Question

in t years, the population of a certain city grows from 400,000 to a size p given by ( p(t) = 400,000 + 1000t^2 ).
a) find the growth rate, ( \frac{dp}{dt} ).
b) find the population after 15 yr.
c) find the growth rate at ( t = 15 ).
d) explain the meaning of the answer to part (c).

Explanation:

Part (a)

Step1: Differentiate \( P(t) \)

We have \( P(t) = 400000 + 1000t^2 \). The derivative of a constant is 0, and the derivative of \( t^n \) is \( nt^{n - 1} \). So, differentiating term by term:
The derivative of \( 400000 \) (a constant) with respect to \( t \) is \( 0 \).
The derivative of \( 1000t^2 \) with respect to \( t \) is \( 1000\times2t^{2 - 1}=2000t \).
So, \( \frac{dP}{dt}=2000t \).

Step1: Substitute \( t = 15 \) into \( P(t) \)

We know \( P(t)=400000 + 1000t^2 \). Substitute \( t = 15 \):
\( P(15)=400000+1000\times(15)^2 \)

Step2: Calculate the value

First, calculate \( (15)^2 = 225 \). Then, \( 1000\times225 = 225000 \).
Then, \( P(15)=400000 + 225000=625000 \).

Step1: Substitute \( t = 15 \) into \( \frac{dP}{dt} \)

We found in part (a) that \( \frac{dP}{dt}=2000t \). Substitute \( t = 15 \):
\( \frac{dP}{dt}\bigg|_{t = 15}=2000\times15 \)

Step2: Calculate the value

\( 2000\times15 = 30000 \)

Answer:

\( \frac{dP}{dt} = 2000t \)

Part (b)