Related Rates in Calculus

In calculus, related rates refer to problems that involve finding the rate at which one quantity changes with respect to another, given that the two quantities are related. These types of problems often involve real-world situations where different quantities are changing over time, such as the radius and volume of an inflating balloon or the distance between a moving object and a point of reference.

For example, consider a balloon being filled with air. As the volume of the balloon increases, so does its radius. Related rates allow us to determine how quickly the radius is changing at a particular moment in time if we know how fast the volume is increasing.

What are Related Rates?

Related rates in calculus are problems that involve determining how the rate of change of one quantity is related to the rate of change of another quantity. These problems typically arise in situations where two or more variables are linked by an equation, and the rate at which one variable changes is connected to the rate at which another variable changes.

For example, consider a scenario where water is being poured into a conical tank. As the water level rises, both the height of the water and the radius of the water's surface increase. Related rates problems help us determine how fast the water level rises when the radius is increasing at a certain rate, or vice versa.

How to Solve Related Rates?

To solve related rates, we can use the following steps:

Step 1: Identify what quantities are changing and what rates you need to find.

Step 2: Let x, y, r, h, etc., represent the variables that change over time, these rates are typically in the form dx/dt, dy/dt​, etc., which indicate how fast these variables are changing with respect to time.

Step 3: Establish a relationship between the variables using geometry, trigonometry, or physics. For example, in a problem involving a right triangle, you might use the Pythagorean theorem: x² + y² = z².

Step 4: Differentiate both sides of the equation with respect to time t. Use the chain rule when necessary, since each variable is likely a function of time.

Step 5: After differentiating, substitute the known values for the variables and their rates of change. Be sure to include any given information from the problem.

Step 6: Solve the resulting equation for the unknown rate of change.

Solved Examples: Related Rates in Calculus

Problem 1: A balloon is inflated and its radius increases at a rate of the 2 cm/min. How fast is the volume of the balloon increasing when the radius is 5 cm?

Solution:

Identify Given and Required Quantities:

Given: \frac{dr}{dt} = 2 cm/min ,r = 5 cm

Required: \frac{dV}{dt}

Write Down the Known Relationships:

V = \frac{4}{3} \pi r^3

Differentiate with Respect to Time:

\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right)

\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{dr}{dt}

Substitute Known Values:

\frac{dV}{dt} = 4 \pi (5)^2 \cdot 2 = 200 \pi \, \text{cm}^3/\text{min}

Verify the Units:

The units are correct: cm³/min

Problem 2: A 10-foot ladder is leaning against a wall. The bottom of the ladder is sliding away from the wall at a rate of the 1 foot per second. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 feet from the wall?

Solution:

Identify Given and Required Quantities:

Given: \frac{dx}{dt} = 1 ft/s, x = 6 ft

Required: \frac{dy}{dt}

Write Down the Known Relationships:

x² + y² = 10²

Differentiate with Respect to Time:

2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0

x \cdot \frac{dx}{dt} + y \cdot \frac{dy}{dt} = 0

Substitute Known Values:

6² + y² = 10²

y² = 100 - 36

y = 8 ft

6 \cdot 1 + 8 \cdot \frac{dy}{dt} = 0

\frac{dy}{dt} = -\frac{6}{8} = -\frac{3}{4} \, \text{ft/s}

Verify the Units:

The units are correct: ft/s

Problem 3: The Water is leaking out of a conical tank at a rate of the 10 cubic meters per minute. The tank has a height of the 10 meters and a radius of 5 meters. How fast is the water level dropping when the water is 4 meters deep?

