The velocity of a particle moving along a line is given by v(t) = e ^ t * sin(e ^ t) the interval 0<= t<= pi 2 . What is the distance that the particle traveled on this time interval?

Answer:

0.44237 units

Step-by-step explanation:

Given

[tex]v(t) = e^t*sin(e^t)[/tex]

[tex]0 \le t \le \frac{\pi}{2}[/tex]

Required

The distance traveled in this interval

We have:

[tex]v(t) = e^t*sin(e^t)[/tex]

The distance is calculated as:

[tex]d(t) = \int\limits^a_b {v(t)} \, dt[/tex]

So, we have:

[tex]d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(e^t)} \, dt[/tex]

Let:

[tex]u = e^t[/tex]

Differentiate

[tex]\frac{du}{dt} = e^t[/tex]

So:

[tex]dt = e^{-t} du[/tex]

So, we have:

[tex]d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(e^t)} \, dt[/tex]

[tex]d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t*sin(u)} \, e^{-t} du[/tex]

Rewrite as:

[tex]d(t) = \int\limits^{\frac{\pi}{2}}_0 {e^t* e^{-t}*sin(u)} \, du[/tex]

[tex]d(t) = \int\limits^{\frac{\pi}{2}}_0 {sin(u)} \, du[/tex]

Integrate:

[tex]d(t) = -\cos(u)|\limits^{\frac{\pi}{2}}_0[/tex]

Substitute [tex]u = e^t[/tex]

[tex]d(t) = -\cos(e^t)|\limits^{\frac{\pi}{2}}_0[/tex]

Split

[tex]d(t) = -\cos(e^\frac{\pi}{2}) - [-\cos(e^0)][/tex]

[tex]d(t) = -\cos(e^\frac{\pi}{2}) +\cos(e^0)[/tex]

[tex]d(t) = -0.09793 + 0.5403[/tex]

[tex]d(t) = 0.44237[/tex]

The distance traveled in this interval is: 0.44237 units

dt =

d(t) = \int\limits^2_0 {e^t*sin(e^t)} \, dt

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