# Find an equation of the tangent plane to the given parametric surface at the specified point. x =...

## Question:

Find an equation of the tangent plane to the given parametric surface at the specified point.

## Tangent plane to a surface

This example illustrates how to find the tangent plane to a surface at a given point.

In this particular case, the equation of the surface is given in terms of two parameters, and we must use partial differentiation and cross products.

Suppose we have a surface with parametric equations {eq}\begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} x(u, v)\\ y(u, v)\\ z(u, v) \end{pmatrix}. {/eq}

Then the vector which is normal to the surface at any particular point is given by {eq}\begin{pmatrix} \dfrac{\partial x}{\partial u}\\ \dfrac{\partial y}{\partial u}\\ \dfrac{\partial z}{\partial u} \end{pmatrix} \times \begin{pmatrix} \dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial v}\\ \dfrac{\partial z}{\partial v} \end{pmatrix}. {/eq}

In this example, {eq}\begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} u^2 + 1\\ v^3 + 1\\ u + v \end{pmatrix} {/eq}

and so the normal vector to the surface is

{eq}\begin{pmatrix} \dfrac{\partial x}{\partial u}\\ \dfrac{\partial y}{\partial u}\\ \dfrac{\partial z}{\partial u} \end{pmatrix} \times \begin{pmatrix} \dfrac{\partial x}{\partial v}\\ \dfrac{\partial y}{\partial v}\\ \dfrac{\partial z}{\partial v} \end{pmatrix} = \begin{pmatrix} 2u\\ 0\\ 1 \end{pmatrix} \times \begin{pmatrix} 0\\ 3v^2\\ 1 \end{pmatrix} (*) {/eq}

We need to find the values of u and v at the point (5, 2, 3), so:

{eq}u^2 + 1 = 5 .....(1) \\ v^3 + 1 = 2 .....(2) \\ u + v = 3 .....(3) {/eq}

From (1), {eq}u = \pm 2. {/eq}

From (2), {eq}v = 1. {/eq}

Checking in (3), we see that only the positive value of u is valid, so we have {eq}u = 2, \: v = 1. {/eq}

Substituting these into (*), the normal vector is

{eq}\begin{pmatrix} 4\\ 0\\ 1 \end{pmatrix} \times \begin{pmatrix} 0\\ 3\\ 1 \end{pmatrix} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 4 & 0 & 1\\ 0 & 3 & 1 \end{vmatrix} \\= \mathbf{i}\begin{vmatrix} 0 & 1\\ 3 & 1 \end{vmatrix} - \mathbf{j}\begin{vmatrix} 4 & 1\\ 0 & 1 \end{vmatrix} + \mathbf{k}\begin{vmatrix} 4 & 0\\ 0 & 3 \end{vmatrix} \\ \displaystyle = \mathbf{i}((0\times 1) - (1 \times 3)) - \mathbf{j}((4\times 1) - (1 \times 0)) + \mathbf{k}((4\times 3) - (0 \times 0)) = \begin{pmatrix} -3\\ -4\\ 12 \end{pmatrix} {/eq}

This is the normal vector to the required tangent plane.

Now, a plane with Cartesian equation {eq}n_{1}x + n_{2}y + n_{3}z = c {/eq} has the normal vector {eq}\displaystyle \begin{pmatrix} n_{1}\\ n_{2}\\ n_{3} \end{pmatrix} {/eq}.

So in this case, the equation of the tangent plane is {eq}-3x - 4y + 12z = c {/eq}.

Substitute (5,2,3) to find the value of c: {eq}-3(5) - 4(2) + 12(3) = c \Rightarrow c = 13 {/eq}.

Hence the tangent plane has equation {eq}-3x - 4y + 12z = 13 {/eq}.