How to find instantaneous rate of change for $y=f(t)=-16(t^2)+59t+39$ when t=1?

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Answer 1 · 2016-12-09T00:19:58

$27$

The instantaneous rate of change of $y=f(t)$ when $t=1$ is given by $f'(1)$

We have

$f(t)=-16t^2+59t+39$

Differentiating wrt $t$ we get:

$f'(t)=-32t+59$

So when $t=1$ we have:

$f'(1)=-32+59 = 27$

Hence, the instantaneous rate of change of $y=f(t)$ when $t=1$ is $27$

How to find instantaneous rate of change for y=f(t)=-16(t^2)+59t+39 y=f(t)=-16(t^2)+59t+39 when t=1?