Chapter 11: Q. 45 (page 872)

Use the given velocity vectors $v (t) = r' (t)$ and initial positions in exercise to find the position function $r (t)$ .

Short Answer

Expert verified

$r (t) = (\frac{t^{2}}{2} + 3) i + (\frac{t^{2}}{3} - 4) j$ is the required position function.

Step by step solution

Given Information

Consider $v (t) = <t, t^{2}>, r (0) = <3, - 4>$

The objective is to find the position function $r (t)$ .

Since $v (t) = r' (t)$ we have $r (t) = \int v (t) d t$

$= \int <t, t^{2}> d t = i \int t d t + j \int t^{2} d t = \frac{t^{2}}{2} i + \frac{t^{3}}{3} j + c,$

where $c$ is a vector constant.

$= (\frac{t^{2}}{2} + c_{1}) i + (\frac{t^{3}}{3} + c_{2}) j$

Where $c_{1}$ and $c_{2}$ are scalars.

Calculation

Check: taking the derivative of $r (t), (\frac{t^{2}}{2} + 3) i + (\frac{t^{3}}{3} - 4) j$ we obtain $r' (t) = t i + t^{2} j$ ,

which is the correct tangent vector function.

Next, we evaluate

$r (0) = (\frac{0^{2}}{2} + 3) i + (\frac{0^{3}}{3} - 4) j = 3 i - 4 j = <3, - 4>$ ,

which is true according to the problem.

Thus $r (t) = (\frac{t^{2}}{2} + 3) i + (\frac{t^{3}}{3} - 4) j$ is the required position function.

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Use the given velocity vectors $v (t) = r' (t)$ and initial positions in exercise to find the position function $r (t)$ .

Short Answer

Step by step solution

Given Information

Calculation

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Use the given velocity vectors v(t)=r'(t)and initial positions in exercise to find the position function r(t).

Short Answer

Step by step solution

Given Information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

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Use the given velocity vectors $v (t) = r' (t)$ and initial positions in exercise to find the position function $r (t)$ .