SOLUTION: Consider the curve given by the parametric equations: x = 3t^2 - 6t y = t^4 - 2t^3 - 2t^2 - 2t + 5 a. There are two sections of this curve that go between the point (−3,0) a

Algebra ->  Test -> SOLUTION: Consider the curve given by the parametric equations: x = 3t^2 - 6t y = t^4 - 2t^3 - 2t^2 - 2t + 5 a. There are two sections of this curve that go between the point (−3,0) a      Log On


   

Question 1183482: Consider the curve given by the parametric equations:
x = 3t^2 - 6t
y = t^4 - 2t^3 - 2t^2 - 2t + 5
a. There are two sections of this curve that go between the point (−3,0) and (9,8). Set up an integral that expresses the area under the upper of those two sections and evaluate it.
b. Set up and simplify the integral that represents the length of the curve from the point (9,8) to the point (24,93).
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
a. The area between the upper curve of the parametric equations x+=+3t%5E2+-+6t, y+=+t%5E4+-+2t%5E3+-+2t%5E2+-+2t+%2B+5 and the x-axis
(i.e., between the point (−3,0) and (9,8)) is given by

A+=+6+%2Aint%28%28t%5E5+-+3t%5E4%2B7t+-+5%29%2C+dt%2C+1%2C-1%29+=+56%2F5.
b. The expression for the arc length from from the point (9,8) to the point (24,93) is given by

int%28sqrt%2816t%5E6-48t%5E5%2B4t%5E4%2B32t%5E3+%2B76t%5E2-56t%2B40%29%2C+dt%2C+3%2C4%29

If you need the complete details on these answers message me.