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#