/*! 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 12: 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
Answer: The intersection points are 鉄�(鈭�2)/2, -(鈭�2)/2, 饾湅/4鉄�, 鉄�-(鈭�2)/2, (鈭�2)/2, (3饾湅)/4鉄�, 鉄�(鈭�2)/2, -(鈭�2)/2, (5饾湅)/4鉄�, and 鉄�-(鈭�2)/2, (鈭�2)/2, (7饾湅)/4鉄�.

Step by step solution

01

Write down the given equations

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

Plug the curve components into the plane equation

Substitute the components of the parametric representation of the curve into the plane equation: $$(\sin t) + (\cos t) = 0$$
03

Solve for t

We need to solve the equation \((\sin t) + (\cos t) = 0\) with the restriction \(0\leq t\leq 4\pi\). This equation can be rewritten as: $$\sin t = -\cos t$$ $$\tan t = -1$$ The solution for this equation is \(t = (2n+1)\frac{\pi}{4}\), where \(n\) is any integer, and within the given range. Now we have to find the solution for t within the interval \(0\leq t\leq 4\pi\).
04

Restrict the solutions for t within the given range

Taking into consideration the restriction on the range of t, namely \(0\leq t\leq 4\pi\), we can determine the possible values of n: For n=0, t=\(\frac{\pi}{4}\) For n=1, t=\(\frac{3\pi}{4}\) For n=2, t=\(\frac{5\pi}{4}\) For n=3, t=\(\frac{7\pi}{4}\) All four of these values of t satisfy the condition, and therefore, are valid values for the points of intersection.
05

Find the intersection points

Now, we can find the intersection points by plugging the value of t back into the parametric equation: For \(t=\frac{\pi}{4}\): \(\mathbf{r}\left(\frac{\pi}{4}\right) = \langle\cos\left(\frac{\pi}{4}\right), \sin\left(\frac{\pi}{4}\right), \frac{\pi}{4}\rangle = \langle\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \frac{\pi}{4}\rangle\) For \(t=\frac{3\pi}{4}\): \(\mathbf{r}\left(\frac{3\pi}{4}\right) = \langle\cos\left(\frac{3\pi}{4}\right), \sin\left(\frac{3\pi^4}\right), \frac{3\pi}{4}\rangle = \langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{3\pi}{4}\rangle\) For \(t=\frac{5\pi}{4}\): \(\mathbf{r}\left(\frac{5\pi}{4}\right) = \langle\cos\left(\frac{5\pi}{4}\right), \sin\left(\frac{5\pi}{4}\right), \frac{5\pi}{4}\rangle = \langle\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \frac{5\pi}{4}\rangle\) For \(t=\frac{7\pi}{4}\): \(\mathbf{r}\left(\frac{7\pi}{4}\right) = \langle\cos\left(\frac{7\pi}{4}\right), \sin\left(\frac{7\pi}{4}\right), \frac{7\pi}{4}\rangle = \langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{7\pi}{4}\rangle\) Thus, the intersection points between the curve and the plane are: $$\langle\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \frac{\pi}{4}\rangle, \langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{3\pi}{4}\rangle, \langle\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}, \frac{5\pi}{4}\rangle, \text{and} \langle-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{7\pi}{4}\rangle$$

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Most popular questions from this chapter

An object moves along a path given by \(\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle\) for \(0 \leq t \leq 2 \pi\) a. What conditions on \(a, b, c, d, e,\) and \(f\) guarantee that the path is a circle (in a plane)? b. What conditions on \(a, b, c, d, e,\) and \(f\) guarantee that the path is an ellipse (in a plane)?

Given a fixed vector \(\mathbf{v},\) there is an infinite set of vectors \(\mathbf{u}\) with the same value of proj\(_{\mathbf{v}} \mathbf{u}\). Find another vector that has the same projection onto \(\mathbf{v}=\langle 1,1,1\rangle\) as \(\mathbf{u}=\langle 1,2,3\rangle\).

Consider the lines $$\begin{aligned} \mathbf{r}(t) &=\langle 2+2 t, 8+t, 10+3 t\rangle \text { and } \\ \mathbf{R}(s) &=\langle 6+s, 10-2 s, 16-s\rangle. \end{aligned}$$ a. Determine whether the lines intersect (have a common point) and if so, find the coordinates of that point. b. If \(\mathbf{r}\) and \(\mathbf{R}\) describe the paths of two particles, do the particles collide? Assume \(t \geq 0\) and \(s \approx 0\) measure time in seconds, and that motion starts at \(s=t=0\).

In contrast to the proof in Exercise \(81,\) we now use coordinates and position vectors to prove the same result. Without loss of generality, let \(P\left(x_{1}, y_{1}, 0\right)\) and \(Q\left(x_{2}, y_{2}, 0\right)\) be two points in the \(x y\) -plane and let \(R\left(x_{3}, y_{3}, z_{3}\right)\) be a third point, such that \(P, Q,\) and \(R\) do not lie on a line. Consider \(\triangle P Q R\). a. Let \(M_{1}\) be the midpoint of the side \(P Q\). Find the coordinates of \(M_{1}\) and the components of the vector \(\overrightarrow{R M}_{1}\) b. Find the vector \(\overrightarrow{O Z}_{1}\) from the origin to the point \(Z_{1}\) two-thirds of the way along \(\overrightarrow{R M}_{1}\). c. Repeat the calculation of part (b) with the midpoint \(M_{2}\) of \(R Q\) and the vector \(\overrightarrow{P M}_{2}\) to obtain the vector \(\overrightarrow{O Z}_{2}\) d. Repeat the calculation of part (b) with the midpoint \(M_{3}\) of \(P R\) and the vector \(\overline{Q M}_{3}\) to obtain the vector \(\overrightarrow{O Z}_{3}\) e. Conclude that the medians of \(\triangle P Q R\) intersect at a point. Give the coordinates of the point. f. With \(P(2,4,0), Q(4,1,0),\) and \(R(6,3,4),\) find the point at which the medians of \(\triangle P Q R\) intersect.

For the given points \(P, Q,\) and \(R,\) find the approximate measurements of the angles of \(\triangle P Q R\). $$P(0,-1,3), Q(2,2,1), R(-2,2,4)$$
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