Question

Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−2, uv+1). These two surfaces intersect in a curve C which passes through the point (x, y, z) = (1, 1, 1). Find the tangent line to C through (1, 1, 1).

Answer 1

The tangent to
`C` through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces
`S` and
`T` at that point.

Let
`f(x,y,z)=x^2+2y^3+3z^4`. Then
`S` is the level curve
`f(x,y,z)=6`. Recall that the gradient vector is perpendicular to level curves; we have

`\\abla f(x,y,z)=(2x,6y,12z^2)`

so that the gradient of
`f` at (1, 1, 1) is

`\\abla f(1,1,1)=(2,6,12)`

For the surface
`T`, we have

`\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0`

so that
`\vec r(1,0)=(1,1,1)`. We can obtain a vector normal to
`T` by taking the cross product of the partial derivatives of
`\vec r(u,v)`, and evaluating that product for
`u=1,v=0`:

`(\partial\vec r)/(\partial u)*(\partial\vec r)/(\partial v)=\left(u-2ve^v,-1,\frac{2e^v}u\right)`

`\left((\partial\vec r)/(\partial u)*(\partial\vec r)/(\partial v)\right)(1,0)=(1,-1,2)`

Now take the cross product of the two normal vectors to
`S` and
`T`:

`(2,6,12)*(1,-1,2)=(24,8,-8)`

The direction of vector (24, 8, -8) is the direction of the tangent line to
`C` at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by
`t\in\Bbb R`. Then adding (1, 1, 1) shifts this line to the point of tangency on
`C`. So the tangent line has equation

`\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)`