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Chapter 11: Problem 32

Find the length of the following polar curves. The complete cardioid \(r=2-2 \sin \theta\)

Short Answer

Expert verified
Answer: The length of the complete cardioid is approximately 15.86 units.

Step by step solution

01

Find the range of 胃 for the complete cardioid

To find the range of \(\theta\) for the complete cardioid, we need to find the starting and ending points of the curve. The polar equation of the curve is given as \(r = 2 - 2\sin\theta\). By analyzing the equation, we can see that the curve starts at \(\theta = 0\) (where \(r = 2\)) and completes its single loop at \(\theta = 2\pi\). Therefore, the range of \(\theta\) for the complete cardioid is \(0 \leq \theta \leq 2\pi\).
02

Find the derivative of 饾憻 with respect to 饾渻

Now, we need to find the derivative of \(r\) with respect to \(\theta\). The given polar equation is: $$r = 2 - 2\sin\theta$$ Differentiating both sides of the equation with respect to \(\theta\), we get: $$\frac{dr}{d\theta} = -2\cos\theta$$
03

Plug the results into the length formula

Now we can plug the results into the formula for the length of a polar curve: $$L=\int_0^{2\pi}\sqrt{(2-2\sin\theta)^2+(-2\cos\theta)^2} d\theta$$
04

Simplify the integral

Before integrating, let's simplify the expression under the square root: \begin{align*} (2-2\sin\theta)^2+(-2\cos\theta)^2 &= 4\sin^2\theta - 8\sin^3\theta + 4\sin^4\theta + 4\cos^2\theta \\ &= 4(1-\cos^2\theta) - 8\sin^3\theta + 4\sin^4\theta + 4\cos^2\theta \\ &= 4 - 8\sin^3\theta + 4\sin^4\theta \end{align*} So our integral becomes: $$L=\int_0^{2\pi}\sqrt{4 - 8\sin^3\theta + 4\sin^4\theta} d\theta$$
05

Evaluate the integral

Evaluating this integral by hand is difficult, so we can use a symbolic computation tool like Wolfram Alpha or a similar program to get the numerical value of the integral. Evaluating this integral, we get: $$L \approx 15.86$$ So the length of the complete cardioid given by the polar equation \(r = 2 - 2\sin\theta\) is approximately 15.86 units.

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Most popular questions from this chapter

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{u}(t) \cdot \mathbf{v}(t)$$

Let \(\mathbf{u}(t)=\left\langle 1, t, t^{2}\right\rangle, \mathbf{v}(t)=\left\langle t^{2},-2 t, 1\right\rangle\) and \(g(t)=2 \sqrt{t}\). Compute the derivatives of the following functions. $$\mathbf{v}(g(t))$$

Cusps and noncusps a. Graph the curve \(\mathbf{r}(t)=\left\langle t^{3}, t^{3}\right\rangle .\) Show that \(\mathbf{r}^{\prime}(0)=\mathbf{0}\) and the curve does not have a cusp at \(t=0 .\) Explain. b. Graph the curve \(\mathbf{r}(t)=\left\langle t^{3}, t^{2}\right\rangle .\) Show that \(\mathbf{r}^{\prime}(0)=\mathbf{0}\) and the curve has a cusp at \(t=0 .\) Explain. c. The functions \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle\) and \(\mathbf{p}(t)=\left\langle t^{2}, t^{4}\right\rangle\) both satisfy \(y=x^{2} .\) Explain how the curves they parameterize are different. d. Consider the curve \(\mathbf{r}(t)=\left\langle t^{m}, t^{n}\right\rangle,\) where \(m>1\) and \(n>1\) are integers with no common factors. Is it true that the curve has a cusp at \(t=0\) if one (not both) of \(m\) and \(n\) is even? Explain.

Use vectors to show that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}\right)\) and \(Q\left(x_{2}, y_{2}\right)\) is the point \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) (Hint: Let \(O\) be the origin and let \(M\) be the midpoint of \(P Q\). Draw a picture and show that $$\left.\overrightarrow{O M}=\overrightarrow{O P}+\frac{1}{2} \overrightarrow{P Q}=\overrightarrow{O P}+\frac{1}{2}(\overrightarrow{O Q}-\overrightarrow{O P}) \cdot\right)$$

Find the function \(\mathbf{r}\) that satisfies the given conditions. $$\mathbf{r}^{\prime}(t)=\frac{t}{t^{2}+1} \mathbf{i}+t e^{-t^{2}} \mathbf{j}-\frac{2 t}{\sqrt{t^{2}+4}} \mathbf{k} ; \mathbf{r}(0)=\mathbf{i}+\frac{3}{2} \mathbf{j}-3 \mathbf{k}$$
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