with you. That looks completely wrong. It should be 0. And it's such a weird mistake that I kept staring at it for some time, wondering if I had misunderstood something. I guess that's what you're doing too.

 

[Eight]

I keep thinking it must be a mistake too, but if it is a mistake then it is deliberate, as it has been repeated in a number of places. And during lectures where we go through the notes it was never picked up on.