What is the equation of the normal line to the surface r(u,v)=(cosu, sinv+sinu ,cosv) at the point (1,1,0)?
What I have done so far:
I have found the partial derivatives of the surface separately with respect to u and respect to v. I know I need to multiply the two to get the i,j,k coordinates. Should I sub in the point given before doing this or after? should I change to polar coordinates? how do I sub these values in to cos and sin so that it is defined?
Some quick corrections and answers:
There is no operation that is referred to as "multiplying vectors". Presumably you mean that you compute the cross-product $\mathbf r_u \times \mathbf r_v$.
It is easiest to substitute in the correct values for $u,v$ before taking the cross-product
I have no clue what you mean by either of these questions.
Here's a complete answer. We are given $\mathbf r(u,v) = (\cos u, \sin v + \sin u, \cos v)$. Presumably, we are also given a domain, i.e. the values of $u$ and $v$ that we consider. I suspect that this was given to you as $u,v \in [0,2\pi)$.
Begin by computing the partial derivatives $$ \mathbf r_u(u,v) = (-\sin u, \cos u, 0), \quad \mathbf r_v(u,v) = (0,\cos v, -\sin v). $$ Now, we want to find these vectors at the point $(1,1,0)$ by plugging in the values of $u$ and $v$ associated with this point. This means that we need to find these values. That is, we need to find values of $u,v$ between $0$ and $2\pi$ for which $$ (\cos u, \sin v + \sin u, \cos v) = (0,1,1). $$ Verify that this only occurs when $u = \pi/2$ and $v = 0$. We compute $$ \mathbf r_u(\pi/2,0) = (-1,0,0), \quad \mathbf r_v(\pi/2,0) = (0,0,1). $$ To find a normal vector to the surface, we take the cross product of these vectors to get $$ (-1,0,0) \times (0,0,1) = (0,1,0). $$ To answer the question, give an equation for the line through $(0,1,1)$ in the direction of $(0,1,0)$. For instance, we can write $$ \ell(t) = (0,1,1) + (0,t,0). $$