U-Substitution and redefining boundaries

Question

My homework sets up the problem accordingly:

An object moves horizontally in one dimension with a velocity given by ​v(t) = $8\cos\left(\frac{\pi \cdot t}{6}\right)$ m/s.

Find the The position of the object is given by ​s(t) = $s\left(t\right)=\int _0^t\:v\left(y\right)\:dy\:$ for $t\ge 0$. Find the position function for all $t\ge 0$.

I find this problem differently worded than any other u-substitution problem I've worked on, and I'm having trouble figuring it out. Apparently I can use this relationship:

$\int_a^b\:f\left(g\left(x\right)\right)g'\left(x\right)dx\:=\:\int_{g\left(a\right)}^{g\left(b\right)}f\left(u\right)du\:$

...Which I've used before. I assume g(x) would equal my u-substitution, which is $\frac{\pi \cdot t}{6}$ I presume - but what confuses me are the boundaries, one of which is a variable. Could someone walk me through this?

There is also a follow up question:

What is the period of the motion - that ​is, starting at any​ point, how long does it take for the object to return to that​ position?

Since the period of the sine function is $2\pi$, do I just set the resulting equation to that and solve?

Answer 1

Well, the task tells you two things:

1. The velocity, $v(t)$, is given as $v(t)=8\cos\frac{\pi-t}{6}$
2. The location is $\int_0^t v(y) dy$.

From $1$, you know that $$v(t)=8\cos\frac{\pi-t}{6}$$ meaning that $$v(y)=8\cos\frac{\pi-y}{6}$$

From the second, you then get

$$s(t)=\int_0^t 8\cos\frac{\pi-y}{6}dy$$

which is a basic integral that you should be able to calculate.

Answer 2

You need find the function position $s(t)$, how you know the functon velocity $v(t)$ your problem is equal to find indefined integral $\int v(t)dt$. Let be $g(t) = \frac{\pi - t}{ 6}$, this implies that $\frac{dg}{dt} = -1/6$ and therefore $$ v(t) = -48 g'\cos{g}$$ this implies that

$$\int v(t) dt = \int 48 g'\cos(g) dt = \int -48 g'\cos(g) dt = -48 \int\cos(g)dg = -48 \sin(\frac{\pi - t}{ 6}). $$

Answer 3

To evaluate $$\int_a^bk\cos cx\; dx$$ Put $$u=cx$$ $$du = c\; dx$$ so $$dx = \frac1c (c\; dx) = \frac1c \; du$$ $$x=a \iff u = ca$$ $$x=b \iff u = cb$$ and $$\int_a^b k\cos cx\; dx = \frac kc\int_{ca}^{cb}\cos u\; du = \left[\frac kc\sin u\right]_{ca}^{cb}= \frac kc(\sin cb - \sin ca)$$ In your case, $a=0$, $b=t$, $k=8$, and $c=\pi/6$, so the value is $$\boxed{\frac{48}{\pi}\sin \frac{\pi t}{6}}$$