Directional Derivatives

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April 7th 2009, 10:28 PM
jffyx

Directional Derivatives

Find equations of the tangent plane and normal line to the surface http://hosted.webwork.rochester.edu/...de0bdaccd1.png at the point (-8, 0, 0).
Tangent Plane: (make the coefficient of z equal to 1).
__________________________http://hosted.webwork.rochester.edu/...ad78d017c1.png.
Normal line: http://hosted.webwork.rochester.edu/...fdbf8d6331.png____,____ http://hosted.webwork.rochester.edu/...a27ce889b1.png
http://hosted.webwork.rochester.edu/...12e32f38d1.png____ ,_____ , http://hosted.webwork.rochester.edu/...2046c386e1.png.

Thanks!
April 7th 2009, 10:49 PM
Jhevon

Quote:

Originally Posted by jffyx

Find equations of the tangent plane and normal line to the surface http://hosted.webwork.rochester.edu/...de0bdaccd1.png at the point (-8, 0, 0).
Tangent Plane: (make the coefficient of z equal to 1).
__________________________http://hosted.webwork.rochester.edu/...ad78d017c1.png.
Normal line: http://hosted.webwork.rochester.edu/...fdbf8d6331.png____,____ http://hosted.webwork.rochester.edu/...a27ce889b1.png
http://hosted.webwork.rochester.edu/...12e32f38d1.png____ ,_____ , http://hosted.webwork.rochester.edu/...2046c386e1.png.

Thanks!

Take $F(x,y,z) = xe^y \cos z - z$

the tangent plane is given by: $F_x(-8, 0, 0)(x + 8) + F_y(-8, 0, 0)(y - 0) + F_z(-8, 0, 0)(z - 0) = 0$

the normal line is given by: $\left< -8,0,0 \right> + t \left< \text{normal vector to the above plane} \right>$
April 7th 2009, 11:26 PM
jffyx

How do you find the normal vector to the plane?
April 7th 2009, 11:45 PM
Jhevon

Quote:

Originally Posted by jffyx

How do you find the normal vector to the plane?

when a plane is written in the form: $ax + by + cz + d = 0$

the normal vector is given by: $\vec n = \left< a,b,c \right>$

remember, you want to write it in the form where c = 1