Interpret a derivative's value and sign
Fish in a lake
A lake holds \(F\) fish, \(w\) weeks after spring. The curve shows \(F\), green where the number of fish rises and red where it falls.
Go to week 5. The orange tangent slopes up with slope 80. The slope is the rate, so the lake gains about 80 fish each week at that moment.
The lower graph shows the rate F′, shaded green where it is positive and red where it is negative. Its orange bar is 80 high at week 5.
Go to week 10. The tangent slopes down with slope \(-40\), so the lake loses about 40 fish each week.
The value and the sign
Drag the point along the curve. Where the tangent slopes up, the curve is green and the derivative is positive. Where the tangent slopes down, the curve is red and the derivative is negative.
At the peak the tangent is flat, so the derivative is zero and the amount does not change at that moment.
A derivative describes the change at one input. Its size tells how fast the amount changes, and its sign tells which way.
Notation
To interpret a derivative, give its input, its size and its units. At week 5 the number of fish changes by about 80 fish each week.
\[\cy{F'(5)} = 80\ \text{fish per week}\]A positive derivative means that the amount rises as the input grows, so the tangent slopes up.
\[F'(5) > 0\]A negative derivative means that the amount falls as the input grows, so the tangent slopes down.
\[F'(10) < 0\]If the input changes by a small step, the amount changes by about the derivative times the step. Over the faint step of 1 week, the orange rise is 80 fish.
\[\Delta F \approx \cy{F'(w)}\,\Delta w\]
When it fails
The dashed orange line continues the tangent from week 5 to week 9. It predicts a gain of 4 times 80, which is 320 fish, but the curve gains only 128 fish. The red segment is the error of 192 fish.
The rate falls as the weeks pass, so the curve bends away from the line. A derivative predicts well only for a small step.
Example 1: A river after a storm
The height of a river is \(L(d) = 150 + 30d - 3d^2\) centimeters, \(d\) days after a storm. Say what \(L'(2)\) and \(L'(6)\) mean.
The height is in centimeters and the time is in days, so the rate \(L'\) is in centimeters per day. Differentiate \(L\) to find it.
\[L'(d) = 30 - 6d\]At day 2 the derivative is positive and the tangent slopes up, so the river rises by about 18 centimeters each day.
\[L'(2) = \cy{18}\]At day 6 the derivative is negative and the tangent slopes down, so the river falls by about 6 centimeters each day.
\[L'(6) = \cr{-6}\]
Example 2: Grain in a silo
A silo holds \(G(k) = 200 + 24k - 3k^2\) tons of grain, \(k\) weeks after harvest. Interpret \(G'(6)\).
Differentiate \(G\). Then substitute 6 into \(G'\).
\[\begin{aligned} G'(k) &= 24 - 6k \\ G'(6) &= \cr{-12} \end{aligned}\]A tempting wrong move: The value is negative, so the silo holds \(-12\) tons of grain.
At week 6 the point is at height 236, far above the red label under the axis. The number \(-12\) is the slope of the tangent, so its sign describes the change, not the amount.
The derivative is negative, so at week 6 the silo loses about 12 tons of grain each week.
The silo still holds 236 tons.
\[G(6) = 200 + 144 - 108 = 236\]
Practice
The depth of snow is \(D(h) = 4 + 9h - 0.75h^2\) centimeters, \(h\) hours after the snow begins. Say what \(D'(2)\) means. Then estimate the change in depth during the next 0.2 hours.
Find the slope of the tangent at \(h = 2\). Then multiply the slope by the length of the step.
Solution
Differentiate \(D\). Then substitute 2 into \(D'\).
\[\begin{aligned} D'(h) &= 9 - 1.5h \\ D'(2) &= \cy{6} \end{aligned}\]The derivative is positive, so the snow becomes deeper by about 6 centimeters each hour.
Multiply the rate by the change in time.
\[\Delta D \approx 6 \times 0.2 = \cy{1.2}\]The depth grows by about 1.2 centimeters. As a check, \(D(2.2) - D(2) = 1.17\).
Questions
Questions 1 and 2 use the number of members \(M\) of a gym, \(m\) months after it opens. The top curve shows \(M\), and the lower curve shows its derivative.
The point is at \(m = 2\), where the tangent slopes up with slope 40. What does this tell you?
- The gym gains about 40 members each month.
- The gym has 40 members.
- The gym loses about 40 members each month.
- The gym gains 40 members in total.
Answer
The gym gains about 40 members each month.
At \(m = 8\) the height is 560 and the tangent slopes down with slope \(-20\). Which statement is correct?
- The gym has 560 members and loses about 20 members each month.
- The gym has fewer members than at month 2.
- The gym has 560 members and gains about 20 members each month.
- The gym has −20 members.
Answer
The gym has 560 members and loses about 20 members each month.
A candle has height \(c(t) = 20 - 0.4t - 0.01t^2\) centimeters, \(t\) hours after it is lit. What does \(c'(5)\) tell you?
- The candle becomes shorter by about 0.5 centimeters each hour.
- The candle becomes taller by about 0.5 centimeters each hour.
- The candle is 0.5 centimeters tall.
- The candle becomes shorter by about 0.4 centimeters each hour.
Answer
The candle becomes shorter by about 0.5 centimeters each hour.
Summary
A derivative has a size and a sign: the size tells how fast the amount changes, and the sign tells which way.
The derivative is the slope of the tangent, so a positive value means that the amount rises and a negative value means that it falls.
You can say what the amount does at one input, and estimate a small change as the derivative times the step.
This is one of four skills in topic 4.1 of Which & Why: AP Calculus.