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Chapter 12: Problem 17

Consider the following position functions. a. Find the velocity and speed of the object. b. Find the acceleration of the object. $$\mathbf{r}(t)=\left\langle 1, t^{2}, e^{-t}\right\rangle, \text { for } t \geq 0$$

Short Answer

Expert verified
Answer: The velocity function is $$\mathbf{v}(t) = \left\langle 0, 2t, -e^{-t} \right\rangle$$, the speed is $$\text{Speed} = \| \mathbf{v}(t) \| = \sqrt{4t^2 + e^{-2t}\ }$$, and the acceleration function is $$\mathbf{a}(t) = \left\langle 0, 2, e^{-t} \right\rangle$$.

Step by step solution

01

Calculate the Velocity function

To calculate the velocity function, we need to find the derivative of the given position function with respect to time. So, the velocity function will be the first derivative of the position function $$ \mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \left\langle \frac{d}{dt}(1), \frac{d}{dt}(t^2), \frac{d}{dt}(e^{-t})\right\rangle $$ Now, we differentiate each component with respect to t: $$ \mathbf{v}(t) = \left\langle 0, 2t, -e^{-t} \right\rangle $$
02

Calculate the Speed of the object

The speed is the magnitude of the velocity vector. To find the magnitude, we take the square root of the sum of squares of the individual components of the velocity vector: $$ \text{Speed} = \| \mathbf{v}(t) \| = \sqrt{(0)^2 + (2t)^2 + (-e^{-t})^2} = \sqrt{4t^2 + e^{-2t}\ } $$
03

Calculate the Acceleration function

To find the acceleration function, we need to take the derivative of the velocity function with respect to time. The acceleration function will be the first derivative of the velocity function: $$ \mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \left\langle \frac{d}{dt}(0), \frac{d}{dt}(2t), \frac{d}{dt}(-e^{-t})\right\rangle $$ Now, we differentiate each component with respect to t: $$ \mathbf{a}(t) = \left\langle 0, 2, e^{-t} \right\rangle $$ The velocity, speed, and acceleration functions have been found as: 1. Velocity function: $$\mathbf{v}(t) = \left\langle 0, 2t, -e^{-t} \right\rangle$$ 2. Speed: $$\text{Speed} = \| \mathbf{v}(t) \| = \sqrt{4t^2 + e^{-2t}\ } $$ 3. Acceleration function: $$\mathbf{a}(t) = \left\langle 0, 2, e^{-t} \right\rangle $$

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Most popular questions from this chapter

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 \(s=t=0\).

An object on an inclined plane does not slide provided the component of the object's weight parallel to the plane \(\left|\mathbf{W}_{\text {par }}\right|\) is less than or equal to the magnitude of the opposing frictional force \(\left|\mathbf{F}_{\mathrm{f}}\right|\). The magnitude of the frictional force, in turn, is proportional to the component of the object's weight perpendicular to the plane \(\left|\mathbf{W}_{\text {perp }}\right|\) (see figure). The constant of proportionality is the coefficient of static friction, \(\mu\) a. Suppose a 100 -lb block rests on a plane that is tilted at an angle of \(\theta=20^{\circ}\) to the horizontal. Find \(\left|\mathbf{W}_{\text {parl }}\right|\) and \(\left|\mathbf{W}_{\text {perp }}\right|\) b. The condition for the block not sliding is \(\left|\mathbf{W}_{\mathrm{par}}\right| \leq \mu\left|\mathbf{W}_{\text {perp }}\right| .\) If \(\mu=0.65,\) does the block slide? c. What is the critical angle above which the block slides with \(\mu=0.65 ?\)

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.

Relationship between \(\mathbf{T}, \mathbf{N},\) and a Show that if an object accelerates in the sense that \(d^{2} s / d t^{2}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). If an object decelerates in the sense that \(d^{2} s / d t^{2}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N}\)

Prove that the midpoint of the line segment joining \(P\left(x_{1}, y_{1}, z_{1}\right)\) and \(Q\left(x_{2}, y_{2}, z_{2}\right)\) is $$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)$$
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