# Thread: Finding the instantaneous rate of change

1. ## Finding the instantaneous rate of change

A soccer ball is kicked into the air such that its height, in m, after t seconds, can be modelled by the function h(t)=-4.9t^2+12t+0.5. Determine the instantaneous rate of change of the height of the ball after 1 s.

I tried doing it using derivatives, but I got this unfactorable numerator.

2. ## Re: Finding the instantaneous rate of change

Find h'(t) then, h'(1)

3. ## Re: Finding the instantaneous rate of change

Originally Posted by Dragon08
A soccer ball is kicked into the air such that its height, in m, after t seconds, can be modelled by the function h(t)=-4.9t^2+12t+0.5. Determine the instantaneous rate of change of the height of the ball after 1 s.

I tried doing it using derivatives, but I got this unfactorable numerator.
$\displaystyle \lim_{t \to 1} \frac{(-4.9t^2+12t+0.5) - (-4.9+12+0.5)}{t-1}$

$\displaystyle \lim_{t \to 1} \frac{-4.9t^2+12t - 7.1}{t-1}$

$\displaystyle \lim_{t \to 1} \frac{(-4.9t+7.1)(t-1)}{t-1}$

$\displaystyle \lim_{t \to 1} (-4.9t+7.1) = 2.2$ m/s

4. ## Re: Finding the instantaneous rate of change

There are other ways to find a derivative than using the definition!

However, the definition is, of course,
$\displaystyle \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}$
Obviously, the denominator is going to 0. In order that the limit exist, the numerator must also go to 0. And if a polynomial goes to 0 as h goes to 0, The polynomial must be 0 at h= 0- and if a p(a)= 0, x- a is a factor.

5. ## Re: Finding the instantaneous rate of change

Oh, okay. Thanks!