# 1. Find an equation of the tangent plane to the parametric surface \mathbf{r}(s, t) = 3 s \cos...

## Question:

1. Find an equation of the tangent plane to the parametric surface

{eq}\mathbf{r}(s, t) = 3 s \cos (t) \mathbf{i} + 3 s \sin (t) \mathbf{j} + 5 t \mathbf{k} {/eq}

at when s = 2 and {eq}t = \frac{3 \pi}{4} {/eq}.

2. Find an equation of the tangent plane to the parametric surface

{eq}x = 2u + 5v, \ y = 3u^2, \ z = 2u - 2v {/eq}

at the point (14, 12, 0).

## Tangent Planes of Parametric Surfaces

To find the tangent plane to a surface given parametrically as a vector-function of two parameters, {eq}\displaystyle \mathbf{r}(u,v), (u,v)\in\mathcal{D} {/eq}

we need to find the normal vector {eq}\displaystyle \mathbf{n}(u,v) {/eq} as the cross vector of the following partial derivatives, {eq}\displaystyle \mathbf{r}_u\times \mathbf{r}_v. {/eq}

Above, the cross product is defined as

{eq}\displaystyle \langle a_1,b_1,c_1 \rangle \times \langle a_2,b_2,c_2\rangle= \begin{array}{|ccc|} \mathbf{i} &\mathbf{j}&\mathbf{k}\\ a_1&b_1&c_1\\ a_2&b_2&c_2 \end{array}. {/eq}

1. To find the tangent plane to the surface given parametrically as {eq}\displaystyle \mathbf{r}(s,t)=\langle 3s\cos t, 3s\sin t, 5t \rangle, {/eq} at the point {eq}\displaystyle \left(s,t\right)= \left( 2, \frac{3\pi}{4}\right), {/eq} we will compute the normal vector to the plane.

{eq}\displaystyle \begin{align} \mathbf{r}_s \times \mathbf{r}_t&= \langle 3\cos t, 3\sin t, 0 \rangle \times \langle -3s\sin t, 3s\cos t, 5 \rangle\\ \\ &= \begin{array}{|ccc|} \mathbf{i} &\mathbf{j}&\mathbf{k}\\ 3\cos t & 3\sin t & 0\\ -3s\sin t & 3s\cos t & 5 \end{array} = \langle 15 \sin t, -15\cos t, 9s\cos^2 t+9s\sin^2 t\rangle = 3\langle 5 \sin t, -5\cos t, 3s\rangle, \left[\text{ using the identity } \cos^2 t+\sin^2 t=1\right]\\ &\text{ at the given point, the normal vector is }\mathbf{n} \left( 2, \frac{3\pi}{4}\right)=\left \langle \frac{5\sqrt{2}}{2}, \frac{5\sqrt{2}}{2}, 6\right\rangle. \end{align} {/eq}

To write the tangent plane equation, we need the Cartesian coordinates of the point {eq}\displaystyle \mathbf{r}\left( 2, \frac{3\pi}{4}\right): x=-3\sqrt{2}, y=3\sqrt{2}, z=\frac{15\pi}{4}, {/eq}

therefore the tangent plane is {eq}\displaystyle \boxed{\frac{5\sqrt{2}}{2} \left(x+3\sqrt{2}\right)+\frac{5\sqrt{2}}{2} \left(y-3\sqrt{2}\right)+6 \left(z-\frac{15\pi}{4}\right)=0}. {/eq}

2. To find the tangent plane to the surface given parametrically as {eq}\displaystyle \mathbf{r}(u,v)=\langle 2u+5v, 3u^2, 2u-2v \rangle, {/eq} at the point {eq}\displaystyle (14,12,0) {/eq} we will compute the normal vector of the plane.

{eq}\displaystyle \begin{align} \mathbf{r}_u \times \mathbf{r}_t&= \langle 2,6u,2 \rangle \times \langle 5,0,-2 \rangle= \begin{array}{|ccc|} \mathbf{i} &\mathbf{j}&\mathbf{k}\\ 2 & 6u & 2 \\ 5 & 0 & -2 \end{array} \\\\ &= \langle -12u, 14, -30u \rangle = 2\langle -6u, 7, -15u \rangle\\\\ &\text{ at the given point } u= v= 2: \mathbf{n} \left( 2, 2\right)=\left \langle -12, 7, -30\right\rangle. \end{align} {/eq}

Therefore the tangent plane is {eq}\displaystyle \boxed{12(x-14)-7(y-12)+30z=0}. {/eq}