Related Rates

How to Solve a Related Rates Problem

Here are a few guidelines to follow when solving a related rates problem.

1. Assign a variable to each quantity that changes and state the known and unknown rates of change with respect to time \(t\).

2. Find an equation relating the variables. Don’t forget about business models.

3. Differentiate both sides of the equation implicitly with respect to \(t\). Treat all variables as functions of \(t\).

4. Replace the variables and their derivatives by the numerical data given. Watch out for words like “increasing” or “decreasing”, “rising” or “falling”, “away” or “towards”, etc. to determine if a derivative is positive or negative.

5. Solve the equation for the unknown rate of change.

Example 1

Related rates problem

Given \(p^2x + 6x = 5\) and \(\dfrac{dp}{dt} = 3\). Compute \(\dfrac{dx}{dt}\) when \(p = 2\).

Step 1:   Assign variables and state the known and unknown rates of change.

\[\begin{align*} \text{Variables (as functions of $t$): } & ~~ x(t) \text{ and } p(t) \\ \text{Known rate of change: } & ~~ \dfrac{dp}{dt} = 3 \\ \text{Unknown rate of change: } & ~~ \dfrac{dx}{dt} \text{ when } p=2 \end{align*}\]

Step 2:   Find an equation which relates the variables.

\[p^2x + 6x = 5\]

Step 3:   Differentiate both sides of the equation implicitly with respect to   \(t\).

\[\begin{align*} \frac{d}{dt}(p^2x + 6x) &= 2pp'x + p^2x' + 6x' && \hbox{implicit differentiation & product rule}\\ &= 0 && \hbox{since $\dfrac{d}{dt} 5 = 0$} \end{align*}\]

Step 4:   Find the value of   \(x\).

Since \(x\) appears in the equation after differentiation, we need to find its value by plugging in \(p=2\) into the given equation \(p^2x + 6x = 5\).

\[(2)^2x + 6x = 5 ~~~~~\Rightarrow~~~~~ 10x = 5 ~~~~~\Rightarrow~~~~~ x = 1/2\]

Step 5:   Plug values and derivatives into the equation found in Step 3.

Plug the known values of the variables and their derivatives (\(p=2\), \(x=1/2\), \(\frac{dp}{dt} = 3\)) into the equation found in Step 3.

\[2(2)(3)\frac{1}{2} + 2^2x' + 6x' = 0 ~~~~~\Rightarrow~~~~~ 6 + 10x' = 0 ~~~~~\Rightarrow~~~~~ x' = -3/5\]

Therefore \(\dfrac{dx}{dt} = -3/5\) when \(p=2\) and \(\dfrac{dp}{dt} = 3\).

Example 2

Related rates problem: demand vs. revenue

A company is decreasing the production of energy drinks at a rate of 100 cases per day. All cases produced can be sold. The daily demand function is given by

\[p(x) = 50 - \frac{x}{200}\]

where \(x\) is the number of cases produced and sold, and \(p(x)\) is the unit price in dollars. Find the rate of change of the revenue with respect to the time in days when daily production is 300 cases.

Step 1:   Assign variables and state the known and unknown rates of change.

\[\begin{align*} \text{Variables (as functions of $t$): } & ~~ x(t) = \text{ the quantity produced and sold per day} \\ & ~~ R(t) = \text{ the daily revenue} \\ \text{Known rate of change: } & ~~ \dfrac{dx}{dt} = -100 \text{ since production is decreasing} \\ \\ \text{Unknown rate of change: } & ~~ \dfrac{dR}{dt} \text{ when } x=300 \end{align*}\]

Step 2:   Find an equation which relates the variables.

\[\begin{align*} R &= x\cdot p(x) && \text{revenue function}\\ &= x \left(50 - \frac{x}{200}\right) && \text{since $p(x) = 50 - \frac{x}{200}$}\\ &= 50x - \frac{1}{200}x^2 && \hbox{simplify} \end{align*}\]

Step 3:   Differentiate both sides of the equation implicitly with respect to   \(t\).

\[\begin{align*} \frac{dR}{dt} &= \frac{d}{dt}\left(50x - \frac{1}{200}x^2 \right)\\ &= 50 \,\frac{dx}{dt} - \frac{1}{200}\,2x\,\frac{dx}{dt} && \text{implicit differentiation & power rule}\\ &= 50 \, \frac{dx}{dt} - \frac{x}{100}\,\frac{dx}{dt} && \text{simplify} \end{align*}\]

Step 4:   Plug values and derivatives into the equation found in Step 3.

Plug the known values of the variables and their derivatives (\(x=300\), \(\frac{dx}{dt} = -100\)) into the equation found in Step 3.

\[\begin{align*} \frac{dR}{dt} &= 50(-100) - \frac{300}{100}(-100)\\ &= -5000 + 300 \\ &= -4700 \end{align*}\]

Therefore revenue is decreasing at $4700 per day when the level of production is 300 cases per day and decreasing at 100 cases per day.

Example 3

Related rates problem: price vs. demand

The weekly demand function is given by

\[2p + x + 4xp = 22\]

where \(x\) is the number of thousands of units demanded weekly and \(p\) is in dollars. If the price is increasing at a rate of 25 cents per week when the weekly demand is 4000 units, at what rate is the demand decreasing?

