Q. 65 - Calculus

Question

(a) Find an integral that represents the length of an elliptical track whose equations are given by the parametric equations $x=\sin \theta ,y=3\cos \theta ,\theta \in [0,2\pi ]$, where $x$and $y$are in kilometers.

(b) approximate the length of the track, using the midpoint method with $20$subintervals.

Step-by-Step Solution

Answer

(a) The integral that represents the length of an elliptical track of given parametric

equations is${\int }_0^{2\pi }\sqrt{(\cos \theta {)}^2+(-3\sin \theta {)}^2}d\theta$.

(b) The value of the integral is$13,4$.

Part (a) Step 1: Given information

The parametric equations, $x=\sin \theta ,y=3\cos \theta ,\theta \in [0,2\pi ]$. 

Part (a) Step 2: Calculation

Consider parametric equations, $x=\sin \theta ,y=3\cos \theta ,\theta \in [0,2\pi ].$

The objective is to find the integral that represents the length of an elliptical track of given parametric equations.

The arc length of the given parametric equations represents the length of an elliptical track. The objective is to draw the parametric curve and find the arc length of the curve.

The formula to find the arc length of the curve is,

The length of the curve $={\int }_a^b\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+{\left(g^{\text{'}}\left(\theta \right)\right)}^2}dt$

Here $f\left(\theta \right)=\sin \theta ,g\left(\theta \right)=3\cos \theta ,\theta \in [0,2\pi ]$

First, find the derivative of the parametric equations.

Take $f\left(\theta \right)=\sin \theta$

Differentiate with respect to $\theta$then

$$\begin{gathered} f^{\text{'}}\left(\theta \right)=\frac{d}{d\theta }\left(\sin \theta \right) \\ {f}^{\text{'}}\left(\theta \right)=\cos \theta \end{gathered}$$

Now take $g\left(\theta \right)=3\cos \theta$

Differentiate with respect to $\theta$ then

$$\begin{gathered} g^{\text{'}}\left(\theta \right)=3\frac{d}{d\theta }\left(\cos \theta \right) \\ {g}^{\text{'}}\left(\theta \right)=-3\sin \theta \end{gathered}$$

Substitute the values of $f^{\text{'}}\left(\theta \right)=\cos \theta ,g^{\text{'}}\left(\theta \right)=-3\sin \theta$in the arc length formula.

Thus,

The length of the curve $={\int }_0^{2\pi }\sqrt{(\cos \theta {)}^2+(-3\sin \theta {)}^2}d\theta$

Since the length of the curve $={\int }_0^{2\pi }\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+{\left(g^{\text{'}}\left(\theta \right)\right)}^2}d\theta$

Therefore, the integral that represents the length of an elliptical track of a given parametric

equations is ${\int }_0^{2\pi }\sqrt{(\cos \theta {)}^2+(-3\sin \theta {)}^2}d\theta$

Part (b) Step 1: Given information

The  integral for the arc length of the curve ${\int }_0^{2\pi }\sqrt{(\cos \theta {)}^2+(-3\sin \theta {)}^2}d\theta$

Part (b) Step 2: Calculation

Consider the integral for the arc length of the curve.${\int }_0^{2\pi }\sqrt{(\cos \theta {)}^2+(-3\sin \theta {)}^2}d\theta$

The objective is to find the value of the integral.

${\int }_{\theta }^{2\pi }\sqrt{(\cos \theta {)}^2+(-3\sin \theta {)}^2}d\theta ={\int }_0^{2\pi }\sqrt{{\cos }^2\theta +9{\sin }^2\theta }d\theta$

The approximate value of the integral is ${\int }_0^{2\pi }\sqrt{{\cos }^2\theta +9{\sin }^2\theta }d\theta =13.4$

Therefore, the value of the integral is $13,4 .$