Subject: Calculus

Related Rates

Related rates are differential calculus problems which involves rates of change of related variables. Specifically, these are problems that involve finding a rate that a quantity changes by relating that quantity to other quantities whose rates of change are known. For example, say we have function f that describes the number of calculus articles in the website Math^2 at a given time. But when the number of calculus articles changes in time, it is then denoted mathematically as \frac{df}{dt}.Rates are usually associated with respect to time. Thus, in this section, we will be dealing more with derivatives with respect to time.

Don't Worry.. I will give you steps in Solving Related Rates Problems

Some very useful guidelines in solving related rates problems are given below. Take a look and it might help you a lot.

Step 1.
Sketch the figure of the problem whenever possible and try to keep it as simple as possible.

Step 2.
Put and label the correct variables that corresponds to the problem.

Step 3.
Identify and write down the known values of the variables involving constants and their derivative with respect to time t.

Step 4.
Relate the variables that depend on t through an equation.

Step 5.
Find the derivative with respect to time the both sides of equation in step 4 and relate the rates of change of the variables.

Step 6.
Then solve for the value of the desired quantity by substituting the known values obtained in step 3 to the resulting equation in step 5.

Conventions

To further help us in assigning appropriate values of the variables in the problem, it is helpful to put negative (-) sign to a decreasing rate and plus (+) sign to an increasing rate.

Illustration

Suppose we are asked to find the rate of how the height of a water inside a cubic tank (volume=50 m^{3}) changes when it is initially filled with water and as time goes by, a hole is drawn at the bottom, thus discharging water at constant rate of -1.2 m^{3} per minute (note the negative sign). So how do we solve this? Well, let us try to follow above tips. First we draw and then try to identify and label the known values (see the figure at the right.)
Looks like we're finished with step 1 and 2. Next, we will proceed with step 3 and 4 simultaneously. We know that the volume of a cubic tank as given by V=s^{3}, where s is the length of each edge of the cube. But since we have the initial volume to be 50 meters cube, then we can solve for s, thus,

V=s^{3}
50 m^{3}=s^{3}
s=\sqrt[3]{50}m

Now, note that we can't just use the relation V=s^{3} in proceeding to differentiation because obviously, as the tank discharges water, only the height of the water changes, not its sides! So, we have to take another volume equation that relates each box's dimension independently.
Remember that the volume is "length times width times height" and taking into account that length equals width, then we can express the function of the volume to be,

V=l \cdot w \cdot h=s^{2}h

Then, we proceed to step 5. Differentiating the above equation with respect to time upon holding s constant as it should be, we get,

dV=s^{2}\frac{dh}{dt}
\frac{dh}{dt}=\frac{dV}{s^{2}}

Plugging the values of s and dV into the equation above we get

\frac{dh}{dt}=\frac{-1.2m^{3}/min}{(\sqrt[3]{50}m)^{2}}
\frac{dh}{dt}=-13.57m/min

This is where the importance of the sign convention takes place. Even though its obvious that the height of the water decreases as the tank discharges, we still get a "mathematically sound" result telling us that the height decreases because of the negative sign. Thus, as the tank discharges water, the height of the water decreases at an amount of 13.57 meters per minute.

Example #1

A spherical balloon is being pumped with gas at the rate of 10 cubic inches per second. What is the rate of change of the balloon's radius when its diameter is exactly 30 inches?

\frac{dr}{dt}=\frac{\pi}{90} inches per second.

Example #2

An expanding rectangle has an area that is increasing at the rate of 48 square centimeters per second. The rectangle's length is always equal to the square of its width. At what rate is the length increasing at that instant when the width is 2 centimeters?

\frac{dl}{dt}=16 centimeters per second.

Example #3

A bicycle with a velocity of 60 kilometers per hour (km/hr) alongside a long street passes an aircraft travelling upwards with a speed of 20 kilometers per hour. When the aircraft is exactly a kilometer up the moment the bicycle is directly below it, determine the rate of change of the distance between the bike and the aircraft increasing after one minute?

Answer: 52 km/hr

Example #4

A hot air balloon is moving upwards with a speed of 550 mi/hr. A camera man who is taking footage of the rising balloon is 25 miles away from a tower which is located directly below the hot air balloon (click below to see the configuration). Determine the rate at which the angle of elevation of his video device changing when the hot air balloon is 25 miles away from the tower?

The angle of inclination of his camera is changing at 11 radians per hour.

Example #5

In example #4, what is the angle of inclination of the camera when the hot air balloon is exactly 50 miles away from the ground?

60 degrees.

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