Question

Find the length of the following polar curves. The complete cardioid \(r=4+4 \sin \theta\)

Short answer

Answer: Since the integral's limits of integration are the same after substitution, the arc length is 0.

Step-by-step solution

Find the range of \(\theta\) for the cardioid

A cardioid has one loop, so we need to find the range of \(\theta\) where the curve completes one full loop. Let's analyze the given polar equation \(r = 4 + 4\sin\theta\): When \(\theta = 0\), we have \(r = 4 + 4\sin(0) = 4\). As we increase the value of \(\theta\), the value of \(r\) increases until it reaches its maximum at the turning point when \(\sin\theta = 1\). This occurs at \(\theta = \frac{\pi}{2}\), where \(r = 4 + 4\sin\left(\frac{\pi}{2}\right) = 4 + 4 = 8\). Now, as \(\theta\) keeps increasing, we can see that the value of \(r\) will again reach \(4\) when \(\theta\) is equal to \(\pi\). At this point, the curve has completed one full loop. Therefore, the range of \(\theta\) for the cardioid is \(0 \le \theta \le \pi\).

Compute the derivative \(\frac{dr}{d\theta}\)

Now, we need to find the derivative of \(r\) with respect to \(\theta\). The given polar equation is \(r = 4 + 4\sin\theta\). Using the chain rule, we find: \(\frac{dr}{d\theta} = \frac{d(4+4\sin\theta)}{d\theta} = 4\cos\theta\)

Calculate the arc length using the integral formula

Using the formula for the arc length of polar curves, we plug our function values and the derivative back into the formula and compute the integral: \(L = \int_{0}^{\pi} \sqrt{r^2 + (\frac{dr}{d\theta})^2} d\theta = \int_{0}^{\pi} \sqrt{(4+4\sin\theta)^2 + (4\cos\theta)^2} d\theta\) Now we simplify the expression inside the square root before calculating the integral: \(L = \int_{0}^{\pi} \sqrt{16\sin^2\theta + 32 \sin\theta + 16 + 16\cos^2\theta} d\theta\) Since \(\sin^2\theta + \cos^2\theta = 1\), we can simplify the expression as: \(L = \int_{0}^{\pi} \sqrt{16 + 32\sin\theta} d\theta\)

Evaluate the integral

To evaluate the integral, we will use substitution: Let \(u = 32\sin\theta + 16\). Then, \(du = 32\cos\theta d\theta\). We need to change the limits of integration according to the \(u\) substitution: When \(\theta = 0\), we have \(u = 32\sin(0) + 16 = 16\). When \(\theta = \pi\), we have \(u = 32\sin(\pi) + 16 = 16\). Now the integral becomes: \(L = \frac{1}{32} \int_{16}^{16} \sqrt{u} du\) Since the limits of integration are the same, the value of the integral is \(0\). Therefore, the arc length of the cardioid \(r = 4 + 4 \sin \theta\) is \(L = 0\).