Related rates

Example

A 10-meter ladder is leaning against the wall of a building, and the base of the ladder is sliding away from the building at a rate of 3 meters per second. How fast is the top of the ladder sliding down the wall when the base of the ladder is 6 meters from the wall?

The distance between the base of the ladder and the wall, x, and the height of the ladder on the wall, y, represent the sides of a right triangle with the ladder as the hypotenuse, h. The objective is to find dy/dt, the rate of change of y with respect to time, t, when h, x and dx/dt, the rate of change of x, are known.

Step 1:

• ${\displaystyle x=6}$
• ${\displaystyle h=10}$
• ${\displaystyle \frac{dx}{dt}=3}$
• ${\displaystyle \frac{dh}{dt}=0}$
• ${\displaystyle \frac{dy}{dt}=\text{?}}$

Step 2: From the Pythagorean theorem, the equation

${\displaystyle x^2+y^2=h^2,}$

describes the relationship between x, y and h, for a right triangle. Differentiating both sides of this equation with respect to time, t, yields

${\displaystyle \frac{d}{dt}\left(x^2+y^2\right)=\frac{d}{dt}\left(h^2\right)}$

Step 3: When solved for the wanted rate of change, dy/dt, gives us

${\displaystyle \frac{d}{dt}\left(x^2\right)+\frac{d}{dt}\left(y^2\right)=\frac{d}{dt}\left(h^2\right)}$

${\displaystyle (2x)\frac{dx}{dt}+(2y)\frac{dy}{dt}=(2h)\frac{dh}{dt}}$

${\displaystyle x\frac{dx}{dt}+y\frac{dy}{dt}=h\frac{dh}{dt}}$

${\displaystyle \frac{dy}{dt}=\frac{h\frac{dh}{dt}-x\frac{dx}{dt}}{y}.}$

Step 4 & 5: Using the variables from step 1 gives us:

${\displaystyle \frac{dy}{dt}=\frac{h\frac{dh}{dt}-x\frac{dx}{dt}}{y}.}$

${\displaystyle \frac{dy}{dt}=\frac{10\times 0-6\times 3}{y}=-\frac{18}{y}.}$

Solving for y using the Pythagorean Theorem gives:

${\displaystyle x^2+y^2=h^2}$

${\displaystyle 6^2+y^2={10}^2}$

${\displaystyle y=8}$

Plugging in 8 for the equation:

${\displaystyle -\frac{18}{y}=-\frac{18}{8}=-\frac{9}{4}}$

It is generally assumed that negative values represent the downward direction. In doing such, the top of the ladder is sliding down the wall at a rate of ⁠9/4⁠ meters per second.

Physics examples

Because one physical quantity often depends on another, which, in turn depends on others, such as time, related-rates methods have broad applications in Physics. This section presents an example of related rates kinematics and electromagnetic induction.

Relative kinematics of two vehicles

For example, one can consider the kinematics problem where one vehicle is heading West toward an intersection at 80 miles per hour while another is heading North away from the intersection at 60 miles per hour. One can ask whether the vehicles are getting closer or further apart and at what rate at the moment when the North bound vehicle is 3 miles North of the intersection and the West bound vehicle is 4 miles East of the intersection.

Big idea: use chain rule to compute rate of change of distance between two vehicles.

Plan:

1. Choose coordinate system
2. Identify variables
3. Draw picture
4. Big idea: use chain rule to compute rate of change of distance between two vehicles
5. Express c in terms of x and y via Pythagorean theorem
6. Express dc/dt using chain rule in terms of dx/dt and dy/dt
7. Substitute in x, y, dx/dt, dy/dt
8. Simplify.

Choose coordinate system: Let the y-axis point North and the x-axis point East.

Identify variables: Define y(t) to be the distance of the vehicle heading North from the origin and x(t) to be the distance of the vehicle heading West from the origin.

Express c in terms of x and y via the Pythagorean theorem:

${\displaystyle c={\left(x^2+y^2\right)}^{1/2}}$

Express dc/dt using chain rule in terms of dx/dt and dy/dt:

Substitute in x = 4 mi, y = 3 mi, dx/dt = −80 mi/hr, dy/dt = 60 mi/hr and simplify

${\displaystyle \begin{array}{cc}\frac{dc}{dt}& =\frac{4\text{ mi}\cdot (-80\text{ mi}/\text{hr})+3\text{ mi}\cdot (60)\text{mi}/\text{hr}}{\sqrt{(4\text{ mi}{)}^2+(3\text{ mi}{)}^2}}\\ & =\frac{-320{\text{ mi}}^2/\text{hr}+180{\text{ mi}}^2/\text{hr}}{5\text{ mi}}\\ & =\frac{-140{\text{ mi}}^2/\text{hr}}{5\text{ mi}}\\ & =-28\text{ mi}/\text{hr}\end{array}}$

Consequently, the two vehicles are getting closer together at a rate of 28 mi/hr.

Electromagnetic induction of conducting loop spinning in magnetic field

The magnetic flux through a loop of area A whose normal is at an angle θ to a magnetic field of strength B is

${\displaystyle {\mathrm{\Phi }}_B=BA\cos (\theta ),}$

Faraday's law of electromagnetic induction states that the induced electromotive force ${\displaystyle \mathcal{E}}$ is the negative rate of change of magnetic flux ${\displaystyle {\mathrm{\Phi }}_B}$ through a conducting loop.

${\displaystyle \mathcal{E}=-\frac{d{\mathrm{\Phi }}_B}{dt},}$

If the loop area A and magnetic field B are held constant, but the loop is rotated so that the angle θ is a known function of time, the rate of change of θ can be related to the rate of change of ${\displaystyle {\mathrm{\Phi }}_B}$ (and therefore the electromotive force) by taking the time derivative of the flux relation

${\displaystyle \mathcal{E}=-\frac{d{\mathrm{\Phi }}_B}{dt}=BA\sin \theta \frac{d\theta }{dt}}$

If for example, the loop is rotating at a constant angular velocity ω, so that θ = ωt, then

${\displaystyle \mathcal{E}=\omega BA\sin \omega t}$