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Suppose you need to know an equation of the tangent plane to a surface S at the point P(2, 1, 3)....

Question:

Suppose you need to know an equation of the tangent plane to a surface S at the point P(2, 1, 3). You don't have an equation for S but you know that the curves {eq}r_1(t)= \left \langle 2 + 3t, 1 - t^2, 3 - 4t + t^2 \right \rangle,\ r_2(t)= \left \langle 1 + u^2, 2u^3 - 1, 2u + 1 \right \rangle {/eq} both lie on S. Find the equation of the tangent plane at P.

The Tangent Plane to a Surface:

Suppose that {eq}S {/eq} is a smooth surface in {eq}\mathbb{R}^3 {/eq}. If {eq}\vec{v} {/eq} is a vector such that {eq}\vec{v}=r'(t_0) {/eq} for some curve {eq}r {/eq} which is contained entirely in the surface {eq}S {/eq}, then we say that {eq}\vec{v} {/eq} is a tangent vector to {eq}S {/eq} at the point {eq}r(t_0) {/eq}. The tangent vectors to {eq}S {/eq} at a point {eq}p {/eq} form a plane, which is called the tangent plane to {eq}S {/eq} at {eq}p {/eq}.

Answer and Explanation:

Note that {eq}r_1(t)=(2+3t,1-t^2,3-4t+t^2) {/eq} passes through the point {eq}P=(2,1,3) {/eq} at {eq}t=0 {/eq}, while {eq}r_2(u)=(1+u^2,2u^3-1,2u+1) {/eq} passes through the point {eq}P(2,1,3) {/eq} at {eq}u=1 {/eq}.

Differentiating {eq}r_1 {/eq} gives:

{eq}\begin{align*} r_1'(t)&=\left<\frac{d}{dt}(2+3t),\frac{d}{dt}(1-t^2),\frac{d}{dt}(3-4t+t^2)\right>\\ &=\left<3,-2t,-4+2t\right> \end{align*} {/eq}

and so {eq}r_1'(0)=\left<3,0,-4\right> {/eq}. Since {eq}r_1(t) {/eq} lies on the surface {eq}S {/eq} for all {eq}t {/eq}, we know that {eq}r_1'(t) {/eq} is tangent to {eq}S {/eq} at the point {eq}r_1(t) {/eq} for all {eq}S {/eq}. Since {eq}r_1(0)=P {/eq}, it follows that {eq}r_1'(t) {/eq} is tangent to {eq}S {/eq} at {eq}P {/eq}.

Similarly, differentiating {eq}r_2 {/eq} gives:

{eq}\begin{align*} r_2'(u)&=\left<\frac{d}{du}(1+u^2),\frac{d}{du}(2u^3-1),\frac{d}{du}(2u+1)\right>\\ &=\left<2u,6u^2,2\right> \end{align*} {/eq}

and so

{eq}\begin{align*} r_2'(1)&=\left<2(1),6(1^2),2\right>\\ &=\left<2,6,2\right>\, . \end{align*} {/eq}

Since {eq}r_2(u) {/eq} lies on the surface {eq}S {/eq} for all {eq}t {/eq}, and {eq}r_2(1)=P {/eq}, it follows that {eq}r_2'(1) {/eq} is tangent to {eq}S {/eq} at {eq}P {/eq}.

So the vectors {eq}v_1=\left<3,0,-4\right> {/eq} and {eq}v_2=\left<2,6,2\right> {/eq} are both tangent to the surface {eq}S {/eq} at the point {eq}P {/eq}. So their cross product will be normal to {eq}S {/eq} at {eq}P {/eq}. We have:

{eq}\begin{align*} v_1 \times v_2 &=\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 3 & 0 & -4\\ 2 & 6 & 2 \end{vmatrix}&&\text{(by the determinant mnemonic for the cross product)}\\ &=(0(2)-(-4)(6))\mathbf{i}+((-4)(2)-3(2))\mathbf{j}+(3(6)-0(2))\mathbf{k}&&\text{(expanding the determinant)}\\ &=24\mathbf{i}-14\mathbf{j}+18\mathbf{k} \, . \end{align*} {/eq}

So the vector {eq}24\mathbf{i}-14\mathbf{j}+18\mathbf{k} {/eq} is normal to {eq}S {/eq} at the point {eq}P {/eq}. That is, the tangent plane to {eq}S {/eq} at {eq}P {/eq} is the plane which is normal to the vector {eq}24\mathbf{i}-14\mathbf{j}+18\mathbf{k} {/eq} and passes through the point {eq}P=(2,1,3) {/eq}. So the equation of this plane is:

{eq}\begin{align*} (24\mathbf{i}-14\mathbf{j}+18\mathbf{k}) \cdot \left[(x-2)\mathbf{i}+(y-1)\mathbf{j}+(z-3)\mathbf{k}\right]&=0\\ 24(x-2)-14(y-1)+18(z-3)&=0&&\text{(evaluating the dot product)}\\ 24x-48-14y+14+18z-54&=0\\ 24x-14y+18z-88&=0\\ 24x-14y+18z&=88 \, . \end{align*} {/eq}

That is, the desired tangent plane has the equation {eq}\boxed{24x-14y+18z=88}\, {/eq}.


Learn more about this topic:

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Tangent Plane to the Surface

from GRE Math: Study Guide & Test Prep

Chapter 14 / Lesson 3
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