# Thread: Integration of Sine, Pi, and Solving for t

1. ## Integration of Sine, Pi, and Solving for t

My class is doing a test review for homework, and I was doing well so far until I bumped into this:

A particle moves along the x-axis in such a way that its acceleration at time t for t (greater than or equal to) 0 is given by a(t)=2 cos(t). At time t=0, the velocity of the particle is v(0)= -1 and its position is x(0)= 0.

a. Write an equation for the velocity for the velocity v(t) of the particle.
Already found this, it was: v(t)=2 sin t -1

b. Write an equation for the position x(t) of the particle.
Found this also, which was: x(t)= -2 cos t - t +2

c. For what values of t, 0< t < $\displaystyle \pi$, is the particle at rest? (less thans are less than or equal to).
The problem I have with this is isolating t using: -1= -2 cos t - t +2

I already did a similar problem, but it was much simpler to isolate t to one side and solve for it. (2$\displaystyle \pi$-2$\displaystyle \pi$sin 2$\displaystyle \pi$t = 0 ----> t=0.25)

Another little question while I'm on the subject, how do I $\displaystyle \frac{d}{dx} tan (4x + 9x^2)?$ I'm too used to single terms.

Some help would be very much appreciated.

2. Originally Posted by DarkestEvil
My class is doing a test review for homework, and I was doing well so far until I bumped into this:

A particle moves along the x-axis in such a way that its acceleration at time t for t (greater than or equal to) 0 is given by a(t)=2 cos(t). At time t=0, the velocity of the particle is v(0)= -1 and its position is x(0)= 0.

a. Write an equation for the velocity for the velocity v(t) of the particle.
Already found this, it was: v(t)=2 sin t -1

b. Write an equation for the position x(t) of the particle.
Found this also, which was: x(t)= -2 cos t - t +2

c. For what values of t, 0< t < $\displaystyle \pi$, is the particle at rest? (less thans are less than or equal to).
The problem I have with this is isolating t using: -1= -2 cos t - t +2

to find when the particle is at rest, set v(t) = 0, not x(t).

I already did a similar problem, but it was much simpler to isolate t to one side and solve for it. (2$\displaystyle \pi$-2$\displaystyle \pi$sin 2$\displaystyle \pi$t = 0 ----> t=0.25)

Another little question while I'm on the subject, how do I $\displaystyle \frac{d}{dx} tan (4x + 9x^2)?$ I'm too used to single terms.

chain rule ... $\displaystyle \textcolor{red}{\frac{d}{dx}\tan{u} = \sec^2{u} \cdot \frac{du}{dx}}$
...

3. Hmmm... I did what you said, and used v(t)=2 sin t - 1
So: 2 sin t - 1 = 0 ---> sin t = 1/2 ---> t=30
Ah, thanks. I just remembered $\displaystyle \pi$= 180 degrees, thanks.

Do I do the same to this equation?
v(t)=2$\displaystyle \pi$ - 2$\displaystyle \pi$ sin 2$\displaystyle \pi$t = 0
But then I get stuck at sin 2$\displaystyle \pi$t = 1

Sorry, I'm just a high school student. ^^"

4. Originally Posted by DarkestEvil

So: 2 sin t - 1 = 0 ---> sin t = 1/2 ---> t=30 ... degrees are not the correct solution to these types of equations.

note that $\displaystyle \textcolor{red}{t = \frac{\pi}{6}}$ is not the only solution in the given interval

v(t)=2$\displaystyle \pi$ - 2$\displaystyle \pi$ sin 2$\displaystyle \pi$t = 0
But then I get stuck at sin 2$\displaystyle \pi$t = 1
$\displaystyle 2\pi - 2\pi \sin(2\pi t) = 0$

$\displaystyle 2\pi[1 - \sin(2\pi t)] = 0$

$\displaystyle \sin(2\pi t) = 1$

$\displaystyle 2\pi t = \frac{\pi}{2}$

$\displaystyle t = \frac{1}{4}$

5. Ah, I see, because 30 degrees is the same as $\displaystyle \frac{\pi}{6}$. Thanks a lot.