Find Length of Parametrized Curve: x(t), y(t)

[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?

 

[xanthym]

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 )}


~~

 

[thepatientmental]

To find the length of a parametrized curve, we use the formula:

L = \int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2} dt

In this case, a = 0 and b = 1, so we have:

L = \int_{0}^{1} \sqrt{(24*t-24)^2 + (-12*t^2+24*t)^2} dt

To integrate this, we can use the substitution u = t^2, which gives us:

du = 2t dt

And our integral becomes:

L = \int_{0}^{1} \sqrt{(24*u-24)^2 + (-12*u+24*u)^2} \frac{1}{2\sqrt{u}} du

Simplifying, we get:

L = \int_{0}^{1} \frac{1}{2} \sqrt{(24*u-24)^2 + (12*u)^2} du

Now, we can use the trigonometric identity:

\sin^2 \theta + \cos^2 \theta = 1

To rewrite our integral as:

L = \int_{0}^{1} \frac{1}{2} \sqrt{24^2(u-1)^2 + 12^2(u)^2} du

= \int_{0}^{1} \frac{1}{2} \sqrt{12^2(u^2 + (u-1)^2)} du

= \int_{0}^{1} \frac{1}{2} \sqrt{12^2(2u^2 - 2u + 1)} du

= \frac{1}{2} \int_{0}^{1} 12\sqrt{2u^2 - 2u + 1} du

We can now use the substitution v = 2u^2 - 2u + 1, which gives us:

dv = (4u-2) du

And our integral becomes:

L = \frac{1}{2} \int_{0}^{1} 12\sqrt{v} \frac{dv}{4u-2}

= \frac{3}{4} \int_{0}^{1} \sqrt{v} dv

= \frac{3}{4} \left[ \