Instantaneous Rate of Change of Height

Calc12

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Nov 17, 2010
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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?



thanks in advance
Code:
 
Hello, Calc12!

You left out a "square" in the function . . .


A soccor ball is kicked into the air.
The polynomial function: f(t)=4.9t2+16t+1 represents the height in meters at t seconds.\displaystyle \text{The polynomial function: }\:f(t) \:=\: -4.9t^2 +16t + 1\,\text{ represents the height in meters at }t\text{ seconds.}

Determine the instantaneous rate of change of height at 1 second, 2 seconds and 3 seconds.

You want to use: f(t)  =  limh0f(t+h)f(t)h\displaystyle \text{You want to use: }\:f'(t) \;=\;\lim_{h\to0}\frac{f(t+h) - f(t)}{h}

. . And you should get: f(t)  =  9.8t+16\displaystyle \text{And you should get: }\:f'(t) \;=\;-9.8t + 16


Then evaluate: f(1),  f(2),  f(3).\displaystyle \text{Then evaluate: }\:f'(1),\;f'(2),\;f'(3).

 
Thanks again Soroban.. You don't even understand how much of a saviour you and this site truly is!
 
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