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Chapter 12: Problem 16

Consider the following position functions. a. Find the velocity and speed of the object. b. Find the acceleration of the object. $$\mathbf{r}(t)=\langle 3 \sin t, 5 \cos t, 4 \sin t\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Short Answer

Expert verified
Based on the given position function $$\mathbf{r}(t)=\langle 3 \sin t, 5 \cos t, 4 \sin t\rangle$$, what are the expressions for the following: a. Velocity vector b. Speed c. Acceleration vector

Step by step solution

01

Find the velocity vector

To find the velocity vector, we need to differentiate the position function with respect to time (t). Given position function: $$\mathbf{r}(t)=\langle 3 \sin t, 5 \cos t, 4 \sin t\rangle.$$ Differentiate each component with respect to t: $$\frac{d}{dt}(3 \sin t) = 3 \cos t,$$ $$\frac{d}{dt}(5 \cos t) = -5 \sin t,$$ $$\frac{d}{dt}(4 \sin t) = 4 \cos t.$$ So, the velocity vector is given by: $$\mathbf{v}(t)=\langle 3 \cos t, -5 \sin t, 4 \cos t \rangle.$$
02

Find the speed of the object

The speed of an object is the magnitude of its velocity vector. Therefore, we need to find the magnitude of the velocity vector found in step 1. $$\text{Speed} = \|\mathbf{v}(t)\| = \sqrt{(3 \cos t)^2 + (-5 \sin t)^2 + (4 \cos t)^2}.$$ Now, simplifying the expression: $$\text{Speed} = \sqrt{9 \cos^2 t + 25 \sin^2 t + 16 \cos^2 t} = \sqrt{25 \sin^2 t + 25 \cos^2 t}.$$ Using the identity \(\sin^2 t + \cos^2 t = 1\), we get: $$\text{Speed} = \sqrt{25( \sin^2 t + \cos^2 t)} = \sqrt{25} = 5.$$ So, the speed of the object is constant and equal to 5 units.
03

Find the acceleration vector

To find the acceleration vector, we need to differentiate the velocity vector with respect to time (t). Given velocity vector: $$\mathbf{v}(t)=\langle 3 \cos t, -5 \sin t, 4 \cos t \rangle.$$ Differentiate each component with respect to t: $$\frac{d}{dt}(3 \cos t) = -3 \sin t,$$ $$\frac{d}{dt}(-5 \sin t) = -5 \cos t,$$ $$\frac{d}{dt}(4 \cos t) = -4 \sin t.$$ So, the acceleration vector is given by: $$\mathbf{a}(t)=\langle -3 \sin t, -5 \cos t, -4 \sin t \rangle.$$ Thus, the solution to the given exercise is: a. The velocity of the object is given by: $$\mathbf{v}(t)=\langle 3 \cos t, -5 \sin t, 4 \cos t \rangle$$ and the speed is constant and equal to 5 units. b. The acceleration of the object is given by: $$\mathbf{a}(t)=\langle -3 \sin t, -5 \cos t, -4 \sin t \rangle.$$

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Most popular questions from this chapter

Assume that \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) are vectors in \(\mathrm{R}^{3}\) that form the sides of a triangle (see figure). Use the following steps to prove that the medians intersect at a point that divides each median in a 2: 1 ratio. The proof does not use a coordinate system. a. Show that \(\mathbf{u}+\mathbf{v}+\mathbf{w}=\mathbf{0}\) b. Let \(\mathbf{M}_{1}\) be the median vector from the midpoint of \(\mathbf{u}\) to the opposite vertex. Define \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) similarly. Using the geometry of vector addition show that \(\mathbf{M}_{1}=\mathbf{u} / 2+\mathbf{v} .\) Find analogous expressions for \(\mathbf{M}_{2}\) and \(\mathbf{M}_{3}\) c. Let \(a, b,\) and \(c\) be the vectors from \(O\) to the points one-third of the way along \(\mathbf{M}_{1}, \mathbf{M}_{2},\) and \(\mathbf{M}_{3},\) respectively. Show that \(\mathbf{a}=\mathbf{b}=\mathbf{c}=(\mathbf{u}-\mathbf{w}) / 3\) d. Conclude that the medians intersect at a point that divides each median in a 2: 1 ratio.

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