Q. 50. - Calculus

Question

                                                              Annie is designing a kayak with the central cross-section given by $r=1-0.5\sin \theta$, where $r$ is measured in feet. Kayak and central cross-section

She will cover the kayak with an impermeable polyurethane fabric. The fabric comes in rolls 1 meter wide. Annie needs to use the length of the fabric to run the length of her boat, so she hopes to be able to use just two widths of the fabric to cover the boat. Will she be able to?

Step-by-Step Solution

Answer

The required length of her boat is $6.68$

Step 1: Given information

Take a look at the polar function 

$r=1-0.5\sin \theta$

Step 2: The objective is to find the area of the curve.

Formula to find the area is $A={\int }_{\alpha }^{\beta }\frac{1}{2}\left(f\right(\theta ){)}^{2}d\theta$or $A={\int }_{\alpha }^{\beta }\frac{1}{2}{r}^{2}d\theta$

The limits are from $0 to 2\pi$

Step 3: Find the formula for the length of the polar curve

The formula for the length of the polar curve is given by,

$L={\int }_{\alpha }^{\beta }\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+\left(f\right(\theta ){)}^2}d\theta$

To begin, calculate the derivative of the given polar equation.

$$\begin{gathered} f\left(\theta \right)=r=1-0.5\sin \theta \\ \Rightarrow f\left(\theta \right)=1-0.5\sin \theta \end{gathered}$$

Distinguish $f\left(\theta \right)=1-0.5\sin \theta$ now with respect to $\theta$

$$\begin{gathered} f^{\text{'}}\left(\theta \right)=\frac{d}{d\theta }(1-0.5\sin \theta ) \\ \Rightarrow f^{\text{'}}\left(\theta \right)=-0.5\cos \theta \end{gathered}$$

Now, put the values $f\left(\theta \right),f^{\text{'}}\left(\theta \right)$in $L={\int }_{\alpha }^{\beta }\sqrt{{\left(f^{\text{'}}\left(\theta \right)\right)}^2+\left(f\right(\theta ){)}^2}d\theta$

The limits are from $0$ to $2\pi$

Then,

$$\begin{gathered} L={\int }_0^{2\mathrm{\pi }}\sqrt{(0.25{\cos }^2\theta )+(1+0.25 {\sin }^2\theta -\sin \theta )}d\theta \\ \Rightarrow L={\int }_0^{2\pi }\sqrt{1.25-\sin \theta }d\theta \end{gathered}$$

Thus,

${\int }_0^{2\pi }\sqrt{1.25-\sin \theta }d\theta =6.68$

Hence, the length of her boat is $6.68$