# Find the instantaneous velocity for a paticle?

The distance moved by a particle is described by the $s(t)=\sin^2(3t)$ where $t$ is time measured in seconds. Use successive estimates of the average velocity between $t=0.5$ and $t=0.5+h$, for smaller and smaller $h$ values (e.g: $h=0.1,0.01,0.001,0.0001,\ldots$) to obtain an estimate for the instantaneous velocity of the particle at time $t=0.5$.

I know that first you plug in $t=0.5$ and find that answer, and then you put in $t=0.6, 0.51, 0.501, 0.5001,\ldots$ and you subtract, for example, the answer you get for $t=0.5$ from the answer you get for $t=0.6$ and then you divide them by $0.6 - 0.5$. I have been doing this and I seem to be getting strange answers and would like someone else to try it. Also, I know you can get this answer by finding the derivative and plugging in $t=0.5$.

I am rusty at derivatives, but is the derivative of the function $6\sin(3x)\cos(3x)$?

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Your description of the process is correct. If you are getting strange results, it may be simply a matter of using the calculator incorrectly. Perhaps if you give the results the problem can be diagnosed. Note that rounding can mess things up, use the memory feature of your calculator. – André Nicolas Jan 20 '13 at 0:50
Another common problem is having the calculator in degree mode! Switch to radians. – André Nicolas Jan 20 '13 at 0:55
s = 0,948379208; 0,998336583; 0,995410696; 0,995038495; 0,995000481 and at t=0,5 it's 0,994996248 – Juris Jan 20 '13 at 0:57
Corresponding velocities: -0,466170401; 0,334033456; 0,414447578; 0,422469006; 0,423270925 – Juris Jan 20 '13 at 0:58
@FlybyNight: If $s(t)=\sin^2(3t)$, then $s'(t)=6\sin(3t)\cos(3t)=3\sin(6t)$. – Clayton Jan 20 '13 at 1:03