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Leaves of polar curves

As we approach the end of the semester, I will be taking some time to write-up a few more post for my Calculus 2 student’s. In these posts, we will focus on functions which are not represented in the usual \(y\) is a function of \(x\) form. Instead, we will look at parametric, polar and vector representations. If you would like to view more Calculus 2 study materials, please visit them here. You can also find the YouTube videos for most of those posts here.

\(r=3\sin(4\theta)\)

In this example, we will be using the polar curve \(r=3\sin(4\theta)\). With this curve, we will both find the area enclosed by each of the leaves of the graph and the slope of the tangent for any given \(\theta\). In order to help with both of these problems, we begin by given a sketch of the graph. In order to do this, we will first graph the equation \(y=3 \sin(4x)\).

Using this graph, we can now plot points for the polar curve. In particular, we would see that at \(\theta=0\), we have \(r=0\). Then, \(r\) will grow until it reaches a maximum of \(3\) at an angle of \(\frac{\pi}{8}\). After this point, \(r\) will decrease until it is again \(0\) at \(\frac{\pi}{4}\). In the picture below, this will be the first leaf. After this, the radius will decrease until reaching a minimum of \(-3\) at \(\frac{3\pi}{8}\) and will increase until it has reached zero at \(\frac{\pi}{2}\). If you continue this process you will get the following picture. Note that we have labeled the leaves in the order the would be drawn as we let \(\theta\) move from \(0\) to \(2 \pi\).

Slope of tangents

As we look at the graph, we would note that the tangent line, for any given \(\theta\), would be the line that goes through the point on the curve and has a slope the same as \(\frac{dy}{dx}\) at the given point. Therefore, in order to find the slope of the tangent line to \(r=3\sin(\theta)\) for any given \(\theta\), we need to find \(\frac{dy}{dx}\).

In order to find this, we saw in the post Tangents to Parametric Equations that \(\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\) when \(x\) and \(y\) are given as functions of \(\theta\). Furthermore, we know that if we want to convert from polar to Cartesian coordinates, we have that \(y=r\sin(\theta)\) and \(x=r \cos(\theta)\). We then find that
\begin{align*}
\frac{dy}{dx}&=\frac{\frac{d r \sin(\theta)}{d \theta}}{\frac{d r \cos(\theta)}{d \theta}} \\
&=\frac{\frac{d}{d\theta}(3\sin(4\theta)\sin(\theta))}{\frac{d}{d\theta}(3\sin(4\theta)\cos(\theta)}.
\end{align*}
Here we replaced \(r\) with \(3\sin(4\theta)\) since this is the given function \(r\) in terms of \(\theta\). We then need to take these derivatives. In order to find these, we will need to use a product rule (more on the product rule). Doing these one at a time we find,
\begin{align*}
\frac{d}{d\theta}3\sin(4\theta)\sin(\theta) &=SF’+FS’
\end{align*}
where \(S=\sin(\theta)\), \(F=3\sin(4\theta)\), \(S’=\cos(\theta)\) and for \(F’\) we need to use a chain rule (more on chain rule). For this one, we have \(i=4 \theta\), \(o=3\sin(i)\), so \(o’=3\cos(i)\) and \(i’=4\), thus giving \(3\cos(i)*4=12\cos(4\theta)\). Combining these, we find,
\begin{align*}
\frac{d}{d\theta}&3\sin(4\theta)\sin(\theta)\\ &=SF’+FS’ \\
&=\sin(\theta)*12\cos(4\theta)+3\sin(4\theta)\cos(\theta).
\end{align*}

If you follow a similar process for \(\frac{dx}{d\theta}\), you will find that
\begin{align*}
\frac{d}{d\theta}&(3\sin(4\theta)\cos(\theta) \\ &=\cos(\theta)*12\cos(4\theta)-3\sin(4\theta)\sin(\theta).
\end{align*}

We can now find that
\begin{align*}
\frac{dy}{dx}=\frac{\sin(\theta)*12\cos(4\theta)+3\sin(4\theta)\cos(\theta)}{\cos(\theta)*12\cos(4\theta)-3\sin(4\theta)\sin(\theta)}.
\end{align*}

This is indeed what we were looking for. If we were going to use it further, I would suggest simplifying what you could. However, the question only asked to find \(\frac{dy}{dx}\), so I would leave the problem as is.

Areas of the leaves

We now move on to finding the area enclosed by the function \(r=3\sin(4\theta)\) on each of its leaves and as a whole. In order to do this, we will be using the techniques we used in Finding Area using Integrals. In this case; however, we will be cutting triangles instead of a rectangles. I have drawn an example of such a triangle below.

In this case, we see that area of the triangle will be \(\frac{1}{2}bh\), where \(b\) is base and \(h\) is the height. For the given triangle, the base would be the given \(r\) for the respective \(\theta\). Furthermore, the height would be \(r\sin(\Delta \theta)\) where \(\Delta \theta\) is the change in the angle made to create the triangle. Now, since we will be letting \(\Delta \theta\) go to \(0\), we can use the fact the \(\sin(x) \approx x\) for small \(x\). We, therefore, get that the area of the triangle will be
\begin{align*}
A_{T}&=\frac{1}{2}r *r\sin(\Delta \theta) \\
&\approx \frac{1}{2}(3\sin(4\theta))^{2}\Delta \theta.
\end{align*}

We then let \(\Delta \theta \to 0\) and sum up the areas of the triangles to find that the exact total area will be
\begin{align*}
\int_{a}^{b}\frac{9}{2}\sin^{2}(4\theta)d\theta,
\end{align*}
where \(a\) and \(b\) are the smallest and largest \(\theta\) for a given leaf. In order to find \(a\) and \(b\) we will look back at our work from graphing the function. Note that the first leaf started at \(\theta=0\) and ended up back at \(r=0\) when \(\theta=\frac{\pi}{4}\). Therefore, the area of the first leaf will be
\begin{align*}
\int_{0}^{\frac{\pi}{4}}\frac{9}{2}\sin^{2}(4\theta)d\theta.
\end{align*}

In order to finish this, we will now have to find the anti-derivative of the integrand. We have completed an integral similar to this one in Integration Techniques 2-Reduction Formulas, so I will refer you there for an example on this process. In the end, we will find,
\begin{align*}
\int_{0}^{\frac{\pi}{4}}&\frac{9}{2}\sin^{2}(4\theta)d\theta \\ &=\frac{9}{16}(-\sin(4\theta)\cos(4\theta)+4\theta)|_{0}^{\frac{\pi}{4}} \\
&=\frac{9}{16}(\pi).
\end{align*}

Hence, the area of each leaf will be \(\frac{9 \pi}{4}\). If we would like to find the total area enclosed by all the leaves, we could either integrate from \(0\) to \(2\pi\) to enclose all of them, or we could multiply the current answer by \(8\) since there are \(8\) leaves and each will have the same area due to the symmetry of the function.

Conclusion

I hope you learned something about polar curves and computing derivatives and integrals for these types of functions. If this helped you with your studying make sure to like the post and share it with anyone else you know that may benefit from the help. Good luck on the rest of your studying.

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