Step 1:   Assign variables and state the known and unknown rates of change.

\[\begin{align*} \text{Variables (as functions of $t$): } & ~~ x(t) = \text{ demand in units of a thousand} \\ & ~~ p(t) = \text{ unit price in dollars}\\ \text{Known rate of change: } & ~~ \dfrac{dp}{dt} = \dfrac{25}{100} \text{ since price (in dollars) is increasing } \\ \text{Unknown rate of change: } & ~~ \dfrac{dx}{dt} \text{ when } x = 4 \end{align*}\]

Step 2:   Find an equation which relates the variables.

\[2p + x + 4xp = 22\]

Step 3:   Differentiate both sides of the equation implicitly with respect to   \(t\).

\[\begin{align*} \frac{d}{dt}(2p + x + 4xp) &= 2\frac{dp}{dt} + \frac{dx}{dt} + 4\frac{dx}{dt}p + 4x\frac{dp}{dt} && \hbox{implicit differentiation & product rule} \\ &= 0 && \hbox{since $\dfrac{d}{dt}22 = 0$} \end{align*}\]

Step 4:   Find the value of   \(p\).

Since \(p\) appears in the equation after differentiation, we need to find its value by plugging in \(x=4\) into the given demand equation \(2p+x+4xp = 22\).

\[2p + 4 + 4(4)p = 22 ~~~~~\Rightarrow~~~~~ 18p + 4 = 22 ~~~~~\Rightarrow~~~~~ p = 1\]

Step 5:   Plug values and derivatives into the equation found in Step 3.

Plug the known values of the variables and their derivatives (\(x=4\), \(p=1\), \(\frac{dp}{dt} = 25/100\)) into the equation found in Step 3.

\[\begin{align*} 0 &= 2\left(\frac{25}{100}\right) + \frac{dx}{dt} + 4\frac{dx}{dt}(1) + 4(4)\left(\frac{25}{100}\right) \\ &= \frac{1}{2} + 5\frac{dx}{dt} + 4 && \hbox{simplify} \\ &= \frac{9}{2} + 5\frac{dx}{dt} \end{align*}\]

Therefore

\[\frac{9}{2} + 5\frac{dx}{dt} = 0 ~~~~~\Rightarrow~~~~~ 5\frac{dx}{dt} = -\frac{9}{2} ~~~~~\Rightarrow~~~~~ \frac{dx}{dt} = -\frac{9}{10}\]

and since \(x\) is measured in thousands of units, this means that demand is decreasing by \(900\) units per week (i.e., \(9/10\) of one thousand) when the weekly demand is 4000 units and the price is increasing at 25 cents per week.

Example 4

Related rates problem: price vs. supply

The wholesale price \(p\) of an e-tablet writing stylus in dollars is related to the supply \(x\) in thousands by:

\[125p^2 - x^2 = 100\]

If 20,000 styluses are available at the beginning of a week, and the price is falling at 4 cents per week, at what rate is the supply falling?

Step 1:   Assign variables and state the known and unknown rates of change.

\[\begin{align*} \text{Variables (as functions of $t$): } & ~~ x(t) = \text{ supply in units of a thousand} \\ & ~~ p(t) = \text{ unit price in dollars}\\ \text{Known rate of change: } & ~~ \dfrac{dp}{dt} = -\dfrac{4}{100} \text{ since price (in dollars) is falling} \\ \text{Unknown rate of change: } & ~~ \dfrac{dx}{dt} \text{ when } x = 20 \end{align*}\]

Step 2:   Find an equation which relates the variables.

\[125p^2 - x^2 = 100\]

Step 3:   Differentiate both sides of the equation implicitly with respect to   \(t\).

\[\begin{align*} 250p\frac{dp}{dt} - 2x\frac{dx}{dt} = 0 && \text{power rule} \end{align*}\]

Step 4:   Find the value of   \(p\).

Since \(p\) appears in the equation after differentiation, we need to find its value by plugging in \(x=20\) into the given supply equation \(125p^2 - x^2 = 100\).

\[125p^2 - 20^2 = 100 ~~~~~\Rightarrow~~~~~ 125p^2 = 500 ~~~~~\Rightarrow~~~~~ p^2 = 4~~~~~\Rightarrow~~~~~ p = 2\]

Step 5:   Plug values and derivatives into the equation found in Step 3.

Plug the known values of the variables and their derivatives (\(x=20\), \(p=2\), \(\frac{dp}{dt} = -4/100\)) into the equation found in Step 3.

\[\begin{align*} 0 &= 250(2)\left(-\frac{4}{100}\right) - 2(20)\frac{dx}{dt} \\ &= -500\left(\frac{4}{100}\right) - 40 \frac{dx}{dt} && \hbox{simplify} \\ &= -20 - 40 \frac{dx}{dt} \end{align*}\]

Therefore

\[-20 - 40 \frac{dx}{dt} = 0 ~~~~~\Rightarrow~~~~~ -40 \frac{dx}{dt} = 20 ~~~~~\Rightarrow~~~~~ \frac{dx}{dt} = -\frac{1}{2}\]

and since \(x\) is measured in thousands of units, this means that supply is falling by \(500\) units per week (i.e., \(1/2\) of one thousand) when the weekly demand is 20,000 styluses and the price is falling at 4 cents per week.