Consider the equation

Show that satisfies

(Assume the second derivative exists.)

We differentiate (keeping in mind we must use the chain rule to differentiate since it is a function of ),

Thus, .

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Stumbling Robot

A Fraction of a Dot
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Tag: Related Rates

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Establish properties of y” given 3x^2 + 4y^2 = 12

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Show y has fixed sign if x^(1/2) + y^(1/2) = 5

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Find properties of rates of change of x^3 + y^3 = 1

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Find the rate of change of a particle moving along a parabola

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Find the rate of change in volume of a right circular cylinder

The following diagram illustrates the setup:
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Find the rate of change of the area of a triangle as a vertex moves

Here’s a graph of the setup:
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Find the relative rate of change of volume to rate of change of height in a water tank

For this problem we will consider the graph of the following function (the hemisphere tank and water will then be obtained as solids of revolution of this graph about the -axis):
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Use related rates to find the rate at which we need to add water to a leaky tank

The following diagram illustrates the situation:
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Use related rates to find the water level in a tank shaped like a right circular cone

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How fast is a boat approaching a lighthouse

Consider the equation

Show that satisfies

(Assume the second derivative exists.)

We differentiate (keeping in mind we must use the chain rule to differentiate since it is a function of ),

Thus, .

Consider the equation

Assuming exists, show that has a fixed sign without solving for .

We differentiate with respect to , keeping in mind that is a function of so we must use the chain rule to differentiate .

Since , we have implies . Hence, .

Given the equation

This defines as a function of .

- Without solving for show that the derivative satisfies the equation:
(Assume the derivative exists.)

- Show that
whenever . (Assume the second derivative exists.)

- For this part we differentiate each side with respect to (keeping in mind that is a function of , so we need to use the chain rule to differentiate ).
- Using part (a) we differentiate to find ,

Consider a particle moving along the parabola .

- Find the point on the curve at which the rate of change of the abscissa and ordinate are changing at the same rate.
- Find the rate in part (a) if at time we have and .

- Since we have
Thus, when . Then so implies .

- If we have and at time and we have then

Given a right circular cylinder whose radius increases at a constant rate and whose altitude is a linear function of the radius. Also, given that the altitude is increasing at a rate three times that of the radius. The volume is increasing at a rate of 1 cubic feet per second when the radius is 6 feet. When the radius is 36 feet the volume is increasing at a rate of cubic feet per second. Find the value of the integer .

The following diagram illustrates the setup:

Since the altitude (which we denote ) is a linear function of the radius and increases three times as quickly, we have

When , we are given . Thus, we solve for and get .

When , we have

So, since implies and

we have,

Solving for we obtain,

Then, when we have

Let be a right triangle in the plane. The angle at vertex is the right angle, the vertex is fixed at the origin, and the vertex lies on the parabola . At time , the vertex is at the point and moves up the -axis at a constant rate of 2 centimeters per second. What is the rate of change of the area of the triangle at time seconds?

Here’s a graph of the setup:

We are given that the vertex moves upward along the -axis at a constant rate of 2 centimeters per second, this means,

Then, we compute the area of the triangle, call it , in terms of and ,

(Where we used the product rule to differentiate with respect to , recalling that both and are functions of as well.) We are then given the formula for with respect to ,

So, when we have (see the comments for an explanation of how we calculated ) which implies . Therefore,

Consider a water tank shaped like a hemisphere with a radius of 10 feet. At time let,

Find the rate of change of the volume relative to the rate of change of the height () when .

If water is flowing into the tank at a constant rate of cubic feet per second, find when .

For this problem we will consider the graph of the following function (the hemisphere tank and water will then be obtained as solids of revolution of this graph about the -axis):

First, we find a formula for the volume of the water in the tank as a function of . We do this by considering the solid of revolution (for a review of solids of revolution see these exercises) of about the -axis:

Differentiating with respect to we then have,

So, when feet we have

Next, we are given cubic feet per second. We know from above, . So,

Then, to get in terms of we evaluate,

Thus,

So, if , we have

Given a water tank shaped like a right circular cone with radius 15 feet at its base and altitude 10 feet. Water leaks from the tank at a constant rate of 1 cubic foot per second, and water is added to the tank at a constant rate of cubic feet per second. Find the value of so that the water level will be rising at a rate of 4 feet per second when the depth of the water is 2 feet.

The following diagram illustrates the situation:

First, we compute the radius of the water level in terms of the height of the water,

Furthermore, the radius at the water line is 3 feet when the height of the water is 2 feet. So, computing the volume of water in terms of and we have,

(As in the previous exercise, we are careful to use the product rule when evaluating this derivative.)

The problem then gives us the rate of change of volume of the water is

(Since we are adding cubic feet per second, and 1 cubic foot per second is leaking out.) So we set and solve for ,

Given a water tank in the shape of a right circular cone with radius of the base 4 feet and altitude of 10 feet. Water is added to the tank at a constant rate of 5 cubic feet per minute. Find the rate at which the water level is rising when the depth of the water is 5 feet when:

- the vertex of the cone is pointed up;
- the vertex of the cone is pointed down.

- First, we give the general setup for the problem. The following diagram may be helpful,
Next, we compute the radius of the waterline in terms of the height (or altitude) ,

Then, using the formula for the volume of a right circular cone we have and letting be the volume of the water we have is equal to the volume of the entire tank less the volume of the empty (smaller) cone above the water:

This implies

(Here, we needed to use the product rule for derivatives since both and are functions of , so when we differentiate their product we have be careful to use the product rule and get both terms shown above.)

The problem gives us that the rate of change in the volume of water is 5 cubic feet per minute, so,

When , we have . Substituting these values and obtained above we have,

- Part (b) will be almost the same as part (a) with a few minor changes. Here is the diagram we will work from,
Now, the radius of the waterline in terms of the height of the water is

Then, again using the volume of a right circular cone and letting denote the volume of the water (things are a bit simpler this time since the water is a cone, and we don’t have to subtract anything) we have

(As in part (a), we had to be careful to use the product rule since both and are functions of ). The problem then gives us that the rate of change in the volume of water is 5 cubic feet per minute, so again,

When we still have . Substituting these values along with we have

The same answer as in part (a).

Given a boat sailing at a constant speed of 12 miles per hour. The boat is 4 miles offshore, sailing parallel to a straight coast. How fast is it approaching a lighthouse on the shore when it is exactly 5 miles from the lighthouse?

Let

The following diagram illustrates the setup:

The problem gives us that the distance along the shore to the lighthouse is changing at a rate of 12 miles per hour (since the boat is moving at 12 miles per hour and stays parallel to the straight shoreline). Thus, . Then, using the Pythagorean theorem, when we have . Furthermore, solving in terms of and differentiating we have

So, we then have at ,