From a Verbal Rate to a Derivative
A statement such as “the water level is falling at 2 centimeters per minute” describes how one quantity changes as another quantity changes. To write it in derivative notation, identify the changing quantity, the variable it depends on, and whether the stated direction makes the rate positive or negative.
The derivative records a signed rate of change. The number 2 gives the size of the stated rate, while “falling” tells us to use a negative sign. If \(h(t)\) represents the water level at time \(t\), the statement becomes \(h'(t)=-2\) at the time being described.
A Reliable Translation Process
Before writing an equation, make sure the symbols match the situation. If the changing quantity is a height, a population, or a temperature, give it a variable name. If it changes over time, write that it depends on time. Then identify the time or other input at which the sentence reports the rate.
Choose a variable for the quantity described, such as \(h\) for height or \(P\) for population.
If the statement describes change over time, write \(h(t)\) or \(P(t)\). The derivative must be with respect to the variable that is changing.
For a quantity \(h\) depending on time \(t\), write \(h'(t)\) or \(\frac{dh}{dt}\), not just \(h(t)\).
A stated increase or rise gives a positive rate when its magnitude is specified; a stated decrease or fall gives a negative rate. Include the input at which the rate applies.
The phrase “at 5 minutes” identifies where the derivative is evaluated; it does not change the variable being differentiated. For a height \(h(t)\), the rate at 5 minutes is written \(h'(5)\). By contrast, \(h(5)\) is the height itself at 5 minutes.
Why the Direction Determines the Sign
The sign convention follows from the definition of a derivative. At an input \(a\), the derivative is the limit of difference quotients, when the limit exists. For time moving forward, take small positive changes \(\Delta t\). The quotient compares the change in the quantity with a positive change in time.
If the height is falling near \(a\), then for small positive \(\Delta t\), the later height is less than the earlier height. Thus \(h(a+\Delta t)-h(a)<0\), while \(\Delta t>0\). The difference quotient is negative. If the derivative exists, its limit must be nonpositive. It need not be strictly negative: negative quotients can approach zero.
Similarly, if the height is rising near \(a\), the later height is greater, so the difference quotients for small positive \(\Delta t\) are positive. If the derivative exists, their limit must be nonnegative, but it can equal zero. A verbal claim that a quantity is changing strictly over an interval does not by itself guarantee a nonzero derivative at every point of that interval.
Worked Examples
Worked Example: A Water Level Falling at a Specified Time
A tank’s water level is \(h(t)\), where \(t\) is measured in minutes. At \(t=4\), the level is falling at 2 centimeters per minute. Translate the statement into derivative notation.
Step 1: Identify the changing quantity and input. The level is \(h(t)\), so its rate with respect to time is \(h'(t)\).
Step 2: Use the stated time. The phrase “at \(t=4\)” means evaluate the derivative at 4, giving \(h'(4)\).
Step 3: Assign the sign. The rate has magnitude 2, and “falling” means the signed rate is negative.
Check: Read the equation in words: at \(t=4\), the water level changes at a signed rate of \(-2\) centimeters per minute. The negative sign matches “falling,” and the derivative at 4 records a rate, not the level itself.
Worked Example: A Population Increasing Over Time
Let \(P(t)\) be the number of fish in a pond, with \(t\) measured in months. At \(t=6\), the population is increasing at 120 fish per month. Write the rate in derivative notation.
Step 1: Name the rate. Since \(P\) depends on \(t\), the rate is \(P'(t)\).
Step 2: Evaluate at the stated input. The sentence gives the rate at 6 months, so use \(P'(6)\).
Step 3: Translate “increasing.” The stated magnitude is 120, and an increase gives a positive signed rate.
Check: The equation says that at 6 months the population’s rate of change is positive 120 fish per month. This matches “increasing at 120,” and the notation includes the specified time.
Worked Example: Temperature Decreasing at a Given Rate
Let \(T(t)\) be the temperature, in degrees Celsius, \(t\) hours after noon. At \(t=3\), the temperature is decreasing at 0.8 degrees Celsius per hour. Translate the statement.
