# Related rate

## Homework Statement

The amount of water in a storage tank, in gallons, is modeled by a continuous function on the time interval 0<=t<=7, where t is measured in hours. In this model, rates are give as:

(i)The rate at which water enters the tank is f(t)=100t^2*sin(sqrt(t))
(ii) The rate at which water leaves the tank is
g(t)=250 for 0<=t<3 gallons/hour
g(t)=2000 for 3<t<=7 gallons/hour

The graphs of f and g intersect at t=1.617 and 5.076. At time t=0, the amount of water in the tank is 5000 gallons.

a. For 0<=t<=7, find the time intervals during which the amount of water in the tank is decreasing.
b. For 0<=t<=7, at what time t is the amount of water in the tank greatest ? To the nearest gallon, compute the amount of water at this time.

None

## The Attempt at a Solution

a. I got 2 intervals: 3<t<5.076 and 0<t<1.617 because g(t) >f(t) in these two intervals. Am I right ??
b. How should I do part b ?? I kinda get the idea in my head that g(t) should be as small as possible and f(t) has to be as big as possible. Can anyone give me a hint ??

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Here's how I would do it:

First, make one equation. So f(t) - g(t)
Then find the derivative of that equation for both t<3 and t>3. From there you should be able to find the answer to ( a ). Derivatives describe the behavoir of the original equation. They give the slope of the tangent to the equation at a certain point.

for ( b ) find the maximum amount of water in the tank. With differentiable equations continuous for the whole domain which is being tested, the maximum or minimum can be found at the ends of the domain or the points where the derivative = 0 or undefined. Find these points and compare them.

So, mathematically:

f(t)=100t^2 * sin( \sqrt{t}) - 250
or
f(t)=100t^2 * sin( \root{t}) - 2500

The derivative is

f'(t) = [200t * sin(\root(t))] * [100t^2 \cdot cos(\root(t))* \frac{1}{2}t^-\frac{1}{2} ]

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