Rate of Change (check)

[confused_07]

I believe I did it right, just want to make sure:

Suppose that a tank initially contains 2000g of water and the rate of change of its volume after the tank drains for 't' min is '(t)=(.5)t-30 (in gal/min). How much water does the tank contain after it has been draining for 25 min?

int (a=0, b=25) '(t) dt
= int (a=0, b=25) [(.5)t-30] dt
= [(1/3)t^(3/2) - 30t] (a=0, b=25)
= 708.334 gallons

Therefore, 2000 - 708.334 = 1291.666 gallons left in tank after 25 minutes.

 

[skeeter]

please clarify the volume's rate of change. you have posted ...

'(t)=(.5)t-30 (in gal/min)

which actually means \(\displaystyle \L \frac{dV}{dt} = \frac{t}{2} - 30\)

now, could you have possibly meant it to be ...

\(\displaystyle \L \frac{dV}{dt} = \sqrt{t} - 30\)

reason I ask is your antiderivative makes little sense except for the possible rate of change involving the square root of time.

 

[confused_07]

That's how my text book has it (.5)t - 30. So I used (1/2)t when I did my calculations.

 

[skeeter]

then your antiderivative is incorrect ...

\(\displaystyle \L V = 2000 + \int_0^{25} \frac{t}{2} - 30 dt\)

\(\displaystyle \L V = 2000 + \left[\frac{t^2}{4} - 30t \right]_0^{25}\)

\(\displaystyle \L V = 2000 + \left[\frac{25^2}{4} - 30(25)\right] = 1406.25 \, gal\)