Prelim_2_Fall_2007

# Prelim_2_Fall_2007 - Math 192 Prelim 2 Solutions 7:30-9:00...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 192, Prelim 2 - Solutions October 25, 2007. 7:30-9:00 1) a) 3 = 1 2 + 2 Â· 2 + 2 and 3 = 5 Â· 1-- 2 so (1 , , 3) is on both of those surfaces. b) Set f ( x, y, z ) = x 2 + 2 y 2 + 2- z . Then f ( x, y, z ) = 0 is a level surface. Now âˆ‡ f ( x, y, z ) = 2 x i + 4 y j- k and âˆ‡ f (1 , , 3) = 2 i- k is a normal to the tangent plane. Thus the equation of the tangent plane is 2( x- 1) + 0( y- 0)- ( z- 3) = 0 â‡’ 2 x- z =- 1 c) Set g ( x, y, z ) = 5 x- y- 2- z . Then g ( x, y, z ) = 0 is a level surface. Observe that âˆ‡ g ( x, y, z ) = 5 i- j- k and âˆ‡ g (1 , , 3) = 5 i- j- k is a normal to the tangent plane of S 2 at (1 , , 3). Since the line tan- gent to the curve of intersection of S 1 and S 2 is in the tangent planes of both surfaces at (1 , , 3) then âˆ‡ f (1 , , 3) Ã—âˆ‡ g (1 , , 3) =- i- 3 j- 2 k is a vector parallel to the line. Thus the parametric equation for the line is x = 1- t , y =- 3 t and z = 3- 2 t ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

Prelim_2_Fall_2007 - Math 192 Prelim 2 Solutions 7:30-9:00...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online