Tangent Planes

**opiaboy** · October 5th 2009, 07:45 AM

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

**Calculus26** · October 5th 2009, 08:19 AM

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

You have for your normal

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

**CaptainBlack** · October 6th 2009, 01:39 AM

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 [tex][\math] tags rather than [tex][/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

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