1. ## Tangent Planes

Hello math help forums! This is my first post on these forums and I hope to come back here often. Go calc!

Okay, so I just finished a problem on my homework, but for some reason the logic I used in completing this problem seems odd. I don't know why though.

Here is the problem:

Find a point on the surface of $z=8-3x^2-2y^2$ at which the tangent plane is perpendicular to the line $x=2-3t, y=7+8t, z=5-t$. Find the equation of the tangent plane at that point.

First, I know that a plane in space is defined by normal vector to the plane and a point on that plane. Because the given line is perpendicular to the tangent plane, I used the direction numbers of the line to form a scalar equation for the tangent plane:

$-3(x-x_{0}) + 8(y-y_{0}) -1(z-z_{0})$

Where $(x_{0}, y_{0}, z_{0})$ is the point on the surface at which the tangent plane is perpendicular to the line.

The equation of the tangent plane is also given by:

$z-z_{0} = f_{x}(x_{0}, y_{0})(x-x_{0}) + f_{y}(x_{0}, y_{0})(y-y_{0})$

$f_{x}(x_{0}, y_{0}) = -6x_{0}$

$f_{y}(x_{0}, y_{0}) = -4y_{0}$

Thus,

$-6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0}) + z_{0} - z = 0$

Since both equations equal 0, and because the point $(x_{0}, y_{0}, z_{0})$ is the same for each equation, I set the equations equal to each other. For some reason, I feel like I am doing something wrong at this point... I don't know why.

$-6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0}) + z_{0} - z = -3(x-x_{0}) + 8(y-y_{0}) -1(z-z_{0})$

The z terms cancel out, and I am left with:

$-6x_{0}(x-x_{0}) + -4y_{0}(y-y_{0})= -3(x-x_{0}) + 8(y-y_{0})$

Thus,

$-6x_{0}=-3 \Rightarrow x_{0} = \frac{1}{2}$

and

$-4y_{0}=8 \Rightarrow y_{0} = -2$

I then plugged these points into the original equation for the surface to obtain $z_{0} = \frac{-3}{4}$

Does this logic seem right?

Thank you for your help! It is greatly appreciated.

Sincerely,

Opiaboy

2. It is kind of hard to follow because of the latex but you are not quite correct :

N = 6x i + 4y j + k

Further N is parallel to M = -3 i + 8 j - k

Since N and M are parallel N = cM for some c

her c = -1 so the z comp match

-6x = -3 -4y = 8

x= 1/2 y= -2

To find z sub the x and y values into

z=8-3x^2-2y^2 so z = 8 -3/4 -8 = -3/4
so the pt is (1/2,-2, -3/4)

Note for your point N = -3 i + 8j + k which is not parallel to M

3. Originally Posted by opiaboy
*EDIT okay for some reason the Latex thing isn't working and my math symbols look like jank... gotta fix that later..
Your LaTeX problem is that you are using $$[\math] tags rather than [tex]$$ tags.

If you are using the advanced editor you will find a button on the tool bar with a $\Sigma$ which will automatically put the tags around any selected text.

CB