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Chapter 11: Q. 10 (page 901)

Velocity and acceleration vectors: Find the velocity and acceleration vectors for the given vector functions.
$r (t) = \{t, 2 t^{2}, 3 t^{3}\}$

Short Answer

Expert verified

Ans: $v (t) = <1, 4 t, 9 t^{2}> and a (t) = 鉄� 0, 4, 18 t 鉄�$

Step by step solution

01

Step 1. Given information:

$r (t) = \{t, 2 t^{2}, 3 t^{3}\}$

02

Step 2. Denoting the velocity and acceleration of the vector:

The velocity vector, $v (t)$ is given by

$v (t) = \frac{d}{d t} r (t) = <\frac{d}{d t} x (t), \frac{d}{d t} y (t), \frac{d}{d t} z (t)>$ .

That is, we take the derivative of each component function of $r (t)$ .

The acceleration vector, $a (t)$ is given by

$a (t) = \frac{d}{d t} v (t)$ .

That is, $a (t)$ is obtained by taking the derivative of each component function of $v (t)$ .

03

Step 3. Finding the velocity and acceleration of the vector:

Consider

$r (t) = <t, 2 t^{2}, 3 t^{3}>$

$v (t) = \frac{d}{d t} (r (t)) = \frac{d}{d t} <t, 2 t^{2}, 3 t^{3}> = <\frac{d}{d t} (t), \frac{d}{d t} (2 t^{2}), \frac{d}{d t} (3 t^{3})> = <1, 4 t, 9 t^{2}> a (t) = \frac{d}{d t} (v (t)) = \frac{d}{d t} <1, 4 t, 9 t^{2}> = <\frac{d}{d t} (1), \frac{d}{d t} (4 t), \frac{d}{d t} (9 t^{2})> = 鉄� 0, 4, 18 t 鉄�$

thus

$v (t) = <1, 4 t, 9 t^{2}> and a (t) = 鉄� 0, 4, 18 t 鉄�$

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Most popular questions from this chapter

Given a differentiable vector function $r (t)$ defined on $[a, b]$ , explain why the integralrole="math" localid="1649610238144" ${鈭�}_{a}^{b} ||r^{'} (t)|| d t$ would be a scalar, not a vector.

Let $r (t) = (x (t), y (t)), t 鈭� [a, 鈭�),$ be a vector-valued function, where a is a real number. Under what conditions would the graph of r have a horizontal asymptote as $t 鈫� 鈭� ?$ Provide an example illustrating your answer.

In Exercises 19鈥�21 sketch the graph of a vector-valued function $r (t) = (x (t), y (t))$ with the specified properties. Be sure to indicate the direction of increasing values oft.

Domainlocalid="1649578696830" $t 鈮� 0, r (0) = (0, 3), r (1) = (- 2, 1), \lim_{t 鈫� 鈭�} x (t) = - 5, and \lim_{t 鈫� 鈭�} y (t) = - 鈭� .$

Compute the cross product of the vector functions $r_{1} (t) = (x_{1} (t), y_{1} (t)) and (r_{2} (t) = x_{2} (t), y_{2} (t))$ by thinking of ${鈩�}^{2}$ as the xy-plane in ${鈩�}^{3} .$ That is, let $r_{1} (t) = (x_{1} (t), y_{1} (t), 0) and r_{2} (t) = (x_{2} (t), y_{2} (t), 0),$ and take the cross product of these vector functions.

Let $r (t)$ be a differentiable vector function. Prove that role="math" localid="1649602115972" $\frac{d}{d t} (鈥� r 鈥�) = \frac{1}{鈥� r 鈥�} r 路 r^{r} 路$ (Hint: role="math" localid="1649602160237" $鈥� r {鈥�}^{2} = r 路 r)$

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