/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 66 Find the points (if they exist) ... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 66

Find the points (if they exist) at which the following planes and curves intersect. $$y+x=0 ; \mathbf{r}(t)=\langle\cos t, \sin t, t\rangle, \text { for } 0 \leq t \leq 4 \pi$$

Short Answer

Expert verified
Question: Find the points where the given plane and parametric curve intersect (if any points exist). The plane equation is \(y + x = 0\) and the parametric curve is given by \(\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle\) for \(0 \leq t \leq 4\pi\). Answer: The four intersection points between the plane and parametric curve are: 1. \(\mathbf{r}\left(\frac{7\pi}{2} - \frac{\pi}{12}\right) = \left\langle\cos\left(\frac{7\pi}{2} - \frac{\pi}{12}\right), \sin\left(\frac{7\pi}{2} - \frac{\pi}{12}\right), \frac{7\pi}{2} - \frac{\pi}{12}\right\rangle\) 2. \(\mathbf{r}\left(\frac{7\pi}{2} + \frac{\pi}{12}\right) = \left\langle\cos\left(\frac{7\pi}{2} + \frac{\pi}{12}\right), \sin\left(\frac{7\pi}{2} + \frac{\pi}{12}\right), \frac{7\pi}{2} + \frac{\pi}{12}\right\rangle\) 3. \(\mathbf{r}\left(\frac{15\pi}{2} - \frac{\pi}{12}\right) = \left\langle\cos\left(\frac{15\pi}{2} - \frac{\pi}{12}\right), \sin\left(\frac{15\pi}{2} - \frac{\pi}{12}\right), \frac{15\pi}{2} - \frac{\pi}{12}\right\rangle\) 4. \(\mathbf{r}\left(\frac{15\pi}{2} + \frac{\pi}{12}\right) = \left\langle\cos\left(\frac{15\pi}{2} + \frac{\pi}{12}\right), \sin\left(\frac{15\pi}{2} + \frac{\pi}{12}\right), \frac{15\pi}{2} + \frac{\pi}{12}\right\rangle\)

Step by step solution

01

Write down the given plane and curve equations

The given plane equation is \(y + x = 0\), and the parametric curve is given by \(\mathbf{r}(t) = \langle \cos t, \sin t, t \rangle\) for \(0 \leq t \leq 4\pi\).
02

Substitute the parametric curve into the plane equation

We will substitute the components of the parametric curve into the plane equation to find the intersection points. So, we replace \(x\) with \(\cos t\) and \(y\) with \(\sin t\) in order to find the values of \(t\) that satisfy the plane equation: $$\sin t + \cos t = 0$$
03

Solve for \(t\)

To solve for \(t\), we can use the fact that \(\cos^2 t + \sin^2 t = 1\). Then, square the equation from Step 2: $$(\sin t + \cos t)^2 = 0^2$$ This simplifies to: $$\sin^2 t + 2 \sin t \cos t + \cos^2 t = 0$$ And using \(\cos^2 t + \sin^2 t = 1\), we get: $$1 + 2 \sin t \cos t = 0$$ That gives us the equation: $$\sin t \cos t = -\frac{1}{2}$$
04

Use trigonometric identities to find the solutions

Let's use the double angle identity for sine: \(\sin 2t = 2\sin t \cos t\). Substitute this into the equation from Step 3: $$\sin 2t = -\frac{1}{2}$$ Now, we can find the values of \(t\) that satisfy this equation. Since we are looking for solutions in the interval \(0 \leq t \leq 4\pi\), we need to consider the solutions of \(2t\) in the interval \(0 \leq 2t \leq 8\pi\): $$2t = 7\pi - \frac{\pi}{6}, 7\pi + \frac{\pi}{6}, 15\pi - \frac{\pi}{6}, 15\pi + \frac{\pi}{6}$$
05

Divide by 2 to find the solutions for \(t\)

Now, divide the solutions in Step 4 by 2 to find the solutions for \(t\): $$t = \frac{7\pi}{2} - \frac{\pi}{12}, \frac{7\pi}{2} + \frac{\pi}{12}, \frac{15\pi}{2} - \frac{\pi}{12}, \frac{15\pi}{2} + \frac{\pi}{12}$$
06

Find the corresponding points on the curve

Finally, substitute the solutions for \(t\) back into the parametric curve equation \(\mathbf{r}(t)\) to find the corresponding intersection points: $$\mathbf{r}\left(\frac{7\pi}{2} - \frac{\pi}{12}\right) = \left\langle\cos\left(\frac{7\pi}{2} - \frac{\pi}{12}\right), \sin\left(\frac{7\pi}{2} - \frac{\pi}{12}\right), \frac{7\pi}{2} - \frac{\pi}{12}\right\rangle$$ $$\mathbf{r}\left(\frac{7\pi}{2} + \frac{\pi}{12}\right) = \left\langle\cos\left(\frac{7\pi}{2} + \frac{\pi}{12}\right), \sin\left(\frac{7\pi}{2} + \frac{\pi}{12}\right), \frac{7\pi}{2} + \frac{\pi}{12}\right\rangle$$ $$\mathbf{r}\left(\frac{15\pi}{2} - \frac{\pi}{12}\right) = \left\langle\cos\left(\frac{15\pi}{2} - \frac{\pi}{12}\right), \sin\left(\frac{15\pi}{2} - \frac{\pi}{12}\right), \frac{15\pi}{2} - \frac{\pi}{12}\right\rangle$$ $$\mathbf{r}\left(\frac{15\pi}{2} + \frac{\pi}{12}\right) = \left\langle\cos\left(\frac{15\pi}{2} + \frac{\pi}{12}\right), \sin\left(\frac{15\pi}{2} + \frac{\pi}{12}\right), \frac{15\pi}{2} + \frac{\pi}{12}\right\rangle$$ These four points are the intersections of the given plane and parametric curve.

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