Solutions to Test 1,
Math 20C, Oct 22,2008
Particle path problem:
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.
Choose t=Pi/4 in order to get the point x=0, y=5, z=0.
The velocity vector for this value
(-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.
z=y-x^2 = 1-2x -3y.
Solve for y to get
y = (1/4)*(x-1)^2.
Thus a parametric formula for the intersection curve C is
(x, (1/4)*(x-1)^2, (1/4)*(x-1)^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)