##
Section 5.2 Related Rates

###
Example 5.2.1. Change in revenue with respect to expense, doable two ways.

We can buy widgets wholesale for $10 a widget. In the. retail market, the demand price of widgets is $20 minus 0.1 times the quantity to be sold. Find the derivative of revenue with respect to expense.

## Solution 1. Solution A

Note that the derivative is positive for cost between $0 and $1000. This implies that the revenue is rising until the cost is $1000. After we hit a cost of $1000, the derivative becomes negative. This indicates that the revenue will actually decrease.

## Solution 2. Solution B

The alternative method is to differentiate the equations for revenue (\(r\)) and cost (\(c\)) with respect to quantity (\(q\)), and find the two derivatives \(\frac{d r}{d q}\) and \(\frac{d c}{d q}\text{,}\) then treat them as fractions. The derivative we want is the quotient of these fractions.

Substituting \(q =0.1 c\) gives the same solution we had from the first method.

When using the method of related rates, we act as if the derivatives are fractions that we can multiply or divide to obtain the appropriate fraction. We want to use a bit of caution with that approach, because it does not work with higher order derivatives, or with derivatives of functions of several variables. However, for derivatives of one variable the intuition works. Once again, if we zoom in far enough, the curve will look like a straight line and the derivative is the quotient of rise over run.

For the first example we could use both methods. We either use algebra to eliminate the extra variable, or find two rates of change and combine them to find the rate we are interested in. For some problems we will only have one choice, either because the algebra is too hard, or because we have been given partial information and the algebraic method is impossible.

###
Example 5.2.2. Change in revenue with respect to expense, \(q\) elimination hard.

Find the derivative of revenue with respect to cost (i.e. \(\frac{dr}{dc}\) when \(q=50\text{.}\)

## Solution.

This means that when \(\quantity=50\text{,}\) there is an increase of $1.39 for every dollar increase in cost of investment.

###
Example 5.2.3. Change in revenue with respect to expense, long variable names.

Find the derivative of revenue with respect to cost when \(\quantity=100\text{.}\)

Related rates are also useful when we are looking at a two-step process and we are interested in the rate of the combined process.

###
Example 5.2.4. Composition of functions.

Find the derivative of widgets with respect to goop when \(g=10\text{.}\)

## Solution.

Thus the rate of widget production is increasing by 16 units per increase in on unit of goop at that point.

###
Example 5.2.5. Cobb-Douglas.

The manufacturer currently uses 16 units of labor and 81 units of capital. The total production is constant but the manufacturer is investing in automation. The derivative of capital with respect to time is 2. How fast is the amount of labor needed changing?

## Solution.

If capital is increasing at a rate of 2 per unit of time, then labor is decreasing at a rate of \(-0.1317\) per unit of time.

###
Reading Questions Reading Check

####
1. Reading check, Related Rates.

###
Exercises Exercises: Related Rates Problems

#### 1.

Let \(y=3x+5\) and \(z=4y+7\text{.}\) Find \(\frac{dz}{dx}\) when \(x=2\) by solving for \(z\) as a function of \(x\) and taking the derivative, and also by finding \(\frac{dz}{dy}\) and \(\frac{dy}{dx}\) and using related rates to apply the chain rule.

## Solution.

\begin{align*}
z\amp =4y+7= 4(3x+5)+7=12x+27\\
\frac{dz}{dx}\amp=12
\end{align*}

\begin{align*}
z\amp =4y+7 \amp y\amp =3x+5\\
\frac{dz}{dy}\amp =4 \amp \frac{dy}{dx}\amp =3\\
\frac{dz}{dx}\amp =\frac{dz}{dy} \amp \frac{dy}{dx}\amp =4*3=12
\end{align*}

#### 2.

Let \(y=x^2-3x+5\) and \(z=y^2-y+6\text{.}\) Find \(dz/dx\) when \(x=1\) by solving for \(z\) as a function of \(x\) and taking the derivative, and also by finding \(\frac{dz}{dy}\) and \(\frac{dy}{dx}\) and using related rates to apply the chain rule.

#### 3.

Let \(y=1000*1.06^x\) and \(z=200y+3\text{.}\) Find \(\frac{dz}{dx}\) when \(x=5\text{.}\)

## Solution.

First find the derivative

\begin{equation*}
\frac{dz}{dx}=\frac{dz}{dy} \frac{dy}{dx}=(200) (1000*1.06^x \ln(1.06) )= 200,000 \ln(1.06) 1.06^x\text{.}
\end{equation*}

Next plug in our value of \(x\text{:}\)

\begin{equation*}
\text{When }x=5,\quad \frac{dz}{dx}= 200,000 \ln(1.06) 1.06^5\text{.}
\end{equation*}

#### 4.

Let \(y=200*1.08^x+500x\) and \(z=y^2+y\text{.}\) Find \(\frac{dz}{dx}\) when \(x= 3\text{.}\)

#### 5.

Let \(y=3x+5\) and \(z=4x+7\text{.}\) Find \(\frac{dz}{dy}\) when \(x=2\) by solving for \(z\) as a function of \(y\) and taking the derivative, and also by finding \(\frac{dz}{dx}\) and\(\frac{dy}{dx}\) and using related rates to apply the chain rule.

