{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW05 - HW 5 Solutions Math 1920 Spring 2009 Problem 14.6.6...

This preview shows pages 1–3. Sign up to view the full content.

HW 5 Solutions Math 1920 Spring 2009 February 26, 2009 Problem 14.6.6 f = (2 x - y ) i + ( - 2 y - x ) j - k f | (1 , 1 , - 1) = i - 3 j - k The tangent plane has normal f | (1 , 1 , - 1) and contains the point (1,1,-1) and hence has equation 1( x - 1) - 3( y - 1) - 1( z + 1) = 0 x - 3 y - z = - 1 The normal line has direction f | (1 , 1 , - 1) and contains the point (1,1,-1) and hence is given by x = 1 + t y = 1 - 3 t z = - 1 - t Problem 14.6.14 f = yz i + xz j + xy k f (1 , 1 , 1) = i + j + k g = 2 x i + 4 y j + 6 z k g (1 , 1 , 1) = 2 i + 4 j + 6 k . The tangent line to the curve of intersection is orthogonal to both f (1 , 1 , 1) and g (1 , 1 , 1), so we find its direction by taking the cross product of these two vectors. f (1 , 1 , 1) x g (1 , 1 , 1) = i j k 1 1 1 2 4 6 = 2 i - 4 j + 2 k Therefore the equation of the tangent line is given by x = 1 + 2 t y = 1 - 4 t z = 1 + 2 t Problem 14.6.28 a. f x = 3 x 2 y 4 f x (1 , 1) = 3 f y = 4 x 3 y 3 f x (1 , 1) = 4 f (1 , 1) = 1 Therefore L ( x, y ) = 1 + 3( x - 1) + 4( y - 1) = 3 x + 4 y - 6 1

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

View Full Document
b. f x (0 , 0) = 0 f y (0 , 0) = 0 f (0 , 0) = 0 Therefore L ( x, y ) = 0 Problem 14.6.38 First we calculate the partials of f ( x, y, z ) f x = 2 x f y = 2 y f z = 2 z a. f x (1 , 1 , 1) = f y (1 , 1 , 1) = f z (1 , 1 , 1) = 2 and f (1 , 1 , 1) = 3. Therefore at (1,1,1) we have L ( x, y, z ) = 3 + 2( x - 1) + 2( y - 1) + 2( z - 1) = 2 x + 2 y + 2 z - 3. b. f x (0 , 1 , 0) = f z (0 , 1 , 0) = 0, f y (0 , 1 , 0) = 2, and f (0 , 1 , 0) = 1.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 4

HW05 - HW 5 Solutions Math 1920 Spring 2009 Problem 14.6.6...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online