Step 1: Identify the dependent and independent variables. Temperature is \(T\), and time is \(t\), so the rate is \(T'(t)\).
Step 2: Use the time in the statement. Three hours after noon corresponds to \(t=3\), so write \(T'(3)\).
Step 3: Determine the sign. “Decreasing” gives a negative rate, so the signed value is \(-0.8\).
Check: The negative sign represents a decrease, the value 0.8 is the stated rate’s magnitude, and the input 3 identifies when the rate applies.
Worked Example: Position Decreasing While an Object Moves Left
An object’s position on a horizontal number line is \(x(t)\), measured in meters, with positive positions to the right. At \(t=2\) seconds, the object is moving left at 5 meters per second. Write the derivative statement.
Step 1: Identify the position variable. The position is \(x(t)\), so its rate of change is \(x'(t)\).
Step 2: Include the time. The stated rate applies at \(t=2\), giving \(x'(2)\).
Step 3: Relate direction to the chosen coordinate convention. Moving left means the position coordinate decreases, since positive positions point right. The signed rate is therefore \(-5\), not \(+5\).
Check: A negative position rate means the coordinate is changing toward smaller values. That agrees with motion to the left under the stated convention.
Common Mistake: Reporting the Magnitude Without the Sign
A frequent error is to copy the number in a sentence but ignore its direction. For example, suppose someone writes \(h'(4)=2\) for “the water level is falling at 2 centimeters per minute.” The number 2 gives the magnitude, but the positive sign indicates an increasing level. That equation contradicts the word “falling.”
To correct the translation, keep the magnitude 2 and attach the sign determined by the stated direction. Then read the equation back in words. If the reading does not match the sentence, check the sign, variable, and evaluation point.
A related mistake is to write \(h(4)=-2\). This says that the water level itself is \(-2\) at that input; it does not say the level is falling at a rate of 2. The derivative symbol is essential because the sentence describes change.
Careful Language About “Rising” and “Falling”
In ordinary speech, “falling” often describes a trend over a stretch of time, while “falling at 2 centimeters per minute at \(t=4\)” gives a specific instantaneous rate. When the sentence explicitly gives a rate magnitude at an instant, the translation is direct: use a negative derivative for a fall and a positive derivative for a rise.
But a trend by itself does not always specify a nonzero derivative at a particular instant. For example, define \(h(t)=10-(t-5)^3\). For \(t\) near 5, the function is strictly decreasing: if \(t_1<t_2\), then \((t_1-5)^3<(t_2-5)^3\), so \(10-(t_1-5)^3>10-(t_2-5)^3\). Yet its derivative at 5 is zero.
This example is why “falling” does not always justify writing a strictly negative derivative at every point. If a problem says “falling at 2 centimeters per minute at \(t=5\),” it has specified the instantaneous rate’s magnitude, so the derivative is \(-2\). If it only says the level is falling, the trend supports a nonpositive derivative where the derivative exists, but it does not alone determine a strictly negative value.
A Final Translation Checklist
- What quantity is changing, and what symbol represents it?
- What variable does that quantity depend on?
- Does the statement give a rate or a value of the quantity?
- At what input does the rate apply?
- Does the stated direction require a positive or negative sign?
Using this checklist turns a sentence into a precise derivative statement without confusing the quantity itself with its rate or losing the sign that records direction.
Check Your Understanding
For each statement, identify the derivative and include the correct sign and evaluation point.
- Let \(V(t)\) be the volume of water in a container. At \(t=7\) minutes, the volume is increasing at 3 liters per minute. Write the derivative statement.
- Let \(C(t)\) be the concentration of a substance. At \(t=2\) hours, it is decreasing at 0.4 grams per liter per hour. Translate the rate.
- A car’s position is \(x(t)\), with positive direction to the right. At \(t=5\) seconds, it is moving left at 12 meters per second. What is \(x'(5)\)?
- Explain why “the height is falling” does not always imply that its derivative is strictly negative at every instant.
- In one sentence, distinguish what \(P(6)\) represents from what \(P'(6)\) represents when \(P(t)\) is a population.
