# Suppose we have the parametric surface given by x=u+w y=u^{2} z=u-w for -10 \leq u \leq10 and -10...

## Question:

Suppose we have the parametric surface given by

{eq}\displaystyle x = u + w {/eq}

{eq}\displaystyle y = u^{2} {/eq}

{eq}\displaystyle z = u - w {/eq} for {eq}\displaystyle -10 \leq u \leq 10 {/eq} and {eq}\displaystyle -10 \leq w \leq 10 {/eq}

Find an equation of the tangent plane to this surface at the point {eq}\displaystyle (-1, 1, 3). {/eq}

## Equation of Tangent Plane:

The parametric equations {eq}\displaystyle r(u, v) {/eq} are given at the specified point. By using that, we have to find the equation of the tangent plane to the surface. Required equation is {eq}\ \displaystyle \left \langle \vec{n} \right \rangle \cdot \left \langle x-x_{0}, y-y_{0}, z-z_{0} \right \rangle =0 {/eq}. The normal vector {eq}\displaystyle \vec{n} {/eq} has to be evaluated, using the cross product.

## Answer and Explanation:

Let us use the given parametric equations {eq}\displaystyle x = u + w {/eq}, {eq}\displaystyle y = u^{2} {/eq} and {eq}\displaystyle z = u - w {/eq}.

Finding an equation of the tangent plane to this surface at the point {eq}\displaystyle (-1, 1, 3) {/eq}:

{eq}\begin{align*} \displaystyle r(u, w) &=\left \langle u+w, u^{2}, u-w \right \rangle \\ \displaystyle r_{u}(u, w) &=\left \langle 1, 2u, 1 \right \rangle \\ \displaystyle r_{w}(u, w) &=\left \langle 1, 0, -1 \right \rangle \\ \displaystyle r_{u} \times r_{w} &=\begin{vmatrix} i & j & k\\ 1 & 2u & 1\\ 1 & 0 & -1 \end{vmatrix} \\ \displaystyle r_{u} \times r_{w} &=((-1)(2u)-(0)(1))\vec{i}-((-1)(1)-(1)(1))\vec{j}+((0)(1)-(1)(2u))\vec{k} \\ \displaystyle r_{u} \times r_{w} &= -2u \vec{i} +2 \vec{j}-2u \vec{k} \\ \displaystyle \text{Values of u and w}: \\ \displaystyle x &= u + w \Rightarrow -1= u + w &\left[ -----\text{eq(1)} \right] \\ \displaystyle y &= u^{2} \Rightarrow 1= u^{2} \Rightarrow u=1 & \left[ -----\text{eq(2)} \right] \\ \displaystyle z &= u - w \Rightarrow 3= u - w & \left[ -----\text{eq(3)} \right] \\ \displaystyle \text{Substituting value of u in eq(1)}: \\ \displaystyle -1 &= 1+w \\ \displaystyle w &=-1-1 \\ \displaystyle w &=-2 \\ \displaystyle \text{Values are} \ \ u=1 & \ \ \text{and} \ \ w=-2 \\ \\ \displaystyle r_{u} \times r_{w} &= -2u \vec{i} +2 \vec{j}-2u \vec{k} \\ \displaystyle r_{u} \times r_{w} &= -2(1) \vec{i} +2 \vec{j}-2(1) \vec{k} \\ \displaystyle r_{u} \times r_{w} &= -2 \vec{i} +2 \vec{j}-2 \vec{k} \\ \displaystyle \left \langle \vec{n} \right \rangle \cdot \left \langle x-x_{0}, y-y_{0}, z-z_{0} \right \rangle &=0 \\ \displaystyle \left \langle -2, 2, -2 \right \rangle \cdot \left \langle x-(-1), y-1, z-3 \right \rangle &=0 \\ \displaystyle (-2)(x+1)+(2)(y-1)+(-2)(z-3) &=0 \\ \displaystyle -2x-2+2y-2-2z+6 &=0 \\ \displaystyle -2x+2y-2z-2-2+6 &=0 \\ \displaystyle -2x+2y-2z+2 &=0 \\ \displaystyle \left( \div 2 \right) -x+y-z+1 &=0 \end{align*} {/eq}

Equation of the tangent plane is {eq}\ \displaystyle \mathbf{\color{blue}{ -x+y-z+1 =0 }} {/eq}.