Choose the quantity whose rate is needed
A bakery’s cost
A bakery pays \(C\) dollars to bake \(n\) loaves in one day. The curve shows the cost \(C\) for each number of loaves.
Drag the orange point to the right. The orange line touches the curve at the point, and its slope shows how fast the cost changes there.
Go to 100 loaves. The slope is 3, and the horizontal axis counts loaves, so the cost grows by about 3 dollars for each extra loaf.
The output and the input
The vertical axis shows the output y, the quantity that the problem follows. The horizontal axis shows the input x, the quantity that y depends on.
Move the point. The slope dy/dx of the tangent is the orange rise, a change in y, divided by the faint run, a change in x.
Wherever the point is, the slope is in units of the output for each unit of the input. The input is the quantity that the question names, and it is not always time.
Notation
Name the output, the quantity that the question asks about.
\[\cy{C} = f(n)\]Name the input, the quantity that the output depends on. The question gives it after “with respect to” or “as … increases”.
Write the rate. The input goes in the denominator, because a slope divides a change in the output by a change in the input.
\[\cy{\dfrac{dC}{dn}}\]Give the units, which are the units of the output for each unit of the input.
\[\text{dollars per loaf}\]
When it fails
The same cost can depend on another input. The bakery bakes 20 loaves each hour, so after \(t\) hours \(n = 20t\) and \(C = 30 + 40t + 2t^2\). The curve keeps its shape, because each number of loaves matches one time: 100 loaves is 5 hours.
The tangent at 100 loaves is the same line, but its run is now counted in hours. One hour is 20 loaves, so the slope is 3 × 20 = 60 dollars per hour, not 3 dollars per loaf. The rate depends on which input the question names.
Example 1: Fertilizer and yield
A garden gives \(Y = 40 + 6f - 0.2f^2\) kilograms of tomatoes when \(f\) kilograms of fertilizer are used. Find how fast the yield changes with the fertilizer when \(f = 5\).
The yield Y is the output. The fertilizer f is the input, because the question asks how Y changes as f changes.
\[\cy{Y} = 40 + 6f - 0.2f^2\]Differentiate with respect to f. Each term of Y is a power of f, so use the power rule.
\[\cy{\dfrac{dY}{df}} = 6 - 0.4f\]Substitute \(f = 5\) after you differentiate. The tangent at \(f = 5\) has slope 4.
\[6 - 0.4(5) = 4\]At 5 kilograms of fertilizer, the yield rises by about 4 kilograms of tomatoes for each extra kilogram of fertilizer.
Example 2: Heart rate and power
A rider’s heart rate is \(H = 60 + 0.5p + 0.002p^2\) beats per minute when the power is \(p\) watts. Find how fast the heart rate changes as the power increases, when \(p = 150\).
The heart rate H is the output. The horizontal axis shows the power p in watts, so the power is the input.
A tempting wrong move: Rates are about time, so write the rate as \(dH/dt\).
\[\cr{\dfrac{dH}{dt}}\]The horizontal axis shows power, not time. The formula for \(H\) has no \(t\) in it, so \(dH/dt\) has no meaning here.
Write the rate with respect to the power, because the question names the power as the input.
\[\cy{\dfrac{dH}{dp}} = 0.5 + 0.004p\]Substitute \(p = 150\).
\[0.5 + 0.004(150) = 1.1\]At 150 watts, the heart rate rises by about 1.1 beats per minute for each extra watt.
Practice
A store sells \(N = 600/p\) bags of sand each week when the price is \(p\) dollars. Find how fast \(N\) changes with the price when \(p = 20\). Give the units.
The slope that you need is taken along the horizontal axis, which shows the price.
Solution
The output is N, the number of bags, and the input is p, the price in dollars. Write N as a power of p to use the power rule.
\[N = 600p^{-1}\]Differentiate with respect to \(p\).
\[\dfrac{dN}{dp} = -600p^{-2}\]Substitute \(p = 20\). The units are bags per dollar.
\[-\dfrac{600}{400} = \cy{-1.5}\]At 20 dollars, sales fall by about 1.5 bags each week for each extra dollar of price.
Questions
The graph shows the height \(h\) of a plant in centimeters against the dose \(d\) of a growth hormone in milligrams. Question 3 does not use the graph.
The rate \(dh/dd\) is the slope of this curve. What are its units?
- centimeters per milligram
- milligrams per centimeter
- centimeters per day
- centimeters
Answer
centimeters per milligram
At which dose does the height change most quickly as the dose increases?
- \(d = 0\)
- \(d = 2\)
- \(d = 4\)
- \(d = 6\)
Answer
\(d = 0\)
A steel ball of radius \(r\) centimeters has mass \(m = 8r^3\) grams. How fast does the mass change with the radius when \(r = 2\)?
- 96 grams per centimeter
- 64 grams
- 48 grams per centimeter
- 96 grams per second
Answer
96 grams per centimeter
Summary
To find a rate, you name the output, the quantity that the question follows, and the input, the quantity that it depends on.
The rate is the slope dy/dx of the output against the input, in units of the output for each unit of the input.
When a question asks how fast something changes, take the input from its words, because the input is not always time.
This is one of four skills in topic 4.3 of Which & Why: AP Calculus.