# Calculus and Physics Help!

If a particle's position is given by $x = 4-12t+3t^2$ (where $t$ is in seconds and $x$ is in meters):

a) What is the velocity at $t = 1$ s?

Ok, so I have an answer:

$v = \frac{dx}{dt} = -12 + 6t$

At $t = 1$, $v = -12 + 6(1) = -6$ m/s

But my problem is that I want to see the steps of using the formula $v = \frac{dx}{dt}$ in order to achieve $-12 + 6t$...

I am in physics with calc, and calc is only a co-requisite for this class, so I'm taking it while I'm taking physics. As you can see calc is a little behind. We're just now learning limits in calc, and I was hoping someone could help me figure this out.

-
It's too bad English composition isn't a prereq... –  rschwieb Sep 20 '12 at 17:52
it's nice that you want to ascribe units to your stuff, but those units are obviously wrong. –  Jonathan Sep 21 '12 at 4:45

There is a differentiation rule called the power rule for power law functions which states that

$\frac{d}{dx}(x^n)=nx^{n−1}$

You should be able to prove this using the definition of a derivative (in terms of limits) that you have learned. Another rule is that the derivative of a sum of functions is equal to the sum of the derivatives of those individual functions (which is also easy to prove). So you can differentiate each term in your expression separately, and add the results. The only difference is that here your independent variable is time (t) rather than x.

-

You see the problem here is that the question is asking for a velocity at $t=1$. This means that they require and instantaneous velocity which is by definition the derivative of the position function at $t=1$. If you don't want to use derivative rules for some reason and you don't mind a little extra work then you can calculate the velocity from a limit. (In reality this is the same thing as taking the derivative as you will later see. A derivative is just a limit itself.)

To take the instantaneous velocity, we use the formula $\overline{v} = \frac{\Delta x}{\Delta t}$. I use $\overline{v}$ to denote the average velocity. Between time $t$ and $t+\Delta$ we have $$\overline{v} = \frac{x(t+\Delta t) - x(t)}{\Delta t} = \frac{\left(3(t+\Delta t)^2 - 12(t+\Delta t)+4\right)-\left(3t^2-12t+4\right)}{\Delta t}$$ Simplifying a bit $$\overline{v} = \frac{6t\Delta t + (\Delta t)^2 -12\Delta t}{\Delta t}$$ Now comes the calculus. If we take the limit as $\Delta t \rightarrow 0$, that is if we take the time interval to be smaller and smaller so that the average velocity approaches the instantaneous, then the instantaneous velocity $v$ is $$v = \lim_{\Delta t\rightarrow 0}\frac{6t\Delta t + (\Delta t)^2 -12\Delta t}{\Delta t}=\lim_{\Delta t\rightarrow 0}\left(6t-12+\Delta t\right)=6t-12$$ Notice that this is exactly the derivative you had before and the above steps are in fact a calculation of the derivative from first principles.

-

Think it this way. On a $x$ versus $t$ plot. The slope of the line is the rate of change of $x$ with respect to $t$ which is given by,

$lim \ \Delta t\rightarrow0 \ \frac{\Delta x}{\Delta t}=\frac{dx}{dt}=v$