# Math Help - Physics / Rate of Change / Velocity

1. ## Physics / Rate of Change / Velocity

Can't figure this one out! I tried, but I have major doubts about my answer:

The number of gallons of water in a tank t minutes after the tank has started to drain is Q(t) = 200(30 - t)^2 . How fast is the water running out at the end 10 minutes? What is the average rate at which the water flows out during the first 10 minutes?

2. Originally Posted by djalali59
Can't figure this one out! I tried, but I have major doubts about my answer:

The number of gallons of water in a tank t minutes after the tank has started to drain is Q(t) = 200(30 - t)^2 . How fast is the water running out at the end 10 minutes? What is the average rate at which the water flows out during the first 10 minutes?

The exact rate at t=10 is the derviative evaluated at t=10

$\frac{dQ}{dt}=400(30-t)(-1)\bigg|_{t=10}=400(30-10)(-1)=-8000$

The average rate of change is just the slope of the line at the points

$(0,Q(0)), (10,Q(10))$ so we get

$r_{ave}=\frac{Q(10)-Q(0)}{10-0}=\frac{200(30-10)^2-200(30)^2}{10}=$
$20(20^2-30^2)=20(50)(-10)=-10000$

3. Originally Posted by TheEmptySet
The exact rate at t=10 is the derviative evaluated at t=10

$\frac{dQ}{dt}=400(30-t)(-1)\bigg|_{t=10}=400(30-10)(-1)=-8000$

The average rate of change is just the slope of the line at the points

$(0,Q(0)), (10,Q(10))$ so we get

$r_{ave}=\frac{Q(10)-Q(0)}{10-0}=\frac{200(30-10)^2-200(30)^2}{10}=$
$20(20^2-30^2)=20(50)(-10)=-10000$
but i thought for the derivative of the function, you had to multiply everything out, and then figure out each of the derivatives? Or was I mistaken?

4. Originally Posted by djalali59
but i thought for the derivative of the function, you had to multiply everything out, and then figure out each of the derivatives? Or was I mistaken?
you can do it both ways ... easier to do it the way ES did it using the chain rule for derivatives.