Cosine Integration Problem (at least I think)?

Question

Okay, I give up. I believe that to actually solve this problem is beyond my current abilities. I've approximated the solution to be 304.64, but now I want to know the exact answer.

The problem is: An object moves at a rate of [20+7cos(x)] m/s for 15 s. How far did it travel?

I know I'm close, because if it had traveled a little longer then the rate would have averaged perfectly to 20 m/s, so the answer should be just a little over 15*20, which mine is.

Is this problem a lot easier to do than I think? I started calculus last week, and I just picked up this problem as a challenge. If integrating a trig function is actually what I have to do, and is as complex as it looks, then all I need is to be told so.

Answer 1

distance = integration of speed with respect to x

distance
= integral of (20 + 7 cos x) dx
= integral of 20 dx + integral of (7 cos x) dx
= 20 x + 7 sin x

It travels for 15 s so
distance
= (20 x 15) + (7 sin 15) - [(20 x 0) + (7 sin 0)]
301.18117333 meter

Answer 2

If the velocity speed depends on the *time*, as v = 20 + 7 cos t, the problem is fairly easy:

[A] ... x = INT (20 + 7 cos t) dt
(x from 0 to 15)
[B] ... = 20 t + 7 sin t
[C] ... = (20 * 15 + 7 * sin 15) - 0
[D] ... = 300 + 4.552 = 304.552 m

(Antila's answer is incorrect because it calculated the sine of 15 degrees, but when integrating trig functions you must work with radians.)

HOWEVER

Your formula suggests that the speed depends on the *position* (x), and that makes the problem MUCH more difficult.

For the integral
[1] ... t = INT dx/(a + b cos x)
calculate D = sqrt(a^2 - b^2) and S = a + b,
then
[2] ... t = INT = (2/D) * inv tan {(D/S) tan (x/2)}
In your case, a = 20, b = 7, D = 18.735, S = 27, t = 15. We must solve for x.

The problem is that the inv tan is multi-valued. When x grows from 0 to pi radians, the tan (x/2) goes from 0 to infinity, and therefore the inv tan from 0 to pi/2. This shows that every distance of pi takes time
[3] ... T = (2/D) * pi/2 = pi / D = 0.1677 s.
The total time is
[4] ... t = 15 = 89.5430 T = 90 T - 0.5470 T
so that the distance D is covered 90 pi minus the distance corresponding to the inv tan of the integral running from -0.5470 T to 0 seconds.

That distance is solved as follows:
-0.5470 (2/D) * pi/2 = (2/D) inv tan {(D/S) tan (x/2)}
(D/S) tan (x/2) = tan (-0.5470 pi/2) = 1.1597
x = 2 inv tan (-1.6713) = -2.0632 m

The total distance is therefore
x = 90 pi - 2.0632 = 280.68 m


This is precisely the value I found when I run a computer simulation with steps of milliseconds.

Answer 3

Well, integrating trig functions actually isn't hard at all. The integral of cos(x)dx is just sin(x) + C. Of course things change a bit when you have a coefficient in front of the trig function, but it's not too hard of a conceptual leap to make. I hope that hint helps you.

Answer 4

Does the x in the cosine represent time? Then it is correct.

Answer 5

yep, that's what you need to do