Calculate the contextual derivative
Ticket revenue
A school sells \(q\) tickets at a price of \(p\) dollars, and fewer sell as the price rises: \(q = 200 - 10p\). The teal rectangle has width \(p\) and height \(q\), so its area is the revenue \(R = pq\).
Drag the slider to raise the price. The rectangle gains a green strip on its right because it is wider and loses a red strip from its top because fewer tickets sell. Both strips are drawn for each dollar of the rise.
At \(p = 8\), the height is 120, so the green strip adds \(1 \times 120 = 120\) for each dollar. The height falls by 10, so the red strip removes \(8 \times 10 = 80\), and the revenue grows by about 40 dollars for each dollar of price. At \(p = 10\), the two strips are equal, so the revenue does not change.
Two factors that both change
A rectangle has width \(a\) and height \(b\), and both depend on \(x\), so its area is the product \(ab\).
Drag the slider to increase \(x\). The width changes, so a green strip with area \(a'b\) appears for each unit of \(x\). The height changes too, so a red strip with area \(ab'\) appears, where \(b'\) is negative here.
At every \(x\), the rate of the area is the green strip plus the red strip.
Notation
The product rule gives the rate of a product: the rate of each factor multiplies the other factor. With \(a = p\) and \(b = q\), the green strip is \(q\) and the red strip removes \(10p\).
\[\dfrac{d}{dx}\big(\cb{ab}\big) = \cg{a'b} + \cr{ab'}\]The quotient rule gives the rate of a quotient. The bottom \(b\) must not be 0, because \(a/b\) is not defined there.
\[\dfrac{d}{dx}\left(\cy{\dfrac{a}{b}}\right) = \dfrac{a'b - ab'}{b^2}\]The top of the result has a minus sign, so the order of the two terms matters.
Differentiate the model first. Substitute the values of the instant last, because a value substituted early becomes a constant with derivative 0.
When it fails
In one step from 8 to 10 dollars, the height falls from 120 to 100, so the green strip reaches above the dashed new rectangle. This faint corner is not revenue and must be taken off: \(2 \times (-20) = -40\).
The green strip adds 240 and the red strip removes 160, so the strips alone give 80, but the revenue grows by only 40. The corner is the product of two changes, so it shrinks faster than the strips as the step shrinks. The derivative keeps only the strips.
Example 1: Drug concentration
The concentration of a drug in the blood is \(c = \dfrac{5t}{t^2 + 4}\) milligrams per liter, \(t\) hours after a dose. Find how fast \(c\) changes when \(t = 1\).
The concentration is a quotient with top \(a = 5t\) and bottom \(b = t^2 + 4\). Differentiate each part.
\[a' = 5, \quad b' = 2t\]Use the quotient rule.
\[\cy{c'} = \dfrac{5(t^2 + 4) - 5t(2t)}{(t^2 + 4)^2}\]Simplify the top.
\[c' = \dfrac{20 - 5t^2}{(t^2 + 4)^2}\]Substitute \(t = 1\). The tangent at \(t = 1\) has the same slope.
\[c'(1) = \dfrac{15}{25} = 0.6\]At 1 hour, the concentration rises by about 0.6 milligrams per liter each hour.
Example 2: Harvest from planted acres
A farmer plants \(a\) acres. The yield per acre falls as the farmer plants more: \(y = 50 - 0.4a\) bushels per acre. The harvest is \(H = ay\) bushels. Find how fast \(H\) changes with \(a\) when \(a = 60\).
The harvest is the product of \(a\) and \(y\), and both depend on \(a\). At \(a = 60\), \(y = 26\). The rates with respect to \(a\) are:
\[a' = 1, \quad y' = -0.4\]A tempting wrong move: Multiply the two rates.
\[\cr{H' = a'\,y' = -0.4}\]The product of the rates is only the tiny red dashed box at the corner. One more acre adds a green strip of 26 and removes a red strip of 24, so the harvest rises by about 2.
Use the product rule.
\[\begin{aligned} H' &= a'y + ay' \\ &= (1)(26) + (60)(-0.4) \end{aligned}\]Planting one more acre raises the harvest by about 2 bushels. As a check, \(H(61) - H(60) = 1.6\).
\[H' = 26 - 24 = 2\]
Practice
A charity event has \(v = 80 + 6d\) visitors on day \(d\). Each visitor gives \(g = 5 - 0.1d\) dollars. The daily donations are \(T = vg\). Find how fast \(T\) changes on day 10.
The rectangle has width v and height g. Compare the green strip that new visitors add with the red strip that smaller gifts remove.
Solution
On day 10, find both factors and both rates.
\[\begin{aligned} v &= 140, \; g = 4 \\ v' &= 6, \; g' = -0.1 \end{aligned}\]Use the product rule.
\[\begin{aligned} T' &= v'g + vg' \\ &= 6(4) + 140(-0.1) \end{aligned}\]Simplify. On day 10, the donations rise by about 10 dollars each day.
\[T' = 24 - 14 = \cy{10}\]
Questions
A team has \(w\) workers. Each worker makes \(o = 30 - 0.5w\) units. The rectangle shows the total output \(Q = wo\), and its strips show the change for one more worker. Question 3 does not use the graph.
At \(w = 20\), what does the red strip measure?
- The output lost because each worker makes less
- The output of the new worker
- The total output \(Q\)
- The output per worker \(o\)
Answer
The output lost because each worker makes less
At \(w = 40\), does adding one worker raise or lower the total output?
- Lower, because the red strip is larger than the green strip
- Raise, because each new worker adds output
- Raise, because the rectangle is wider
- It stays the same
Answer
Lower, because the red strip is larger than the green strip
A tank holds \(6t\) grams of salt in \(100 + 2t\) liters after \(t\) hours, so the concentration is \(c = \dfrac{6t}{100 + 2t}\). Find \(c'(10)\) in grams per liter each hour.
- \(\dfrac{1}{24}\)
- 3
- 0.5
- 5
Answer
\(\dfrac{1}{24}\)
Summary
When a model multiplies or divides two quantities that both change, you differentiate it with the product rule or the quotient rule.
The rate of a product has two parts, one for each strip, because the change in each factor acts on the other factor.
You differentiate the model first and substitute the values of the instant last.
This is one of four skills in topic 4.3 of Which & Why: AP Calculus.