3dmultivariable-calculus
Let $C$ be the curve of intersection of the ellipsoid $x^2+2y^2+3z^2=39$ and the plane $3x+y-7z=0$. Find the parametric equations for the tangent line to $C$ at $(5,-1,2)$.
I don't know how to find the curve $C$.. I know the normal vector of the plane is $(3,1,-7$), but what can I do with the ellipsoid?
The tangent line at $(x_0,y_0)$ has equation $$ (x-x_0)\frac{\partial f}{\partial x}+ (y-y_0)\frac{\partial f}{\partial y}=0 $$ where the partial derivatives are computed at $(x_0,y_0)$ and $f(x,y)=x^2y-y^2+x-11$.
Since \begin{align} \frac{\partial f}{\partial x}&=2xy+1 \\[6px] \frac{\partial f}{\partial y}&=x^2-2y \end{align} the tangent line has equation $$ 7(x-3)+7(y-1)=0 $$ that is, $x+y-4=0$.
You have made some error in computing the derivative of $x(t)$.
Solving the equation $x^2t+x-t^2-11=0$ with respect to $x$ yields $$ x=\frac{-1\pm\sqrt{1+44t+4t^3}}{2t} $$ If $t=1$, we get $$ x=\frac{-1\pm7}{2} $$ so we need to take the branch with $+$.
The derivative of $x(t)$ is $$ x'(t)=\frac{1}{2}\cdot\frac{\dfrac{22+6t^2}{\sqrt{1+44t+4t^3}}-(\sqrt{1+44t+4t^3}-1)}{t^2} $$ For $t=1$ we get $$ x'(1)=\frac{1}{2}\frac{\dfrac{28}{7}-(7-1)}{1}=-1 $$
Yes. Your intuition is correct. For a parametrically specified surface, find $\dfrac{\partial \vec{r}}{\partial u} $ and $\dfrac{\partial \vec{r}}{\partial v} $, then find their cross product,
$ \vec{n} = \dfrac{\partial \vec{r}}{\partial u} \times \dfrac{\partial \vec{r}}{\partial v} $
then $\vec{n}$ is a normal vector to the surface, at $\vec{a}$, which you can use to find the equation of the tangent plane, as follows:
$\vec{n} \cdot (\vec{r} - \vec{a}) = 0 $
Best Answer
The tangent plane to the ellipsoid $r(y,z)=(\sqrt{39-2y^2-3z^2},y,z)$ has the basis $$u=r_y=(\frac{-2y}{\sqrt{39-2y^2-3z^2}},1,0)=(2/5,1,0) \\v=r_z=(\frac{-3z}{\sqrt{39-2y^2-3z^2}},0,1)=(-6/5,0,1)$$ Any tangent vector to the ellipsoid is a linear combination of $u,v$. We need one of these vectors perpendicular to $(3,1,-7)$, so we solve for $w=au+bv$ where $$w\cdot (3,1,-7)=0 \\ ((2a-6b)/5,a,b)\cdot(3,1,-7)=0 \\(6a-18b)/5+a-7b=0 \\11a-53b=0$$ Now we can take any value $b=1\implies a=53/11$ and so $w=(53/11)u+v$(evaluate it). Finally, the equation of the tangent line is $$\alpha(t)=(5,-1,2)+tw$$