Population Prob: growth rate of P'(t) = 4500 sqrt(t) + 1000

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Jade

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The development of Astroworld will increase the city's population (P(t)) at the rate of P'(t) = 4500 sqrt(t) + 1000 people/year t years from the start of construction. The population before construction is $30,000. Determine the projected population 9 years after construction of the park has begun.

I am thinking that you need to find the antiderivative then plug in 9 for t?

P(t) = something + 1000t + c

Am I on the right track?
 
Antiderivative of 4500 sqrt (t)

I believe I can change sqrt (t) into t^-1 so then it would be 4500t^-1 then I can take the antiderivative correct?
 
Re: Antiderivative of 4500 sqrt (t)

Jade said:
I believe I can change sqrt (t) into t^-1 so then it would be 4500t^-1 then I can take the antiderivative correct?
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No. Why do you think that?

sqrt(t) = t^(½)

That would NOT be what you have written.

t^(-1) = 1/t -- Most certainly not a square root.
 
Sorry

I am confusing it with 1/x

the derivative of t^(1/2) would be 2/3t^(3/2)

you need to figure in the 4500 as well so it should be 3000t^(3/2)
 
I'm wrong

if you plug in zero to find the constant then i am coming up with zero as the constant which is not right. Doesn't the 30,000 fit into this somehow.
 
This is my conclusion

the antiderivative of 4500 sqrt(t) + 1000 is 3000t^(2/3) + 1000t + C

I plugged in 30,000 for the constant because that was the population before the construction and then plugged in 9 for t

P(9)=120,000
 
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