Chapter 11: Q. 14 (page 851)

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

Short Answer

Expert verified

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

Step by step solution

Part (a) Step 1. Given Information

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

Part(a) Step 2. Finding the displacement vector

r(t) = t, t² a = 2, b = 3.

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

Part (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 $\sqrt{26}$ .

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

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

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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

Short Answer

Step by step solution

Part (a) Step 1. Given Information

Part(a) Step 2. Finding the displacement vector

Part (b) Step 1. Magnitude of displacement vector

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

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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

Short Answer

Step by step solution

Part (a) Step 1. Given Information

Part(a) Step 2. Finding the displacement vector

Part (b) Step 1. Magnitude of displacement vector

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Logic and Functions

Decision Maths

Calculus

Probability and Statistics

Discrete Mathematics

Study anywhere. Anytime. Across all devices.

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