Solution:

Identify Given and Required Quantities:

Given: \frac{dV}{dt} = -10 \, \text{m}^3/\text{min}, h = 10 \, \text{m}, r = 5 \, \text{m}, h_{\text{water}} = 4 \, \text{m}

Required: \frac{dh_{\text{water}}}{dt}

Write Down the Known Relationships:

V = \frac{1}{3} \pi r^2 h

\frac{r}{h} = \frac{5}{10} \implies r = \frac{h_{\text{water}}}{2}

V = \frac{1}{3} \pi \left( \frac{h_{\text{water}}}{2} \right)^2 h_{\text{water}}

V = \frac{1}{3} \pi \cdot \frac{h_{\text{water}}^3}{4}

Differentiate with Respect to Time:

\frac{dV}{dt} = \frac{1}{3} \pi \cdot \frac{3 h_{\text{water}}^2}{4} \cdot \frac{dh_{\text{water}}}{dt}

\frac{dV}{dt} = \frac{1}{4} \pi h_{\text{water}}^2 \cdot \frac{dh_{\text{water}}}{dt}

Substitute Known Values:

-10 = \frac{1}{4} \pi (4)^2 \cdot \frac{dh_{\text{water}}}{dt}

-10 = \frac{1}{4} \pi \cdot 16 \cdot \frac{dh_{\text{water}}}{dt}

\frac{dh_{\text{water}}}{dt} = -\frac{10}{4 \pi \cdot 16} = -\frac{5}{32 \pi} \, \text{m/min}

Verify the Units:

The units are correct: m/min

Problem 4: A car is moving away from a streetlight at a speed of the 30 feet per second. If the streetlight is 20 feet tall and the length of the car’s shadow is increasing how fast is the length of the shadow increasing when the car is 40 feet from the base of the streetlight?

Solution:

Identify Given and Required Quantities:

Given: \frac{dx}{dt} = 30 \, \text{ft/s} height of the streetlight = 20 feet distance of the car from the base = 40 feet

Required: \frac{dL}{dt}

Write Down the Known Relationships:

\frac{\text{height of streetlight}}{\text{height of car}} = \frac{\text{length of shadow}}{\text{distance of car from the base}}

\frac{20}{h} = \frac{L}{40 + L}

Differentiate with Respect to Time:

\frac{d}{dt} \left( \frac{20}{h} = \frac{L}{40 + L} \right)

Use implicit differentiation considering \frac{dL}{dt} and \frac{dx}{dt}.

Substitute Known Values:

h = \frac{20 \cdot (40 + L)}{L}

\frac{dL}{dt} = Solve using similar triangles and differentiation

Verify the Units:

The units of shadow length are correct: ft/s

Practice Questions: Related Rates in Calculus

Q1: At what rate is the shadow of a 10 ft pole lengthening if a light source moves away at 5 ft/s?

Q2: How fast is the radius of a circle increasing if the area is increasing at 8 square units per second?

Q3: At what rate is the water level rising in cone-shaped tank if the volume of water is increasing at 3 cubic meters per minute?

Q4: How fast is the ladder's top sliding down the wall if the bottom is pulled away at 2 m/s?

Q5: How quickly is the distance between the two cars changing if one is moving north at 50 km/h and other east at 60 km/h?

Q6: At what rate is the angle of elevation of a camera changing as a rocket ascends vertically at 100 m/s?

Q7: How fast is the volume of a sphere changing when the radius is increasing at 4 cm/s?

Q8: At what rate is the distance between the tips of the hands of a clock changing at 3:00?

Q9: How fast is the balloon's radius decreasing when it's deflating at a rate of 2 cubic feet per second?

Q10: How quickly is the depth of oil increasing in a cylindrical tank when oil is pumped in at 5 liters per minute?

Read More,

• Calculus in Maths
• Differential Calculus
• Application of Calculus: Real-life Uses
• Average Vs Instantaneous Rate of Change

Conclusion

The Solving related rates problems involves careful application of the calculus principles particularly the chain rule and implicit differentiation. By following a systematic approach—identifying given and required quantities establishing the relationships differentiating with the respect to the time and solving—we can tackle these problems effectively. With practice, we'll become proficient at determining how changing quantities affect each other in the various scenarios.