Question

I have this question on my homework and I need help solving it. I do have part of it done, but I need to improvise as far as expressing the problem goes.

Consider the area under one arch of the curve
y(t) = ksin(wt) for t ≥ 0 where k and w are positive constants. Note: Use lower case k and w when needed.

(a) Using the form (Integral from A to B of ksin(wt) dt, find A and B.

Part b asked to find u and du, which I got on my own: u = wt and du = w dt

(c) What are the new endpoints for the integral in terms of u?

I'm pretty sure I can figure out part d, which is asking for the exact area.

Answer 1

you need to solve y(t) = 0

y(t) = ksin(wt)
0 = ksin(wt)
ksin(wt) = 0
k = 0 or sin(wt) = 0

let's assume k is nonzero, that would have to mean that sin(wt) = 0

what is t if sin(wt) = 0 ?

Answer 2

I've done a bit more work since I posted this earlier. I'm on the last couple parts now, c and d. A I found to be zero, since it says t ≥ 0. Then I found B to be pi/w. So, substituting in u and du, A stays 0, but what is B?

Answer 3

u = w*t


if t = 0 (first root where t >= 0), then
u = w*t
u = w*0
u = 0
so A = 0


if t = pi/w (second root after t = 0)
u = w*t
u = w*pi/w
u = pi
and B = pi

Answer 4

Alright, so let's see... k is pulled out of that equation since it is a constant, I'm assuming. So that turns out looking like this? \[k \int\limits_{0}^{\pi}\sin(u)du\]
Sorry, I'm not that great at calculus, so I'm probably going to be coming here for help a lot in the future.

Answer 5

you're missing one thing though

Answer 6

u = w*t
du = w*dt ----> dt = du/w

so you should have du/w and not just du

Answer 7

what you can do is think of du/w as du*(1/w) then pull out the constant 1/w

Answer 8

you should really have this integral

\[\Large \frac{k}{w} \int_{0}^{\pi} \sin(u)du\]

Answer 9

Oh, I was closer than I thought I'd be. According to the new equation, that results in something like this? \[k/w * (1-\cos(\pi*t))/t\]
I got the (1-cos(pi*t))/t as the result of the integral without the k/w.

Answer 10

don't forget about the limits of integration

Answer 11

The limits? You mean from 0 to pi?

Answer 12

yes

Answer 13

Let \[\LARGE g(u) = \int \sin(u)du\]
you should find that \[\LARGE g(u) = -cos(u) + C\]
to compute \[\LARGE \frac{k}{w}\int_{0}^{\pi} \sin(u)du\], you'll need to compute \[\LARGE \frac{k}{w}*\left(g(\pi) - g(0)\right)\]

Answer 14

g(pi) I get to be 1, and g(0) I get to be -1, so g(pi) - g(0) is 2, correct? So that comes out to 2k/w.

Answer 15

yep,
\[\LARGE \frac{k}{w}\int_{0}^{\pi} \sin(u)du=\frac{2k}{w}\]
is correct

Answer 16

Weird, I kept getting something else for that one part of the equation for some odd reason. But, I ended up with the right answer, so I'll go back and look through this once I've got some free time on my hands after classes tomorrow.

I still have a couple more problems for this homework assignment that I need help on. They have multiple parts, but it is the last two problems of my homework. Should I start a new thread?

Answer 17

yes it's always a good idea to start a new post for a new problem (to avoid clutter and lag)

Answer 18

Alright, thank you very much for the help. And your patience. :)

Answer 19

you're welcome