Find (a) the displacement vectors from r(a) tor(b), (b) the magnitude of the displacement vector, and (c) the distance travelled by a particle on the curve from a to b. r(t) = t, t2 a = 2, b = 3

(a) The displacement vector from r(a) to r(b) is 1,5.(b) Magnitude of displacement vector is 26.(c) Thus the distance travelled by the particle from a to b is&#16...

Q. 14 - Calculus | StudyQuestionHub

Q. 14

Question

Find

(a) the displacement vectors from r(a) tor(b),

(b) the magnitude of the displacement vector, and

r(t) = t, t² a = 2, b = 3

Step-by-Step Solution

Verified

Answer

$(a) The displacement vector from r (a) t o r (b) i s <1, 5> . (b) Magnitude of displacement vector is \sqrt{26} . (c) Thus the distance travelled by the particle from a to b is \frac{3 \sqrt{37}}{2} - \sqrt{17} + \frac{1}{4} \ln (\frac{6 + \sqrt{37}}{4 + \sqrt{17}}) .$

1Part (a) Step 1. Given Information

The given function is r(t) = t, t² a = 2, b = 3.

2Part(a) Step 2. Finding the displacement vector

r(t) = t, t² a = 2, b = 3.

$= r (b) - r a (a) The displacement vector from r (a) t o r (b) = r (b) - r a (a) = r (3) - r (2) = <3, 3^{2}> - <2, 2^{2}> = <3, 9> - <2, 4> = <3, - 2, 9 - 4> = <1, 5> The displacement vector from r (a) t o r (b) i s <1, 5>$

3Part (b) Step 1. Magnitude of displacement vector

Magnitude of displacement vector = $||<1, 5>||$

$= \sqrt{1^{2} + 5^{2}} = \sqrt{26}$

Magnitude of displacement vector is $\sqrt{26}$ .

4Part (c) Step 1. The distance travelled by a particle on the curve from a to b.

$r (t) = <t, t^{2}> r' (t) = <1, 2 t> ||r' (t)|| = \sqrt{1 + {(2 t)}^{2}} = \sqrt{1 + 4 t^{2}} The distance travelled by the particle from a to b is \int_{a}^{b} ||r' (t)|| d t = \int_{2}^{3} \sqrt{1 + 4 t^{2}} d t = \int_{2}^{3} \sqrt{t^{2} + {\frac{1}{2}}^{2}} d t = \int \sqrt{a^{2} + x^{2}} d x = \frac{x}{2} \sqrt{a^{2} + x^{2}} + \frac{a^{2}}{2} \sin h^{- 1} (\frac{x}{a}) = 2 {[\frac{t}{2} \sqrt{t^{2} + {(\frac{1}{2})}^{2}} + \frac{1}{8} \sinh^{- 1} (\frac{t}{\frac{1}{2}})]}_{2}^{3} = [3 \sqrt{9 + \frac{1}{4}} + \frac{1}{4} \sinh^{- 1} (6)] - [2 \sqrt{4 + \frac{1}{4}} + \frac{1}{4} \sinh^{- 1} (4)] = [\frac{3 \sqrt{37}}{2} + \frac{1}{4} \ln (6 + \sqrt{37})] - [\sqrt{17} + \frac{1}{4} \ln (4 + \sqrt{17})] \sinh^{- 1} (x) = \ln (x + \sqrt{x^{2} + 1}) \frac{3 \sqrt{37}}{2} - \sqrt{17} + \frac{1}{4} \ln (\frac{6 + \sqrt{37}}{4 + \sqrt{17}}) Thus the distance travelled by the particle from a to b is \frac{3 \sqrt{37}}{2} - \sqrt{17} + \frac{1}{4} \ln (\frac{6 + \sqrt{37}}{4 + \sqrt{17}}) .$

Q. 1

Other exercises in this chapter

Q. 1

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a count

View solution

Q. 1TB

Parametric equations for the unit circle: Find parametric equations for the unit circle centered at the origin of the xy-plane that satisfy the given conditions

View solution

Q. 2 Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) Parametric equations with three components for acircular helix winding around the x-axis

View solution