math203fall2010sols - CCNY 203 Fall 2010 Final Solutions...

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

View Full Document Right Arrow Icon
CCNY 203 Fall 2010 Final Solutions 1a: We take the cross product of the vector along the line and a displacement vector from the point to a point on the line to get a normal to plane P: Cross @8 3, - 1, - 2 < , 8 2, - 2, 1 < - 8 2, - 3, 1 <D 8 2, 0, 3 < That gives an equation of the plane: 2 H x - 2 L + 0 H y + 3 L + 3 H z - 1 L ã 0 1b: The distance from the plane to the origin we can get by projecting a vector from the origin to a point on the plane onto the normal to the plane: d = 8 2, 0, 3 < 4 + 0 + 9 . 8 2, - 3, 1 < 7 13 2a: We take the gradient: f = x 2 y 3 + ‰ x + z - 2 y Sin @ z D x + z + x 2 y 3 - 2 y Sin @ z D gradf = 8 D @ f, x D , D @ f, y D , D @ f, z D< 9 x + z + 2 x y 3 , 3 x 2 y 2 - 2 Sin @ z D , x + z - 2 y Cos @ z D= At the desired point, we substitute to get: gradf ê . 8 x Ø 0, y Ø 1, z Ø 0 < 8 1, 0, - 1 < The directional derivative is obtained by dotting the gradient with a unit vector in the desired direction: 8 1, 0, - 1 < . 8 3, 4, - 12 < 3 2 + 4 2 + 12 2 15 13 2b: Implicit differentiation with respect to x gives: eqn = 2 x y 3 + ‰ H x + z L H 1 + x z @ x DL - 2 y Cos @ z D x z @ x D ã 0 2 x y 3 - 2 y Cos @ z D z £ @ x D + ‰ x + z H 1 + z £ @ x DL ã 0 Solving gives:
Background image of page 1

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

View Full DocumentRight Arrow Icon
Solve @ eqn, z' @ x DD :: z £ @ x D Ø -‰ x + z -
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/29/2012 for the course MATH 203 taught by Professor Snell during the Spring '11 term at City College of San Francisco.

Page1 / 5

math203fall2010sols - CCNY 203 Fall 2010 Final Solutions...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online