Find the length of parametrized curve given by x(t)=18t2−36t, y(t)=−4t3+12t2+15t where t goes from 0 to 1.

Answer 1

Arc length of a curve can be calculated using the formula: L =∫sqrt[(dx/dt)^2+(dy/dt)^2]dt.

First, find the derivatives dx/dt and dy/dt: 36t - 36 and -12t^2 +24t + 15. Square them, combine like terms, then evaluate the integral from t=0 to t=1 using the basic x^(n+1)/(n+1) integration rules. Note that when evaluating, every term will have at least one t in it, to the F(a) term in F(b) - F(a) goes away (from the Fundamental Thm. of Calculus). The final answer is 31