Chapter 11.3, Problem 3QQ

To determine

To determine: The correct answer about the integral which represents the length of the path described by the parametric equations $x=t^2$ and $y=t$ from $t=0$ to $t=4$ .

Answer to Problem 3QQ

The required correct option is (D).

Explanation of Solution

Given information:

The options:

(A) ${\displaystyle {\int }_0^4\sqrt{4t+1}dt}$

(B) $2{\displaystyle {\int }_0^4\sqrt{t^2+1}dt}$

(C) ${\displaystyle {\int }_0^4\sqrt{2t^2+1}dt}$

(D) ${\displaystyle {\int }_0^4\sqrt{4t^2+1}dt}$

(E) $2\pi {\displaystyle {\int }_0^4\sqrt{4t^2+1}dt}$

Calculation:

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

Differentiate the expressions $x=t^2$ and $y=t$ with respect to t .

  $\begin{array}{c}\frac{dx}{dt}=\frac{d}{dt}\left(t^2\right)\\ =2t\end{array}$

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left(t\right)\\ =1\end{array}$

Consider that the length of the parametric curve be L and $\frac{dx}{dt}$ , $\frac{dy}{dt}$ be continuous functions of t ; so the value L is,

  $L={\displaystyle {\int }_{t_1}^{t_2}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}$

Write the length of the given curve as:

  $\begin{array}{c}L={\displaystyle {\int }_{t_1}^{t_2}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}}\\ L={\displaystyle {\int }_0^4\sqrt{{\left(2t\right)}^2+{\left(1\right)}^2}}\\ L={\displaystyle {\int }_0^4\sqrt{4t^2+1}}\end{array}$

Hence, the required correct option is (D).