## Solution.

If \(y=3x+5\) then \(x=\frac{y-5}{3}\text{.}\) Hence \(z=4x+7=\frac{4}{3} (y-5)+7=\frac{4}{3} y+\frac{1}{3}\text{,}\) so \(\frac{dz}{dy}=\frac{4}{3}\text{.}\)

\(\displaystyle \frac{dz}{dx}=4, \quad \frac{dy}{dx}=3,\quad\frac{dz}{dy}= \frac{dz/dx}{dy/dx}=\frac{4}{3}\)

#### 6.

Let \(y=x^2-3x+5\) and \(z=x^2+4x+5\text{.}\) Find \(\frac{dz}{dy}\) when \(x=3\text{.}\)

#### 7.

Let \(y=1000*1.05^x\) and \(z=100*(1+.5x)\text{.}\) Find \(\frac{dz}{dy}\) when \(x=10\text{.}\)

## Solution.

First find the derivative:

\begin{equation*}
\frac{dz}{dy}= \frac{dz/dx}{dy/dx}=\frac{50}{1000*1.05^x*\ln(1.05)}\text{.}
\end{equation*}

Next plug in our value of \(x\text{:}\)

\begin{equation*}
\text{When }x=10, \frac{dz}{dy}=\frac{50}{1000*1.05^{10}*\ln(1.05)}\text{.}
\end{equation*}

#### 8.

Let \(y=100*1.08^x\) and \(z=100*1.02^x\text{.}\) Find \(\frac{dz}{dy}\) when x=10.

#### 9.

Let \(y=x^2-3x+5\) and \(\frac{dz}{dy} =3\text{.}\) Find \(\frac{dz}{dx}\) when \(x=7\text{.}\)

## Solution.

We want to find \(\frac{dz}{dx}\text{.}\) We know \(\frac{dz}{dy}\) and we can compute \(\frac{dy}{dx}\text{,}\) hence we compute:

\begin{align*}
\frac{dy}{dx}\amp =2x-3,\text{ hence at }x=7\text{ we have }\frac{dy}{dx}=11\\
\frac{dz}{dx}\amp=\frac{dz}{dy}\frac{dy}{dx}=(3)(11)=33
\end{align*}

#### 10.

Let \(y=500*.96^x\) and \(\frac{dz}{dy}=5\text{.}\) Find \(\frac{dz}{dx}\) when \(x=4\text{.}\)

#### 11.

Let \(y=x^2-5x+7\) and \(\frac{dz}{dx} =8\text{.}\) Find \(\frac{dz}{dy}\) when \(x=7\text{.}\)

## Solution.

We want to find \(\frac{dz}{dy}\text{.}\) We know \(\frac{dz}{dx}\) and we can compute \(\frac{dy}{dx}\text{,}\) hence we compute:

\begin{align*}
\frac{dy}{dx}\amp =2x-5,\text{ hence at }x=7\text{ we have }\frac{dy}{dx}=9\\
\frac{dz}{dy}\amp =\frac{dz/dx}{dy/dx}=\frac{8}{9}
\end{align*}

#### 12.

Let \(z=x^2-3x+5\) and \(\frac{dy}{dx} =3\text{.}\) Find dy/dz when x=7.

#### 13.

The revenue and expense equations for gizmos are

\begin{align*}
\revenue\amp=30*\quantity-0.1*\quantity^2\\
\expense\amp =500+10*\quantity
\end{align*}

Find the derivative of revenue with respect to expense when \(\quantity=100\text{.}\)

## Solution.

If \(R= \revenue=30q-0.1q^2\text{,}\) then \(\frac{dR}{dq}=30-0.2q\text{.}\)

If \(E= \expense=500+10q\text{,}\) then \(\frac{dE}{dq}=10\text{.}\)

Combining the two rates we get

\begin{equation*}
\frac{dR}{dE}=\frac{dR/dq}{dE/dq}=\frac{30-0.2q}{10}\text{.}
\end{equation*}

Hence at \(q =100\) we have that

\begin{equation*}
\frac{dR}{dE}=\frac{30-0.2(100)}{10}=\frac{10}{10}=1\text{.}
\end{equation*}

