Related rates is the process of relating two quantities with their respective rates. A change in one quantity will cause change in another quantity. If you know the rate of change of one quantity, you can determine the rate of change of another quantity. Typically, the rates of change are with respect to time (i.e. 2m/s, 5ft/min, 20 miles/hr).
Examples of related rates:
 If we pour water in a vase, the height of the water in the vase increases per unit of time. At the same time, the volume of the water in the vase also increases per unit of time. In this case, the height of the water and the volume of the water is changing at the same time and are related to each other.
 If you pour water into a small puddle, the area of the puddle will increase over a unit of time. At the same time, the radius of the puddle will also increase over a unit of time. Therefore, the radius of the puddle and the area of the puddle are changing at the same time and are related to each other.
There will be three types of related rates problems:
 Area and volume: These problems will give the variables in terms of things such as radius, area, volume, etc.
Below are some common formulas for areas and volumes of common shapes:
2. SOHCAHTOA: These problems will include variables in terms of a side and an angle.
3. Pythagorean Theorem: These types of problems will include variables that represent 2 sides of a right triangle.
Related rates is an application of implicit differentiation. Therefore, before we cover the topic of related rates, we need to briefly go over the concept of implicit differentiation.
Brief Overview of Implicit Differentiation
In implicit differentiation, each term is treated as a function in itself. For example, if I have the equation A + B = C, and A, B, and C are treated as functions, I will differentiate each term as A’, B’ and C’:
The derivative of A + B = C is A’ + B’ = C’
If you have an equation Sin(x) = 3z + 2y^{2}, you will differentiate each term and treat each variable as a function:
By differentiating Sin(x), we obtain x’*Cos(x)
If we take the derivative of 3z, or 3*z, we need to treat z as a function. Therefore, 3 is being multiplied by a function, so we will use the product rule:
f = 3
f’ = 0
g = z
g’ = z’
3*z’ + 0*z = 3z’
Differentiating 2y^{2}, using the chain rule,
u = y
u’ = y’
f = 2(u)^{2},
f’ = 2*2(u) = 4(u)
f’(u)*u’ = 4(y)*y’
Therefore, the derivative of Sin(x) = 3z + 2y^{2} is x’*Cos(x) = 3z’ + 4(y)*y’
For more information on how to derivate by implicit differentiation, visit the article implicit differentiation and watch the accompanying video here.
How To Solve Related Rates Problems
We will use the following mnemonic to solve related rates problems:
Guidelines for Solving RelatedRates Problems:

Example 1:
Suppose we have a ladder with a length of 5m leaning against a wall. The base of the ladder slides away from the wall at the rate of 2m/s. When the top of the ladder is 2m from the floor, find how fast the ladder is coming down.
Solution:
Draw a picture of the problem.
List all the given variables and rates as well as the variables and rates to be determined.
Let’s visit the problem again to determine the given information and the information to be determined.
Suppose we have a ladder with a length of 5m leaning against a wall. The base of the ladder slides away from the wall at the rate of 2m/s. When the top of the ladder is 2m from the floor, find how fast the ladder is coming down.
Given information:
C = 5m (the length of the ladder is 5 m)
A = 2 m (the top of the ladder is 2 m from the floor)
dB/dt = +2 m/s (the base of the ladder is moving to the right of the wall at a rate of 2m/s, so the rate of change is positive).
= 4.582 m (B was determined by using the Pythagorean theorem – c^{2}=a^{2}+b^{2})
dC/dt = 0 m/s (the length of the ladder is not changing)
dA/dt = ? (we need to solve for this quantity).
Write an equation that relates all of the variables and rates together.
If you look at the picture, we essentially have a right triangle. Also, given all three sides of the triangle, we can use the Pythagorean theorem to connect all the variables and rates together.
A^{2}+B^{2}=C^{2}
Perform implicit differentiation on both sides of the equation, treating each variable as a function.
Performing implicit differentiation both sides of the equation, treating A, B, and C as functions,
Each term was differentiated using the chain rule.
Substitute the given variables and rates into the equation.
