/*! 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 64 Find the points (if they exist) ... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 64

Find the points (if they exist) at which the following planes and curves intersect. $$y=1 ; \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Short Answer

Expert verified
Question: Find the points of intersection between the plane \(y=1\) and the curve given by the parametric equation \(\mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle\) for \(0\leq t \leq 2 \pi\). Answer: The points of intersection are \((5\sqrt{3}, 1, 1)\) and \((-5\sqrt{3}, 1, 1)\).

Step by step solution

01

Set the y-coordinate equal to 1

Since our plane is given by the equation \(y=1\), we want to find the points on the curve where the \(y\)-coordinate of \(\mathbf{r}(t)\) is equal to 1. For our curve, the \(y\)-coordinate is given by \(2 \sin t\). So, we want to solve the equation: $$2 \sin t = 1$$
02

Solve for t

To solve the equation \(2 \sin t = 1\), first divide both sides by 2: $$\sin t = \dfrac{1}{2}$$ Now, find the values of \(t\) that satisfy this equation in the given range \(0\leq t \leq 2 \pi\). We know that \(\sin^{-1}(1/2) = \dfrac{\pi}{6}\) and \(\sin^{-1}(-1/2) = \dfrac{5\pi}{6}\). Thus, the values of \(t\) that satisfy this equation are: $$t = \dfrac{\pi}{6}, \quad \dfrac{5\pi}{6}$$
03

Substitute back into the parametric equation

Now that we have the \(t\) values at which the \(y\)-coordinate of the curve is equal to 1, we will substitute these values back into the curve's parametric equation \(\mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle\) to find the corresponding points of intersection. For \(t = \dfrac{\pi}{6}\): $$\mathbf{r}\left(\dfrac{\pi}{6}\right) = \left\langle 10 \cos \dfrac{\pi}{6}, 2 \sin \dfrac{\pi}{6}, 1\right\rangle = \left\langle 5\sqrt{3}, 1, 1\right\rangle$$ For \(t = \dfrac{5\pi}{6}\): $$\mathbf{r}\left(\dfrac{5\pi}{6}\right) = \left\langle 10 \cos \dfrac{5\pi}{6}, 2 \sin \dfrac{5\pi}{6}, 1\right\rangle = \left\langle -5\sqrt{3}, 1, 1\right\rangle$$
04

State the points of intersection

The two points at which the plane \(y=1\) intersects the curve \(\mathbf{r}(t) = \langle 10 \cos t, 2 \sin t, 1\rangle\) are: $$(5\sqrt{3}, 1, 1) \quad \text{and} \quad (-5\sqrt{3}, 1, 1)$$

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