Use your calculator to find the length of the arc from t = 0 to t = 1 of x = t^3 + 1, y = t^2.

Question

Type your answer in the space below and give 2 decimal places. If your answer is less than 1, place a leading "0" before the decimal point (ex: 0.48)

Answer 1

ds^2 = dx^2 + dy^2
so, the arc length is
∫[0,1] √((3t^2)^2 + (2t)^2) dt
= ∫[0,1] t√(9t^2 + 4) dt
If u = 9t^2+4
du = 18t
and so we have
1/18 ∫[4,13] √u du
= 1/18 (2/3 u^(3/2)) [4,13]
= 1/27 (13√13 - 8)