Solve for the missing variable or rate.
Solving for dA/dt,
The ladder slides down the wall at a rate of 4.582 m/s. Notice that dA/dt is negative, which shows that the ladder is sliding down.
Example 2: The angle is increasing at a rate of 6 radians/hour. At what rate is the side length of x increasing when x is 6 feet?
Solution:
Draw a picture of the problem.
In this case, the picture is already drawn for us.
List all the given variables and rates, as well as the variables and rates to be determined.
Let’s visit the problem again to determine the given information and the information to be determined.
The angle is increasing at a rate of 6 radians/hour. At what rate is the side length of x increasing when x is 6 feet?
Given information:
x = 6 ft
y = 10 ft (hypotenuse of triangle, which is shown in the diagram)
= sin^{1}(6/10) = 0.6435rad (this was determined by using trigonometry)
d/dt = 6 radians/second
dx/dt = ? (we need to determine this quantity)
Write an equation that relates all of the variables and rates together.
If you look at the picture, we essentially have a right triangle. Since we are asked to find the rate of change of and a side length x, we know this is a SOHCAHTOA problem.
Which equation do we use – sin, cos, or tan?
From trigonometry, x is on the opposite side of the angle, and the hypotenuse (which is 10 ft) is given. Therefore, we use SOH – Sin() = opposite/hypotenuse to obtain our equation:
Sin() = x/10
Perform implicit differentiation on both sides of the equation, treating each variable as a function.
Performing implicit differentiation both sides of the equation, treating and x as functions.
Differentiating the left side of the equation and performing the chain rule,
Differentiating the right side of the equation, we can use the quotient rule:
f = x
f’ = x’
g = 10
g’ = 0
Now that we have differentiated both sides of the equation,
Substitute the given variables and rates into the equation.
Solve for the missing variable or rate.
48 ft/hr = x’
The side length x is increasing at a rate of 48ft/hr. Notice that the rate of change of x is positive because the length is increasing.
Example 3: The radius of a circle is decreasing at a rate of 4 cm/min. How fast is the area and circumference of the circle changing when the radius is 8 cm?
Solution:
Draw a picture of the problem.
List all the given variables and rates, as well as the variables and rates to be determined.
Let’s visit the problem again to determine the given information and the information to be determined.
The radius of a circle is decreasing at a rate of 4 cm/min. How fast is the area and circumference of the circle changing when the radius is 8 cm?
Given information:
r = 8 cm
dr/dt = 4 cm/min (the radius is decreasing, hence it is negative)
dC/dt = ? (we need to determine this quantity)
dA/dt = ? (we need to determine this quantity).
Write an equation that relates all of the variables and rates together.
If you look at the picture, we essentially have a circle. Since we are asked to find the rate of change of the circumference and area, we can use the circumference and area equations to relate the variables together.
C = 2*π*r
A = π*r^{2}
We will solve each equation separately.
Perform implicit differentiation on both sides of the equation, treating each variable as a function.
Let’s calculate the circumference. Treating C and r as functions, let’s perform implicit differentiation on both sides of the equation.
On the left side,
On the right side, d/dt(2*π*r), since 2π is being multiplied by a function, r, we can do the product rule:
f = 2π
f’ = 0 (2π is a constant; the derivative of a constant is zero)
g = r
g’ = dr/dt, or r’.
Using the product rule formula,
f’*g + g’*f
0 * r + (dr/dt)* 2π
2π*(dr/dt)
Our differentiated equation is:
Let’s calculate the Area. Treating A and r as functions, let’s perform implicit differentiation on both sides of the equation.
On the left side,
On the right side,
d/dt(π*r^{2})
We can also use the product rule, since π is being multiplied by r^{2}, a function:
f = π
f’ = 0 (π is a constant; the derivative of a constant is zero)
g = r^{2}
g’ = 2r * dr/dt (chain rule)
Using the product rule formula,
f’*g + g’*f
Our differentiated equation is:
Substitute the given variables and rates into the equation and solve for the missing variable or rate.