#### 14.

The revenue and expense equations for widgets are

\begin{align*}
\revenue\amp =200*\quantity-0.1*\quantity^2+.005*\quantity^3\\
\expense\amp =1000+20*\quantity\text{.}
\end{align*}

Find the derivative of revenue with respect to expense when \(\quantity=50\text{.}\)

#### 15.

The production of gadgets is a two step process:

\begin{align*}
\text{productA}\amp =50*\text{RawMaterial}+.01*\text{RawMaterial}^2\\
\text{gadgets}\amp =4*\text{productA}-0.0001*\text{productA}^2\text{.}
\end{align*}

Find the derivative of gadgets with respect to RawMaterial when \(\text{productA} =20\text{.}\)

## Solution.

Let’s simplify the notation and write the two equations as

\begin{align*}
p\amp =50r+0.01 r^2\\
g\amp =4 p-0.0001 p^2\text{.}
\end{align*}

We want to find \(\frac{dg}{dr}\) when \(p = 20\text{.}\)

What do we have?

\begin{align*}
\frac{dp}{dr}\amp =50+0.02 r\\
\frac{dg}{dp}\amp =4 -0.0002 p\\
\text{Then }\frac{dg}{dr}\amp =\frac{dg}{dp} \frac{dp}{dr}=(4 -0.0002 p) (50+0.02 r)\text{.}
\end{align*}

We know \(p = 20\) (given), but we need \(r\) to plug into the second part of the equation. The only function that tells us anything about \(r\) is \(p=50r+0.01 r^2\text{.}\) If \(p = 20\) this means \(20=50r+0.01 r^2\text{.}\) So \(0.01 r^2+50r-20=0\text{.}\)

By the quadratic equation we have \(r= \frac{-50\pm \sqrt{2500-4(0.01)(-20)}}{0.02}\text{.}\)

Using Wolfram Alpha we have that \(r = 0.4\text{.}\) Then

\begin{equation*}
\frac{dg}{dr}=(4 -0.004 ) (50+0.008 )\approx 199.83\frac{gadgets}{raw\ material}\text{.}
\end{equation*}

#### 16.

The production of whatchamacallits is a three step process:

\begin{align*}
\text{productA}\amp =10*\text{RawMaterial}-7\\
\text{productB}\amp=15*\text{productA}-20\\
\text{whatchamacallits}=6*\text{productB}-5\text{.}
\end{align*}

Find the derivative of \(\text{whatchamacallits}\) with respect to \(\text{RawMaterial}\) when \(\text{productA} =15\text{.}\)

#### Exercise Group.

Find the indicated derivative for the production function.

##### 17.

Our production function is \(960=5L^{0.25} K^{0.75}\text{.}\) Find \(\frac{dL}{dt}\) if \(L=81\text{,}\) \(K=256\text{,}\) \(\frac{dK}{dt}=5\text{.}\)

## Solution.

Take the derivative with respect to \(t\text{:}\)

\begin{align*}
\frac{d}{dt} 960\amp =5 \frac{d}{dt}[L^{0.25} K^{0.75}]\\
0\amp =5[\frac{d}{dt} (L^{0.25}) K^{0.75}+L^{0.25} \frac{d}{dt} (K^{0.75}) ] \\
0\amp =[\frac{d}{dt} (L^{0.25}) K^{0.75}+L^{0.25} \frac{d}{dt} (K^{0.75}) ] \\
0\amp =[0.25 L^{-0.75} \frac{dL}{dt} K^{0.75}+L^{0.25} 0.75K^{(-0.25)} \frac{dK}{dt}]\text{.}
\end{align*}

Solve for \(\frac{dL}{dt}\text{:}\)

\begin{align*}
0.25 L^{-0.75} \frac{dL}{dt} K^{0.75}\amp =-L^{0.25} 0.75K^{-0.25} \frac{dK}{dt}\\
\frac{dL}{dt}\amp =\frac{-L^{0.25} 0.75K^{-0.25}}{ 0.25 L^{-0.75} K^{0.75}} \frac{dK}{dt}=-3\frac{L}{K}\frac{dK}{dt}\text{.}
\end{align*}

If \(L=81\text{,}\) \(K=256\text{,}\) \(\frac{dK}{dt}=5\text{,}\) then we have \(\frac{dL}{dt}=-3 *\frac{81}{256}*5 \approx -4.75\text{.}\)

##### 18.

Our production function is \(4000=4L^.25 K^.75\text{.}\) Find \(\frac{dK}{dt}\) if \(L=1000\text{,}\) \(K=1000\text{,}\) \(\frac{dL}{dt}=10\text{.}\)