Calculus III - Parametric Surfaces

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Section 17.2 : Parametric Surfaces

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6. The tangent plane to the surface given by the following parametric equation at the point $\left( {8,14,2} \right)$ .

$\vec r\left( {u,v} \right) = \left( {{u^2} + 2u} \right)\vec i + \left( {3v - 2u} \right)\vec j + \left( {6v - 10} \right)\vec k$

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In order to write down the equation of a plane we need a point, which we have, $\left( {8,14,2} \right)$ , and a normal vector, which we don’t have yet.

However, recall that ${\vec r_u} \times {\vec r_v}$ will be normal to the surface. So, let’s compute that.

${\vec r_u} = \left( {2u + 2} \right)\vec i - 2\vec j\hspace{0.5in}{\vec r_v} = 3\vec j + 6\vec k$

${\vec r_u} \times {\vec r_v} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{2u + 2}&{ - 2}&0\\0&3&6\end{array}} \right| = - 12\vec i - 6\left( {2u + 2} \right)\vec j + 3\left( {2u + 2} \right)\vec k$ Show Step 2

Now having ${\vec r_u} \times {\vec r_v}$ is all well and good but it is really only useful if we also know the point, $\left( {u,v} \right)$ for which we are at $\left( {8,14,2} \right)$ so we next need to set the $x$ , $y$ and $z$ coordinates of our point equal to the $x$ , $y$ and $z$ components of our parametric equation to determine the value of $u$ and $v$ we need.

Here are the equations we get if we do that.

$\begin{array}{*{20}{c}}{8 = {u^2} + 2u}\\{14 = 3v - 2u}\\{2 = 6v - 10}\end{array}\hspace{0.5in} \Rightarrow \hspace{0.5in}\begin{array}{l}{0 = {u^2} + 2u - 8 = \left( {u + 4} \right)\left( {u - 2} \right)}\\{14 = 3v - 2u}\\{12 = 6v}\end{array}$ Show Step 3

From the third equation above we can see that we must have $v = 2$ and from the first equation we can see that we must have either $u = - 4$ or $u = 2$ .

Plugging our only choice for $v$ and both choices for $u$ into the second equation we can see that we must have $u = - 4$ .

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Plugging $u = - 4$ and $v = 2$ into the equation for ${\vec r_u} \times {\vec r_v}$ we will arrive at the following normal vector to the surface at $\left( {8,14,2} \right)$ .

$\vec n = {\left( {{{\vec r}_u} \times {{\vec r}_v}} \right)_{u = - 4,v = 2}} = - 12\vec i + 36\vec j - 18\vec k$

Note that, in this case, the normal vector didn’t actually depend on the value of $v$ . That won’t happen in general, but as we’ve seen here that kind of thing can happen on occasion so don’t get excited about it when it does.

The equation of the tangent plane to the surface at $\left( {8,14,2} \right)$ with normal vector $\vec n = - 12\vec i + 36\vec j - 18\vec k$ is,

$- 12\left( {x - 8} \right) + 36\left( {y - 14} \right) - 18\left( {z - 2} \right) = 0\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\, - 12x + 36y - 18z = 372$ Show Step 5

To get a set of parametric equations for the tangent plane all we need to do is solve the equation for $z$ to get,

$z = - \frac{{62}}{3} - \frac{2}{3}x + 2y$

We can then plug this into the vector $\left\langle {x,y,z} \right\rangle$ to get the following set of parametric equations for the tangent plane.

$\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( {x,y} \right) = \left\langle {x,y, - \frac{{62}}{3} - \frac{2}{3}x + 2y} \right\rangle }}$

Note that there will be no restrictions on $x$ and $y$ because we wanted the full tangent plane.