# Instantaneous Rate of Change of Height

• Nov 17th 2010, 07:47 PM
Instantaneous Rate of Change of Height
A soccor ball is kicked into the air. The polynomial function f(t) = -4.9^2 +16t + 1, where the f(t) represents the height in metres at t seconds models this scenario.
Determine the instantaneous rate of change of height at 1 second, 2 seconds and 3 seconds.

Im thinking I use this:
f ' (1) = lim_____ f(1 + h) - f(1)
______ h→0______h

and so forth for 2 seconds, and 3 seconds..

But I got 87.71 m/s for 1 second - doesn't seem right?

• Nov 17th 2010, 08:16 PM
Educated
Is it supposed to be: $\displaystyle f(t) = -4.9t^2 +16t + 1$

$\displaystyle \lim \limits_{h \rightarrow 0} \dfrac{f(1+h) - f(1)}{h}$

$\displaystyle \lim \limits_{h \rightarrow 0} \dfrac{[-4.9(1+h)^2 +16(1+h) +1] - [-4.9(1)^2 +16(1) +1]}{h}$

$\displaystyle \lim \limits_{h \rightarrow 0} \dfrac{-4.9 -4.9h^2 - 9.8h +16+16h +1 + 4.9 -16 -1}{h}$

$\displaystyle \lim \limits_{h \rightarrow 0} -4.9h +6.2 = 6.2$

Now can you do it for 2 seconds and 3 seconds?

The easier way would be to straight differentiate it to get the rate of change.

$\displaystyle f(x) = x^n$
$\displaystyle f'(x) = nx^{n-1}$
• Nov 18th 2010, 08:03 AM