# parametrized curve

by ILoveBaseball
Tags: curve, parametrized
 P: 30 Find the length of parametrized curve given by $$x(t) = 0t^3 +12t^2 - 24t$$, $$y(t) = -4t^3 +12t^2+0t$$, where t goes from zero to one. Hint: The speed is a quadratic polynomial with integer coefficients. it's an arclength question right? x' = 24*t-24 y' = -12*t^2+24*t $$\int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2}$$ i get 6.49 when i use a math program to integrate it which is incorrect. anyone know where i went wrong?
P: 412
 Quote by ILoveBaseball Find the length of parametrized curve given by $$x(t) = 0t^3 +12t^2 - 24t$$, $$y(t) = -4t^3 +12t^2+0t$$, where t goes from zero to one. Hint: The speed is a quadratic polynomial with integer coefficients. it's an arclength question right? x' = 24*t-24 y' = -12*t^2+24*t $$\int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2}$$ i get 6.49 when i use a math program to integrate it which is incorrect. anyone know where i went wrong?
Note that:

$$1: \ \ \ \ \int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2} \, dt \ = \ \int_{0}^{1} \sqrt{ \left ( 12t^{2} \ - \ 24t \ + \ 24 \right )^{2} } \, dt \ =$$

$$2: \ \ \ \ = \ \int_{0}^{1} \left | \left ( 12t^{2} \ - \ 24t \ + \ 24 \right ) \right | \, dt \ \color{red} = \ \mathbf{\left (16 \right )}$$

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