Question

Find the tangent plane to the surface with parametric equations x = u2, y = v2, z = u + 9v at the point (1, 1, 10).

Answer 1

Let $\vec r(u,v)=u^2\,\vec\imath+v^2\,\vec\jmath+(u+9v)\,\vec k$ be the parameterization for the surface.

The normal vector to the plane is

$\vec n=\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}$

$\begin{cases}\frac{\partial\vec r}{\partial u}=2u\,\vec\imath+\vec k\\\frac{\partial\vec r}{\partial v}=2v\,\vec\jmath+9\,\vec k\end{cases}\implies\vec n=-2v\,\vec\imath-18u\,\vec\jmath+4uv\,\vec k$

The point (1, 1, 10) occurs for

$\begin{cases}u^2=1\\v^2=1\\u+9v=10\end{cases}\implies(u,v)=(1,1)$

Then the equation for the plane tangent to the surface at (1, 1, 10) is

$((x\,\vec\imath+y\,\vec\jmath+z\,\vec k)-\vec r(1,1))\cdot\vec n=0$

$((x-1)\,\vec\imath+(y-1)\,\vec\jmath+(z-10)\,\vec k)\cdot(-2\,\vec\imath-18\,\vec\jmath+4\,\vec k)=0$

$-2(x-1)-18(y-1)+4(z-10)=0$

$-2x-18y+4z=20$

$\boxed{x+9y-2z=-10}$