Answered: A particle starts at point (1, 0, 1)…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Question

A particle starts at point (1, 0, 1) and moves 2 units along the curve defined by the vector equation r(t)=< e^t, sqrt(2)t, e^(-t) > positively. To find where the particle is after moving 2 units, we need to find the time T that the particle takes to move 2 units. What's the integral to find the time T?

Which of the following is the correct integral set up to find the value of T?
(input a number from below)
(e'+ e¯)dt
= 2
2 (e'+ vZ + e-")dt = 23 ) 17()| dt
= 2
(4) The answer is not given.

Expert Solution

Step 1

Given : $\vec{r (t)} = < x (t), y (t), z (t) > = < e^{t}, \sqrt{2} t, e^{- t} >$

Distance traveled by the particle =2 units

The initial position of particle = $(1, 0, 1)$

To find:- Time taken by particle to move 2 units in the integral form.

As we know that velocity is a vector function and velocity is given by $\vec{r' (t)} = < x' (t), y' (t), z' (t) >$

and the speed is given by $|\vec{r' (t)}| = \sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}}$

• Distance is the integral of speed so, $\int_{t_{i}}^{t_{f}} [\sqrt{{(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2} + {(\frac{d z}{d t})}^{2}}] d t$