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Solutions to Test 1,
Math 20C, Oct 22,2008
Particle path problem:
(A)
x = 3cos(2t), y=5sin(2t), z=4cos(2t) implies the derivatives are
x' = 6sin(2t), y' = 10cos(2t), z' = 8sin(2t).
Thus x'^2 + y'^2 + z'^2 = 100 sin(2t)^2 + 100 cos(2t)^2 = 100.
The square root of this is 10, so we integrate 10 from t=0 to t=60
to get the arclength.
The answer is 600 feet.
(B)
Choose t=Pi/4 in order to get the point x=0, y=5, z=0.
The velocity vector for this value
t=Pi/4 is
(6, 0, 8).
The tangent line at (0,5,0) has parametric formula
(0,5,0) + s*(6, 0 , 8),
where s is the parameter.
Intersection problem:
z=yx^2 = 12x 3y.
Solve for y to get
y = (1/4)*(x1)^2.
Thus a parametric formula for the intersection curve C is
(x, (1/4)*(x1)^2, (1/4)*(x1)^2  x^2), where x is the parameter.
There are other possible answers as well.
Area of the triangle:
The vector AB is (2,1,1) and the vector AC is (0,1,1).
The cross product of these two vectors is (2, 2, 2)
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 Fall '08
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