Question

A certain cylindrical tank holds 20,000 gallons of water which can be drained from the bottom of the tank in 20 minutes the volume tea of water remaining in the tank after T minutes is given by the function (v)= 20,000(1-(t/20)^2 where B is in gallons zero is less than or equal to T which less than or equal to 20 is in a minutes and T equals zero represent the instant the tank starts draining. The average rate of change in volume of water in the tank from time t=0 to t=20 is (v(20)-v(0))/20-0=-1000 gallons/minute. At what time t is the instantaneous rate of the water draining from the tank at 1000 gallons/minute

Answer 1

Answer:

Impossible. t=30 minutes.

Step-by-step explanation:

We have the function:

$v(t)=20000(1-\frac{t}{20})^2$

Where v(t) represents the amount of gallons remaining after t minutes.

We want to find at what time t is the instanteous rate of change from the tank 1000 gallons per minute.

In order to determine the instantaneous rate of change, let's find the derivative of our function. So, take the derivative of both sides with respect to our time t:

$\frac{d}{dt}[v(t)]=\frac{d}{dt}(20000(1-\frac{t}{20})^2]$

On the right, let's move the coefficient outside:

$v'(t)=20000\frac{d}{dt}[(1-\frac{t}{20})^2]$

To differentiate, let's use the chain rule, which is:

$u(v(x))=u'(v(x))\cdot v'(x)$

Our u(x) is x² and v(x) is (1-t/20). So, u'(x) is 2x and v'(x) is -1/20. Therefore:

$v'(t)=20000(2(1-\frac{t}{20})\cdot -\frac{1}{20})$

Simplify:

$v'(t)=-1000(2(1-\frac{t}{20}))$

Simplify:

$v'(t)=-1000(2-\frac{t}{10})$

Distribute:

$v'(t)=100t-2000$

So, the instantaneous rate of change after time t is given by the above function.

To find when the instantaneous rate of change of the water is 1000 gallons per minute, substitute 100 for v'(t) and solve for t. So:

$1000=100t-2000$

Solve for t. Add 2000 to both sides:

$3000=100t$

Divide both sides by 100:

$t=30$

So, after 30 minutes, the instantaneous rate of change will be 1000 gallons per minute.

However, if we go back to our original function, our domain t is only defined between 0 and 20 minutes.

So, it is impossible for our instantaneous rate of change to ever reach 1000 gallons per minute.