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# 1701 - final07sols.nb M203 Fall 2007 Exam solutions Q1a...

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M203 Fall 2007 Exam solutions Q1a: find a normal to the plane by taking cross product of two vectors in the plane. vector along line: <2,3,4> point on line: (1,2,3), displacement vector from line to point: <1,2,3>-<1,2,0>= <0,0,3> Cross product: <2,3,4> x <0,0,3> = <9,-6,0> giving an equation: 9 ( x-1) - 6 (y-2) + 0 (z-3) = 0 Q1b: Find a normal vector to the plane by putting into a standard form: x+2y-z=4 gives normal vector <1,2,-1>. That gives parameterization: x=1+1t y=2+2t z=3-t Q2a: find tangent vector by computing the derivative and evaluating it at the proper time r = 9 è!!! t, 4 ÅÅÅÅ t , t 2 ÅÅÅÅÅÅ 2 = 9 è!!! t, 4 ÅÅÅÅ t , t 2 ÅÅÅÅÅÅ 2 = rp = D @ r,t D 9 1 ÅÅÅÅÅÅÅÅÅÅÅÅ 2 è!!! t , - 4 ÅÅÅÅÅÅ t 2 , t = rp ê . t Ø 4 9 1 ÅÅÅÅ 4 , - 1 ÅÅÅÅ 4 ,4 = gives the parameterization: x = 2 + t ÅÅÅÅ 4 y = 1 - t ÅÅÅÅ 4 z = 8 + 4 t Q2b: find the gradient of the implict description of the surface: 9 D A x 3 ÅÅÅÅÅÅ z 2 + 4 H y - 1 L 2 , x E ,D A x 3 ÅÅÅÅÅÅ z 2 + 4 H y - 1 L 2 ,y E , D A x 3 ÅÅÅÅÅÅ z 2 + 4 H y - 1 L 2 ,z E= 9 3x 2 ÅÅÅÅÅÅÅÅÅ z 2 ,8 H - 1 + y L , - 2x 3 ÅÅÅÅÅÅÅÅÅ z 3 = Evaluate at the point to get a normal vector: 9 3x 2 ÅÅÅÅÅÅÅÅÅ z 2 ,8 H - 1 + y L , - 2x 3 ÅÅÅÅÅÅÅÅÅ z 3 = ê . 8 x Ø 1, y Ø 2, z Ø - 1 < 8 3,8, 2 < gives the equation 3 H x - 1 L + 8 H y - 2 L + 2 H z + 1 L = 0 Q3a: find the gradient at the point and use the method D u f = “ f ÿ u final07sols.nb 1

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f = x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 2 x + 3 y x ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 2x + 3y grad = 8 D @ f,x D ,D @ f,y D< 9 - 2x
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1701 - final07sols.nb M203 Fall 2007 Exam solutions Q1a...

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