# Thread: Antiderivative/Integral Problem

1. ## Antiderivative/Integral Problem

Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data

Find a function describing the position of the particle.

If I could get help on this I would greatly appreciate it. Basically, I found the v(t) equation and then the s(t) equation through integrals/antiderivatives, but my solution isn't right when I try to submit it. s(t) = -(10sint)-(3cost)-((17/2pi)t)+12

2. Originally Posted by cdlegendary
Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data

Find a function describing the position of the particle.

If I could get help on this I would greatly appreciate it. Basically, I found the v(t) equation and then the s(t) equation through integrals/antiderivatives, but my solution isn't right when I try to submit it. s(t) = -(10sint)-(3cost)-((17/2pi)t)+12
$a(t) = 10\sin{t} + 3\cos{t}$

$v(t) = \int{10\sin{t} + 3\cos{t}\,dt}$

$= -10\cos{t} + 3\sin{t} + C$.

$s(t) = \int{-10\cos{t} + 3\sin{t} + C\,dt}$

$= -10\sin{t} - 3\cos{t} + Ct + D$.

Now using the intital conditions:

$9 = -10\sin{0} - 3\cos{0} + c(0) + D$

$9 = -3 + D$

$D = 12$.

$-8 = -10\sin{2\pi} - 3\cos{2\pi} + 2\pi C + D$

$-8 = -3 + 2\pi C + 12$

$-17 = 2\pi C$

$C = -\frac{17}{2\pi}$.

Therefore $s(t) = -10\sin{t} - 3\cos{t} - \frac{17t}{2\pi} + 12$.

3. Originally Posted by Prove It
$a(t) = 10\sin{t} + 3\cos{t}$

$v(t) = \int{10\sin{t} + 3\cos{t}\,dt}$

$= -10\cos{t} + 3\sin{t} + C$.

$s(t) = \int{-10\cos{t} + 3\sin{t} + C\,dt}$

$= -10\sin{t} - 3\cos{t} + Ct + D$.

Now using the intital conditions:

$9 = -10\sin{0} - 3\cos{0} + c(0) + D$

$9 = -3 + D$

$D = 12$.

$-8 = -10\sin{2\pi} - 3\cos{2\pi} + 2\pi C + D$

$-8 = -3 + 2\pi C + 12$

$17 = 2\pi C$

$C = \frac{17}{2\pi}$.

Therefore $s(t) = -10\sin{t} - 3\cos{t} + \frac{17t}{2\pi} + 12$.
Thanks! That's the same answer I came up with, but my Math homework is entered on an online site, and it seems to not like my notation I suppose..

4. Originally Posted by cdlegendary
Let s(t) denote the position of a particle at time t, and let v and a be the velocity and acceleration respectively. The particle is moving according to the data

Find a function describing the position of the particle.

If I could get help on this I would greatly appreciate it. Basically, I found the v(t) equation and then the s(t) equation through integrals/antiderivatives, but my solution isn't right when I try to submit it. s(t) = -(10sint)-(3cost)-((17/2pi)t)+12

$v(t)=\int a(t)\,dt=\int\left(10\sin t+3\cos t\right)\,dt=-10\cos t+3\sin t+C_1\,,\,\,s(t)=$ $\int v(t)\,dt\int\left(-10\cos t+3\sin t+C_1\right)\,dt$ $=-10\sin t-3\cos t+C_1t+C_2$ ,

$C_1\,,\,C_2=$ constants wrt the variable t. Now:

$9=s(0)=-3+C_2\,\Longrightarrow,C_2=12$

$-8=s(2\pi)=-3+2\pi C_1+12\,\Longrightarrow\,C_1=-\frac{17}{2\pi}$

And, as far as I can see, your solution is correct...check there isn't a minus sign or something like that in the definition of $a(t)$ ...

Tonio

5. Originally Posted by Prove It
$a(t) = 10\sin{t} + 3\cos{t}$

$v(t) = \int{10\sin{t} + 3\cos{t}\,dt}$

$= -10\cos{t} + 3\sin{t} + C$.

$s(t) = \int{-10\cos{t} + 3\sin{t} + C\,dt}$

$= -10\sin{t} - 3\cos{t} + Ct + D$.

Now using the intital conditions:

$9 = -10\sin{0} - 3\cos{0} + c(0) + D$

$9 = -3 + D$

$D = 12$.

$-8 = -10\sin{2\pi} - 3\cos{2\pi} + 2\pi C + D$

$-8 = -3 + 2\pi C + 12$

$17 = 2\pi C$

$C = \frac{17}{2\pi}$.

Therefore $s(t) = -10\sin{t} - 3\cos{t} + \frac{17t}{2\pi} + 12$.

I think a minus sign is missing in $C = \frac{17}{2\pi}$

Tonio

6. Originally Posted by tonio
I think a minus sign is missing in $C = \frac{17}{2\pi}$

Tonio
Yes, edited. Thanks.

7. Thanks again you two. I figured it out, the application accepted the answer when I entered it as -(10sin(t))-(3cos(t))-(((17t)/(2pi)))+12