Let’s start with the circumference:
Now with the area,
Notice that the rate of change for both the circumference and area are negative, which makes sense because the area and circumference are both decreasing.
Example 4: A spherical snowball is melting in such a way that the diameter is decreasing at a rate of 0.3 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 9 cm?
Solution:
Draw a picture of the problem.
List all the given variables and rates as well as the variables and rates to be determined.
Let’s visit the problem again to determine the given information and the information to be determined.
A spherical snowball is melting in such a way that the diameter is decreasing at a rate of 0.3 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 9 cm?
Given information:
D = 9 cm
dD/dt = 0.3 cm/min (the diameter is decreasing, thus the rate is negative)
dV/dt = ? (we need to determine this quantity)
Write an equation that relates all of the variables and rates together.
If you look at the picture, we essentially have a sphere. Since we are asked to find the rate of change of volume of the sphere, we need to use the equation of the volume of a sphere:
In the equation, we are given the diameter D, not r. So, we need everything in terms of the diameter. We know from geometry that r is simply D/2, so we will use that instead of r:
simplifying the expression,
Perform implicit differentiation on both sides of the equation, treating each variable as a function.
Perform implicit differentiation to both sides of the equation, treating V and D as functions,
On the left side, d/dt V is simply dV/dt, of V’.
Let’s differentiate the right side,
Notice that dV/dt is negative, which makes sense because the volume is decreasing.
Example 5: An inverted cone is 10 cm tall, has an opening radius of 5 cm, and was initially full of water. The water now drains from the cone at the constant rate of 15 cm^{3} each second. The water’s surface level falls as a result. At what rate is the water level falling when the water is halfway down the cone?
Solution:
Draw a picture of the problem.
List all the given variables and rates as well as the variables and rates to be determined.
Let’s visit the problem again to determine the given information and the information to be determined.
An inverted cone is 10 cm tall, has an opening radius of 5 cm, and was initially full of water. The water now drains from the cone at the constant rate of 15 cm^{3 }each second. The water’s surface level falls as a result. At what rate is the water level falling when the water is halfway down the cone?
Given information:
h = 10 cm (when the water was full)
r = 5 cm (radius of cone)
h = 5 cm (when the water is halfway down the cone)
dV/dt = 15 cm^{3}/sec (the volume is decreasing, thus the rate is negative)
dh/dt = ? (we need to determine the change in height of the water level)
Write an equation that relates all of the variables and rates together.
If you look at the picture, we essentially have an inverted cone. Since we are asked to find the rate of change of the water level (height) of the cone, we need to use the equation of the volume of a sphere:
Looking at the right side of the equation, we have two variables, r and h. We need to turn the right side into a function of one variable to make the differentiation simpler.
Looking at the diagram, the diagram has a radius of 5 for a height of 10. In other words, the radius is half the height of the cone:
r = ½ h
Therefore, the new equation is
Perform implicit differentiation on both sides of the equation, treating each variable as a function.
Performing implicit differentiation on both sides of the equation, treating V and D as functions,
Notice that dh/dt is negative, which makes sense because the height is decreasing.
Below are some more example problems to familiarize yourself with the process of solving related rates problems.
Ex 2: Related Rates: Determine the Rate of Change of the Area of a Circle With Respect to Time
Ex: Related Rates – Volume of a Melting Snowball
Related Rates – Area of Triangle
Ex: Related Rates – Right Circular Cone
Ex: Related Rates – Rotating Light Projecting on a Wall
Ex: Related Rates Problem — Rate of Change of a Shadow from a Light Pole
Summary of Section
Relates Rates: Using implicit differentiation to find the rate of change of one quantity given the rate of change of another quantity.

References:
http://mathonline.wikidot.com/relatedratesapplications
https://www.matheno.com/calculus1/relatedrates/waterdrainsfromcone/
http://www.chaoticgolf.com/worksheets/calc/academy/4_5band4_6bnotes.pdf
https://calcworkshop.com/triangletrig/sohcahtoa/
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_2_9_ImplicitDiff
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_2_6_RelatedRates
http://www.opentextbookstore.com/appcalc/Chapter211.pdf
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