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

Answer 1

Christopher Allers

#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#

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Answer 2

Luke Alvarez

To find the instantaneous rate of change for (y = f(t) = -16t^2 + 59t + 39) when (t = 1), we need to find the derivative of the function (f(t)) with respect to (t) and then evaluate it at (t = 1).

First, find the derivative of (f(t)):

[f'(t) = \frac{d}{dt}(-16t^2 + 59t + 39)]

[f'(t) = -32t + 59]

Now, evaluate (f'(t)) at (t = 1):

[f'(1) = -32(1) + 59]

[f'(1) = -32 + 59]

[f'(1) = 27]

Therefore, the instantaneous rate of change of (y = f(t)) at (t = 1) is (27).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.