Q 9. - Calculus

Question

Find the arc lengths of the curves defined by the parametric equations on the specified intervals.

$x=t^2$, $y=t^3$, $t\in \left[-2,2\right]$

Step-by-Step Solution

Answer

The required length of the curve is $\begin{array}{rcl}& & \frac{2}{27}\left\{{40}^{\frac{3}{2}}-4^{\frac{3}{2}}\right\}\end{array}$.

Step 1. Given information.

The given parametric equations are $x=t^2$, $y=t^3$.

Step 2. Find the derivative of the parametric equations.

$\begin{array}{rcl}x& =& t^2\\ f^{\text{'}}\left(t\right)& =& 2t\end{array}$     and    $\begin{array}{rcl}y& =& t^3\\ g\text{'}\left(t\right)& =& 3t^2\end{array}$

Step 3. Substitute the value of f ' ( t ) and g ' ( t ) in the formula.

The length of the curve is:

${\int }_a^b\sqrt{{\left(f\text{'}\left(t\right)\right)}^2+{\left(g\text{'}\left(t\right)\right)}^2}dt$, where $\left[a,b\right]=\left[-2,2\right]$.

So, the length of the curve is:

$\begin{array}{rcl}{\int }_a^b\sqrt{{\left(f\text{'}\left(t\right)\right)}^2+{\left(g\text{'}\left(t\right)\right)}^2}dt& =& {\int }_{-2}^2\sqrt{{\left(2t\right)}^2+{\left(3t^2\right)}^2}dt\\ & =& {\int }_{-2}^2\sqrt{4t^2+9t^4}dt\\ & =& {\int }_{-2}^2\sqrt{t^2\left(4+9t^2\right)}dt.................\left(1\right)\end{array}$

Step 4. Now solve the indefinite integral.

Substitute $4+9t^2=z$.

$\begin{array}{rcl}18t dt& =& dz\\ dt& =& \frac{dz}{18t}\end{array}$

$\begin{array}{rcl}\int \sqrt{t^2\left(4+9t^2\right)}dt& =& \int t\sqrt{\left(4+9t^2\right)}dt\\ & =& \int t\sqrt{z}\frac{dz}{18t}\\ & =& \int \sqrt{z}\frac{dz}{18}\\ & =& \frac{1}{18}\int \sqrt{z}dz\\ & =& \frac{1}{18}·\frac{2}{3} z^{\frac{3}{2}}\\ & =& \frac{1}{27}{\left(4+9t^2\right)}^{\frac{3}{2}}\end{array}$

Step 5. Apply limits to solve definite integral.

$\begin{array}{rcl}{\int }_{-2}^2\sqrt{t^2\left(4+9t^2\right)}dt& =& 2{\int }_0^2\sqrt{t^2\left(4+9t^2\right)}dt\\ & =& 2{\left[\frac{1}{27}{\left(4+9t^2\right)}^{\frac{3}{2}}\right]}_0^2\\ & =& 2\left[\frac{1}{27}{\left(4+9{\left(2\right)}^2\right)}^{\frac{3}{2}}\right]-2\left[\frac{1}{27}{\left(4+9{\left(0\right)}^2\right)}^{\frac{3}{2}}\right]\\ & =& 2\left[\frac{1}{27}{\left(40\right)}^{\frac{3}{2}}\right]-2\left[\frac{1}{27}{\left(4\right)}^{\frac{3}{2}}\right]\\ & =& \frac{2}{27}\left\{{40}^{\frac{3}{2}}-4^{\frac{3}{2}}\right\}\end{array}$

Step 6. Simplified answer.

Hence, the required arc length is $\begin{array}{rcl}& & \frac{2}{27}\left\{{40}^{\frac{3}{2}}-4^{\frac{3}{2}}\right\}\end{array}$.