math203fall2011sols

math203fall2011sols - CCNY 203 Fall 2011 Final Solutions 1a...

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CCNY 203 Fall 2011 Final Solutions 1a: Take cross products of the displacement vectors to get a normal to plane P: Cross @8 0, 1, 2 < - 8 1, 0, - 1 < , 8 1, 2, 3 < - 8 0, 1, 2 <D 8 - 2, 4, - 2 < That gives an equation of the plane from the normal vector and a point: - 2 H x L + 4 H y - 1 L - 2 H z - 2 L ã 0 - 2 x + 4 H - 1 + y L - 2 H - 2 + z L ã 0 1b: The direction of the line is the same as the normal vector above, so parameterizing the line as P + t V gives l = 8 0, 1, 2 < + t 8 - 2, 4, - 2 < 8 - 2 t, 1 + 4 t, 2 - 2 t < If we write this in standard form, we get x = - 2 t , y = 1 + 4 t , z = 2 - 2 t . 1c: We use dot product to measure the angle: 8 0, 2, - 2 < . 8 - 2, 4, - 2 < 12 The dot product is not zero and the vectors are not scalar multiples, so the vectors are neither perpendicular nor parallel. 2a: We take the gradient: f = z + z Log A x 2 + y 2 E z + z Log A x 2 + y 2 E gradf = 8 D @ f, x D , D @ f, y D , D @ f, z D< : 2 x z x 2 + y 2 , 2 y z x 2 + y 2 , 1 + Log A x 2 + y 2 E> At the desired point, we substitute to get: gradf ê . 8 x Ø 1, y Ø 0, z Ø 2 < 8 4, 0, 1 < The directional derivative is obtained by dotting the gradient with a unit vector in the desired direction: u = 8 4, 4, 7 < - 8 1, 0, 2 < 3 2 + 4 2 + 5 2 : 3 5 2 , 2 2 5 , 1 2 > Together @8 4, 0, 1 < . u D 17 5 2 2b: This the direction of the gradient: <4,0,1>.

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2c: The appropriate version of the chain rule here is E s = E x x s + E y y s + E z z s which at the relevant points gives 2 μ 1 + 0 μ 1 + H 1 + ln H 4 LL 2 = 4+2 ln(4). Or you can substitute but that is more tedious and prone to error. 3: Taking the first partials and setting them to zero and solving gives:
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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.

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math203fall2011sols - CCNY 203 Fall 2011 Final Solutions 1a...

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