math203fall2010sols

# math203fall2010sols - CCNY 203 Fall 2010 Final Solutions 1a...

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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 - 2ySin @ z D x + z + x 2 y 3 - 2ySin @ z D gradf = 8 D @ f,x D ,D @ f,y D ,D @ f,z D< 9 x + z + 2xy 3 ,3x 2 y 2 - 2Sin @ z D , x + z - 2yCos @ 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 = 2xy 3 + ‰ H x + z L H 1 + ¶ୢ x z @ x DL - 2yCos @ z D ¶ୢ x z @ x D ã 0 2xy 3 - 2yCos @ z D z £ @ x D + ‰ x + z H 1 + z £ @ x DL ã 0 Solving gives:

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Solve @ eqn,z' @ x DD :: z £ @ x D Ø -‰ x + z - 2xy 3 x + z - 2yCos @ z D >> More simply, since we already have the gradient from 2a, we can divide to use: ¶ୢ z ¶ୢ x = - ¶ୢ f ¶ୢ x ¶ୢ f ¶ୢ x in this case.
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