The number of toy kangaroos, K, in a toy box after 't' days is given by #K=t^2+20t#. Estimate the rate at which the number of kangaroos is changing after 3 days?

#K=t^2+20t#

Answer 1

After 3 days, the number of kangaroos is increasing by 26 kangaroos per day.

The derivative of a function represents its rate of change.

First take the derivative of #K=t^2+20t#. The derivative of #t^n# is #nt^(n-1)# by the power rule. So the derivative of #t^2# is #2t#. The derivative of #at# is just #a#, so the derivative of #20t# is just #20#. You should end up with #K'=2t+20# when you add them together.
The question wants to know the rate of change after 3 days, so just plug in 3: #K'=2(3)+20# #K'=26# So there you have it- after 3 days, the number of kangaroos is increasing by 26 kangaroos per day.
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Answer 2

To estimate the rate at which the number of kangaroos is changing after 3 days, we can use the derivative of the function ( K(t) = t^2 + 20t ) with respect to time ( t ) and evaluate it at ( t = 3 ).

( \frac{dK}{dt} = 2t + 20 )

At ( t = 3 ), ( \frac{dK}{dt} = 2(3) + 20 = 6 + 20 = 26 ) kangaroos per day.

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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.

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.

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.

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.

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