ee022d…: “As for the sad ring”

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Find the equation of the tangent plane at the point on the surface $z=e^{x^2+3y-4}$ where $x=-1$ and $y=1$ .

$\answer {-2} \cdot \left (x-\answer { -1}\right ) + \answer { 3 } \cdot \left (y-\answer { 1 }\right ) + \big (-1\big ) \cdot \left (z-\answer {1}\right ) =0$ Writing this in the form $ax+by+cz=d$ gives

$\answer {-2} \cdot x + \answer {3} \cdot y -z = \answer {4}.$

The point $(-1,1)$ lies on the circle $x^2+y^2 = \answer {2}$ , so we find the curve on the surface associated to this curve in the domain.

Let’s use sines and cosines to parameterize the circle.

$\begin{align*} x(t) &= \answer{\sqrt{2}} \cos(t) \\ y(t) &= \answer{\sqrt{2}} \sin(t) \\ \end{align*}$

A parameterization of the desired circle in the $xy$ - plane is

$\vec {r}(t) = \vector {\answer {\sqrt {2}\cos (t)},\answer {\sqrt {2} \sin (t)}},$

and a parameterization of the associated curve on the surface is

$\vec {R}(t) = \vector {\answer {\sqrt {2}\cos (t)},\answer {\sqrt {2} \sin (t)}, \answer { e^{2\cos ^2(t)+3\sqrt {2}\sin (t)-4} }}.$

Note that if we set $x=t$ , we will to use both $\vec {r}_1(t) = \vector {t,\sqrt {2-t^2}}, -\sqrt {2} \leq t \leq \sqrt {2}$ and $\vec {r}_2(t) = \vector {t,-\sqrt {2-t^2}}, -\sqrt {2} \leq t \leq \sqrt {2}$ to trace out the entire circle in the $xy$ -plane.

Find a parametric description of the tangent line to the curve in the previous part at the point $(-1,1,1)$ .

$\vec {l}(t) = \vector {\answer {-t-1 },\answer { -t+1 },\answer { -t+1}}$

Suppose that $t=t_0$ is the $t$ -value for which $\vec {r}(t_0) = \vector {-1,1}$ . We do not need to exhibit $t_0$ explicitly; we will only need to use the facts $\begin{align*} x(t_0) &= \sqrt{2} \cos(t_0) & \textrm{ so }& & \cos(t_0) &=-\frac{1}{\sqrt{2}}. \\ y(t_0) &= \sqrt{2} \sin(t_0) & \textrm{ so } & & \sin(t_0) &=\frac{1}{\sqrt{2}}. \\ \end{align*}$

A vector parallel to the tangent line is $\vec {R}'(t_0)$ . $\begin{align*} \vec{R}'(t) &= \vector{\answer{-\sqrt{2}\sin(t)},\answer{\sqrt{2}\cos(t) },\answer{ (-4\cos(t)\sin(t) +3\sqrt{2}\cos(t))\cdot e^{2\cos^2(t)+3\sqrt{2}\sin(t)-4} }} \\ \end{align*}$

To find $\vec {R}'(t_0)$ , note $\cos (t_0)=-\frac {1}{\sqrt {2}}$ and $\sin (t_0) = \frac {1}{\sqrt {2}}$ , so

$\vec {R}(t_0) = \vector {\answer {-1 },\answer { -1 },\answer { -1}} \\$ Use $\vec {l}(t) =\vec {v}t+\vec {P}_0$ where $P_0 = (-1,1,1)$ .

To determine whether the tangent line $\vec {l}(t)$ lies on the tangent plane, we must check whether

$-2\big [x(t)\big ]+3\big [y(t)\big ]-\big [z(t)\big ] = 4$

for all $t$ .

Does the tangent line lie on the tangent plane? YesNo

Note that $\begin{align*} -2\big[x(t)\big]+3\big[y(t)\big]-\big[z(t)\big] &=-2\big[-t-1\big]+3\big[-t+1\big]-\big[-t+1\big] \\ &= 2t+2-3t+3+t-1 \\ &= 4 \end{align*}$