Chapter 10.1, Problem 33E

To determine

To find: The length of the curve.

Answer to Problem 33E

The answer:

  $L=\frac{2\sqrt{2}-1}{3}$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=\frac{1}{3}{t}^3,\\ y=\frac{1}{2}{t}^2,0\le t\le 1\end{array}$

Calculation:

Know that,

If $dx/dt$

And

  $dy/dt$ are continuous functions of $t$ for $t_1\le t\le t_2$ .then the length of a parametric curve that is traced only once for each $t$ between $t_1$ and $t_2$

equal to:

  

Since,

  $\begin{array}{c}x=t^3/3,\\ y=t^2/2\text{,}\end{array}$

And

  $0\le t\le 1\text{,}$

Then,

  $\begin{array}{c}{t}_1=0\\ t_2=1\end{array}$

And,

  $\begin{array}{c}\frac{dx}{dt}=\frac{3t^2}{3}\\ =t^2\\ \frac{dy}{dt}=\frac{2t}{2}\\ =t\end{array}$

The foregoing findings lead to the conclusion that the parametric curve is traced only once for each $t$ between $0$ and $1$ that $dx/dt$ and $dy/dt$ are continuous functions of $t$ for $0\le t\le 1$. utilise the above findings to determine that the length of the parametric curve is equal to:

  

To solve the integral,

  

Use the substitution,

  $u=t^2+1$

Then,

  $\begin{array}{c}du=2tdt\\ \Rightarrow tdt=\frac{1}{2}du\end{array}$

For,

  $t=0$

Then,

  $t=0$

While for,

  $t=1$

Then,

  $u=2$

To use the previous result,

  

The previous findings indicate that the parametric curve's length is equal to:

  

Therefore, the required length of the curve is $L=\frac{2\sqrt{2}-1}{3}$ .