Thread:

1. Position Function?

Hey, I am in MAT270 and I can't find how to do this problem anywhere in my book, can someone help me out?

I am going to post a similar problem because I would like to figure out how to do this by the steps taken to find the answer to the similar problem.

"Use the position function f(t) meters to find the velocity at time t=a seconds."

√(t+16) [square root of (t+16)], (a) a=0; (b) a=2

2. Hello !
Originally Posted by Morpha
Hey, I am in MAT270 and I can't find how to do this problem anywhere in my book, can someone help me out?

I am going to post a similar problem because I would like to figure out how to do this by the steps taken to find the answer to the similar problem.

"Use the position function f(t) meters to find the velocity at time t=a seconds."

√(t+16) [square root of (t+16)], (a) a=0; (b) a=2
I'm not in MAT270, nor do I know what this stands for... but it doesn't matter

But this is something commonly used in physics.

If f(t) is the position function, then the speed function is $f'(t)$. And the velocity (which means speed at an instant) at time t=a is $f'(a)$

You can view it this way :
- speed is a ratio distance/time
- the derivative is a ratio "function displacement"/"units"

A formal way to view the velocity is to see that velocity at a time t=a is the difference of distance that occurred instantaneously. More basically, it is the difference of distance that occurred within a very very very little interval of time.

Let $\Delta t$ be this interval of time.

$f(a+\Delta t)-f(a)$ denotes the difference of distance within this very little time (around t=a). Dividing by $\Delta t$ will make the speed.
Because we want this interval to be very small, like null, we'll make $\Delta t \to 0$

So finally, we're looking for $\lim_{\Delta t \to 0} ~ \frac{f(a+\Delta t)-f(a)}{\Delta t}=f'(a)$, by definition.

Is it clear ?

NB : the acceleration function is $f''(t)$

3. Given the position function $f(t) = \sqrt{t + 16}$, first calculate the derivative. The derivative is $f'(t) = \frac{1}{2\sqrt{t + 16}}$. Now that you have the derivative, you can evaluate f'(0) and f'(2).
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