# Find the tangent plane to the image of $\phi(u,v)=(u^2,u\sin e^v,\frac{1}{3}u\cos e^v)$ at $(13,-2,1)$

It's clear we need to find the normal to the tangent vectors $$T_u,T_v$$, etc. But if we have to compute $$T_u \times T_v$$ at the specific point mentioned, we have to solve for $$u,v$$ in $$(1)$$:

$$\begin{cases} 13 = u^2 \\ -2 = u \sin e^v \\ 1 = \frac{1}{3} u \cos e^v \end{cases} \tag 1$$

Is this right? If there is no way around this, then how do we solve for $$u,v$$ in $$(1)$$?

• Your last equation doesn’t match the parameterization in the title. That aside, surely you can solve the first equation for $u$ and then substitute into the other two to solve for $v$. If you compute $T_u$ and $T_v$ before doing that and see what factors out, you might find that you don’t actually have to solve for $v$ per se. – amd Sep 15 '19 at 5:46
• In this particular case, it’s also not terribly hard to eliminate $u$ and $v$ to get an implicit Cartesian equation for the surface, then find a formula for the normal at a point in terms of $x$, $y$ and $z$. – amd Sep 15 '19 at 5:57
• @amd $T_u \times T_v = (-\frac{1}{3}ue^v,\frac{2}{3}u^2 e^v \sin e^u, 2u^2 e^v \cos e^u)$. I don't see how I can get around solving for $e^u$. – Deep_Television Sep 15 '19 at 5:59
• It’s $e^v$, not $e^u$. Any nonzero scalar multiple of this vector will do for a normal for the equation of the plane. Do you see any annoying common terms that you can factor out? – amd Sep 15 '19 at 6:01

You don’t need to solve explicitly for both $$u$$ and $$v$$ (and in fact you don’t need to solve for either one). As you’ve found, $$T_u\times T_v = \left(-\frac13ue^v,\frac23u^2e^v\sin{e^v},2u^2e^v\cos{e^v}\right).$$ However, any nonzero scalar multiple of this vector will do for finding an equation of the tangent plane, so we pull out a factor of $$ue^v$$ (at the point we’re interested in $$u\ne0$$) and pull out a $$1/3$$ while we’re at it to make things more compact. So, the normal vector is now the simpler-looking $$(-1,2u\sin{e^v},6u\cos{e^v})$$. Setting $$\phi(u,v)=(13,-2,1)$$ gives you three simple equations in $$u$$, $$\sin{e^v}$$ and $$\cos{e^v}$$, which is exactly what you need to compute the scaled normal. Or, you might notice that $$(-1,2u\sin{e^v},6u\cos{e^v})=(-1,2y,18z),$$ so you don’t need to solve any equations at all!
Another approach that’s a bit less work in this particular case is to derive an implicit Cartesian equation by eliminating $$u$$ and $$v$$. We have $$y=u\sin{e^v}$$ and $$z=\frac13u\cos{e^v}$$, so $$y^2+9z^2=u^2=x$$. Differentiating gives $$\nabla\phi = (-1,2y,18z)$$, which is what we ended up with above.