Solving Related Rates Problems

Related rates problems help find changing quantities in real time, commonly used in physics, motion, and engineering.

Neetesh Kumar | February 08, 2025 $\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space$ Share this Page on:

Table of Content:

1. Definition of Related Rates Problems
2. Formula Used in Related Rates Problems
3. Solved Examples
4. Conclusion

Definition of Related Rates Problems:

Related Rates Problems:
Related rates problems involve finding the rate of change of one quantity when another related quantity is changing. These problems often involve using implicit differentiation to relate the rates of change of two or more variables.

The key idea is to differentiate both sides of an equation that relates the variables, then solve for the desired rate of change.

Formula Used in Related Rates Problems:

The formula typically used in related rates problems is:

$\boxed{\frac{d}{dt} \big[ f(x(t), y(t)) \big] = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}}$

Where:

$\frac{d}{dt} \big[ f(x(t), y(t)) \big]$ is the derivative of the function with respect to time.
$\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ are the rates of change of the variables $x$ and $y$ with respect to time.
$\dfrac{\partial f}{\partial x}$ and $\dfrac{\partial f}{\partial y}$ are the partial derivatives of $f$ with respect to $x$ and $y$ , respectively.

Derivative Examples with Solutions:

Example 1: Changing Radius of a Circle

The radius of a circle is increasing at a rate of $\frac{dr}{dt} = 2 \, \text{cm/s}$ . Find the rate of change of the area of the circle when the radius is 5 cm.

Solution:

We know the area of a circle is given by:

$A = \pi r^2$

Using the chain rule, differentiate both sides with respect to time:

$\dfrac{dA}{dt} = 2\pi r \cdot \dfrac{dr}{dt}$

Substitute the known values:

$\dfrac{dA}{dt} = 2\pi (5) \cdot (2) = 20\pi \, \text{cm}^2/\text{s}$

Thus, the rate of change of the area is: $\boxed{20\pi \, \text{cm}^2/\text{s}}$

Example 2: The Distance Between Two Moving Objects

Two cars are moving along straight paths. Car 1 is traveling east at a speed of 60 km/h, and car 2 is traveling north at a speed of 80 km/h. How fast is the distance between the two cars changing when they are 3 km apart?

Solution:

Let $x$ represent the distance traveled by car 1 (east), and $y$ represent the distance traveled by car 2 (north). The distance between the two cars is given by:

$z = \sqrt{x^2 + y^2}$

Differentiating both sides with respect to time:

$\dfrac{dz}{dt} = \dfrac{d}{dt} \left[ \sqrt{x^2 + y^2} \right] = \dfrac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}$

Substitute the known values:

$\dfrac{dx}{dt} = 60 \, \text{km/h}$
$\dfrac{dy}{dt} = 80 \, \text{km/h}$
$x = 3 \, \text{km}$
$y = 4 \, \text{km}$
$z = 5 \, \text{km}$

Now, substitute these values into the equation:

$\dfrac{dz}{dt} = \dfrac{(3)(60) + (4)(80)}{5}$

$\dfrac{dz}{dt} = \dfrac{180 + 320}{5} = \dfrac{500}{5} = 100 \, \text{km/h}$

Thus, the distance between the two cars is changing at a rate of: $\boxed{100 \, \text{km/h}}$

Example 3: Rate of Change of Volume of a Sphere

The radius of a sphere is increasing at a rate of $\dfrac{dr}{dt} = 0.5 \, \text{cm/s}$ . Find the rate of change of the volume of the sphere when the radius is 3 cm.

Solution:

The volume of a sphere is given by:

$V = \dfrac{4}{3} \pi r^3$

Using the chain rule, differentiate both sides with respect to time:

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

Substitute the known values:

$\frac{dV}{dt} = 4\pi (3)^2 \cdot (0.5) = 4\pi (9) \cdot (0.5) = 18\pi \, \text{cm}^3/\text{s}$

Thus, the rate of change of the volume is: $\boxed{18\pi \, \text{cm}^3/\text{s}}$

Example 4: Rate of Change of Distance Between Two Objects

A boat is moving west at 10 m/s, and a plane is flying at 500 m/s. The boat and plane are 1 km apart, and the plane is directly above the boat at a height of 1 km. Find the rate at which the distance between the boat and the plane is changing after 1 second.

Solution:

Let $x$ represent the horizontal distance of the boat from the starting point, and $y$ represent the height of the plane from the ground. The distance between the boat and the plane is given by:

$z = \sqrt{x^2 + y^2}$

Differentiating both sides with respect to time:

$\dfrac{dz}{dt} = \dfrac{d}{dt} \left[ \sqrt{x^2 + y^2} \right] = \dfrac{x \frac{dx}{dt} + y \frac{dy}{dt}}{z}$

Substitute the known values:

$\frac{dx}{dt} = 10 \, \text{m/s}$
$\frac{dy}{dt} = 500 \, \text{m/s}$
$x = 10 \, \text{m}$
$y = 1000 \, \text{m}$
$z = \sqrt{10^2 + 1000^2} = \sqrt{100 + 1000000} = \sqrt{1000100} \approx 1000.05 \, \text{m}$

Now, substitute these values into the equation:

$\dfrac{dz}{dt} = \dfrac{(10)(10) + (1000)(500)}{1000.05}$

$\dfrac{dz}{dt} = \dfrac{100 + 500000}{1000.05} \approx \dfrac{500100}{1000.05} \approx 500 \, \text{m/s}$

Thus, the rate at which the distance between the boat and the plane is changing is: $\boxed{500 \, \text{m/s}}$

Conclusion:

Related rates problems are crucial for finding the rate of change of one variable in relation to another.
These problems are widely used in physics, engineering, and economics, particularly in problems involving motion, volume changes, and electrical circuits.
Understanding how to set up and differentiate these equations is essential for solving real-world applications.