It might be assumed intuitively that the water level will decrease faster as it lowers towards the tip of the cone.

To model this, relate the radius of the plane of the water to the height of the water:

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

To find the rate of change, take the derivative with respect to time:

The term  is unknown and is what we're solving for, but we're told that .

At height :

It might be assumed intuitively that the water level will decrease faster as it lowers towards the tip of the cone.

To model this, relate the radius of the plane of the water to the height of the water:

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

To find the rate of change, take the derivative with respect to time:

The term  is unknown and is what we're solving for, but we're told that .

At height :

The rate of flow for the pool can be found by taking the total amount of water that flowed from the hose (the 500 gallons) divided by the amount of time it took to give those 500 gallons (3 hours or 180 minutes). Therefore, the rate of flow of the hose is determined to be 500/180 = 2.78 gallons/min. Afterwards, in order to develop an expression for the amount of water in the pool at a given time, you can take the integral of the flow. This integral will sum all the water that has been added up to that point and give you the total amount of water in the pool.

This makes sense because 2.78 gallons are being added every minutes, so the total number of minutes passed multipled by the 2.78 gallons gives the total amount of gallons in the pool. Notice that the C constant was not included, this is because the pool is originally empty. If the pool was not originally empty, the intial amount of water in the pool would have been set as the value of C.

To determine rate of flow, we can take the first derivative of our function by using the power rule: 

At , 

Write the volume of the cylinder.

Substitute the radius.

Differentiate the volume with respect to time.

Substitute the given rate.

The height is increasing at a rate of  inches per second.

The position s(t) is equal to ∫v(t) dt = ∫ 4t + 5 dt = 2t² + 5t + C

Now, since we know that the initial position of the particle (at t = 0) is 0, we know C is 0.  Therefore, s(t) = 2t² + 5t + C

To find the time t when at which the velocity is 45, set v(t) equal to 45. 45 = 4t + 5 → 40 = 4t → t = 10

The position of the particle is s(10) = 2 * 10² + 50 = 200 + 50 = 250

The position of a particle is defined by s(t) = 4t³ – 3t² + 2t.  At what time (to the nearest hundreth) is its velocity equal to 384?

Start by finding the velocity function:

v(t) = s'(t) = 12t² – 6t + 2

To find the time, set v(t) equal to 384: 384 = 12t² – 6t + 2

To solve, set equal the equation equal to 0: 12t² – 6t – 382 = 0

Use the quadratic formula:

(–(–6) ± √(36 – 4*12*(–382)))/(2 * 12)

= (6 ± √(36 + 18336))/24 = (6 ± √(18372))/24 = (6 ± 2√(4593))/24

= (3 ± √(4593))/12 

The initial position of a particle is 44.5. If its velocity is described by v(t) = 3t + 12, what is its position (to the nearest hundreth) at the time when the velocity is equal to 8391?

To solve for t, set v(t) equal to 8391: 3t + 12 = 8391; 3t = 8379; t = 2793

Now, the position function is equal to ∫v(t)dt = ∫ 3t + 12 dt = (3/2)t² + 12t + C

Now, since the initial position is 44.5, C is 44.5; therefore, s(t) = (3/2)t² + 12t + 44.5

To solve for the position, solve s(2793) = (3/2)2793² + 12*2793 + 44.5 = 1.5 * 7800849 + 33516 + 44.5 = 11701273.5 + 33516 + 44.5

Take the derivative of the position function to obtain the velocity function.

We want to know the time when the velocity is -8.  Substitute v into the equation to find t.

To find the time when the ball hits the ground, set  and solve for .

Separate each term and solve for t.

Since negative time does not exist, the only possible answer is .