Related Rates

Related Rates involve finding the rate at which one quantity is changing by relating it to to other quantities with known rates. They apply concepts that we have previously covered, such as Implicit Differentiation and Chain Rule, in order to find the results.

Steps for Finding Related Rates

  1. Identify what's given and what you need to find (pay attention to units)
  2. Find an equation to differentiate
  3. Identify what variables change with time \(\cfrac{d?}{dt}\) and which are actually constant (\(0\))
  4. Substitute all the known values after you determine the derivative
  5. Use another equation if there are still too many unknowns

Example

Air leaks out of a ballon at a rate of \(3\; \left[\cfrac{\text{ft}^3}{\text{min}}\right]\). How fast is the Surface Area shrinking when the radius is \(10\;[\text{ft}]\)? Note: \(SA = 4\pi r^2\) and \(V = \cfrac{4}{3}\pi r^3\).

First, we can determine that as air leaks from the balloon that \(V\) (Volume), \(r\) (radius) and \(SA\) (Surface Area) all change with time. They are all functions of time and can be written as \(V(t)\), \(r(t)\), and \(SA(t)\) respectively:

Their Related Rates are expressed as such:

\(V'(t) = \cfrac{dV}{dt}\)

\(SA'(t) = \cfrac{dSA}{dt}\)

\(r'(t) = \cfrac{dr}{dt}\)

Next, we can identify the rate at which the balloon loses Volume:

\(\cfrac{dV}{dt} = -3\;\left[\cfrac{\text{ft}^3}{\text{min}}\right]\)

We can also determine what we need to find:

\(\cfrac{dSA}{dt} = ?\) at \(r = 10\)

We can create a table listing all of the known values and what we still need to determine:

Variable Quantity Units Rate Rate Units
Volume (V) \(V = \cfrac{4}{3}\pi r^3\) \([\text{ft}^3]\) \(\cfrac{dV}{dt} = -3\) \(\left[\cfrac{\text{ft}^3}{\text{min}}\right]\)
Surface Area (SA) \(SA = 4\pi r^2\) \([\text{ft}^2]\) \(\cfrac{dSA}{dt} = ?\) \(\left[\cfrac{\text{ft}^2}{\text{min}}\right]\)
Radius (r) \(r = 10\) \([\text{ft}]\) \(\cfrac{dSA}{dt} = ?\) \(\left[\cfrac{\text{ft}}{\text{min}}\right]\)

Then, we can set the Surface Area, \(\cfrac{dSA}{dt}\), equal to its equation. We can then determine its derivative using Implicit Differentiation:

\(\cfrac{dSA}{dt} = 4\pi r^2\)

\(\cfrac{dSA}{dt} = (2)4\pi r \cdot \cfrac{dr}{dt}\)

\(\cfrac{dSA}{dt} = 8\pi r \cdot \cfrac{dr}{dt}\)

Since we don't have all the values necessary to determine \(\cfrac{dSA}{dt}\), we can turn to Volume's Related Rate to solve \(\cfrac{dr}{dt}\).

We can differentiate the original Volume equation using Implicit Differentiation:

\(V = \cfrac{4}{3}\pi r^3\)

\(V' = (3) \left(\cfrac{4}{3}\pi r^2\right)\cdot \cfrac{dr}{dt}\)

\(V' = 4\pi r^2\cdot \cfrac{dr}{dt}\)

After, we can set Volume's Related Rate equal to this equation. We can then simplify this expression to solve \(\cfrac{dr}{dt}\):

\(\cfrac{dV}{dt} = 4\pi r^2\cdot \cfrac{dr}{dt}\)

\(-3 = 4\pi (10)^2\cdot \cfrac{dr}{dt}\)

\(-3 = 400\pi \cdot \cfrac{dr}{dt}\)

\(\cfrac{dr}{dt} = \cfrac{-3}{400\pi}\)

We can now substitute the known values into the equation and simplify to determine the rate of change for the Surface Area:

\(\cfrac{dSA}{dt} = 8\pi (10) \left(\cfrac{-3}{400\pi}\right)\)

\(\cfrac{dSA}{dt} = \cfrac{-240\cancel{\pi}}{400\cancel{\pi}}\)

\(\cfrac{dSA}{dt} = -\cfrac{3}{5}\)

Finally, we can complete our table by filling in the missing values:

Variable Quantity Units Rate Rate Units
Volume (V) \(V = \cfrac{4}{3}\pi r^3\) \([\text{ft}^3]\) \(\cfrac{dV}{dt} = -3\) \(\left[\cfrac{\text{ft}^3}{\text{min}}\right]\)
Surface Area (SA) \(SA = 4\pi r^2\) \([\text{ft}^2]\) \(\cfrac{dSA}{dt} = -\cfrac{3}{5}\) \(\left[\cfrac{\text{ft}^2}{\text{min}}\right]\)
Radius (r) \(r = 10\) \([\text{ft}]\) \(\cfrac{dSA}{dt} = \cfrac{-3}{400 \pi}\) \(\left[\cfrac{\text{ft}}{\text{min}}\right]\)

Therefore, we can determine that the Surface Area shrinks at a rate of \(-\cfrac{3}{5}\;\left[\cfrac{\text{ft}^2}{\text{min}}\right]\).


A small balloon is released at a point \(150\;[\text{ft}]\) away from an observer, who is on level ground. If the balloon goes straight up at a rate of \(8\; \left[\cfrac{\text{ft}}{\text{sec}}\right]\).

  1. How fast is the distance from the observer to the balloon increasing when the balloon is \(50\) feet high?
  2. How fast is the angle of elevation increasing?

A particle \(P\) is moving along the graph of \(y = \sqrt{x^2-4}\), \(x \leq 2\) , so that the coordinate of \(P\) is increasing at the rate of \(5\) units per second. How fast is the \(y\)-coordinate of \(P\) increasing when \(x=3\)?

Try these questions: