From a Formula to an Instantaneous Rate
A formula can describe how one quantity depends on another. For example, a formula might give the volume of water in a tank as a function of time, or the cost of producing a particular number of items. If the formula is differentiable at a specified input, its derivative gives the instantaneous rate of change there.
The calculation has two distinct steps: differentiate the quantity formula, then substitute the specified input into the derivative. Keeping these steps separate helps prevent confusing the quantity itself with its rate of change.
If the input is time \(t\), the derivative is often written \(\frac{dQ}{dt}\). If the input is a different variable, use the corresponding derivative notation, such as \(\frac{dQ}{dx}\). The derivative value is not the same as \(Q(a)\): \(Q(a)\) is the quantity at the input, while \(Q'(a)\) is its rate of change there.
A Reliable Calculation Process
For a contextual rate from a given formula, use the following sequence. The order matters: evaluating the original formula first finds a quantity value, not a rate.
Determine which symbol is the output quantity, which is the input, and what units each has.
Find the derivative with respect to the input variable, keeping the input variable in the result.
Substitute the requested input into the derivative, not merely into the original formula.
Use output units per input unit, and check that the formula is differentiable at the input.
For instance, if \(V(t)\) is measured in liters and \(t\) in minutes, then \(V'(t)\) is measured in liters per minute. This follows because a change in liters is divided by a change in minutes in the rate calculation.
Why Differentiating Produces the Instantaneous Rate
The derivative is defined as the limit of average rates over shorter and shorter input intervals. For a quantity \(Q\), the average rate from \(x=a\) to \(x=a+h\), when \(h\ne 0\), is \(\frac{Q(a+h)-Q(a)}{h}\). If this expression approaches a finite limit as \(h\) approaches zero, that limit is \(Q'(a)\), the instantaneous rate at \(a\).
Here is a direct calculation for the formula \(V(t)=t^3-6t^2+15t+10\), where \(V\) is measured in liters and \(t\) in minutes. At \(t=2\), first calculate the quantity:
Now use the difference quotient to find the instantaneous rate at that input. Expanding each power gives:
Therefore, for \(h\ne 0\), the average rate is:
As \(h\) approaches zero, \(h^2\) approaches zero, so the quotient approaches \(3\). Thus \(V'(2)=3\) liters per minute. Differentiating the formula with the power rule gives \(V'(t)=3t^2-12t+15\), and substituting \(t=2\) also gives \(3(2^2)-12(2)+15=12-24+15=3\), confirming the result.
Worked Examples
Worked Example: Water Volume at a Specified Time
A tank’s water volume is modeled by \(V(t)=t^3-6t^2+15t+10\), where \(V\) is in liters and \(t\) is in minutes. Find the instantaneous rate of change of the volume at \(t=2\).
Differentiate each term with respect to \(t\):
Evaluate the derivative at \(t=2\):
The rate is \(3\) liters per minute. To check, the difference quotient at \(t=2\) simplifies to \(3+h^2\), which approaches \(3\) as \(h\to 0\). The quantity itself is \(V(2)=24\) liters; that is a different value from the rate.
Worked Example: Marginal Cost from a Cost Formula
The cost of producing \(q\) items is \(C(q)=0.04q^2+12q+500\) dollars. Find the derivative at \(q=50\), including its units.
Differentiate with respect to \(q\). The constant term has derivative zero, and the power rule gives:
Substitute \(q=50\):
Since \(C\) is measured in dollars and \(q\) in items, the derivative has units of dollars per item. Check the value with the difference quotient. First, \(C(50)=0.04(2500)+12(50)+500=100+600+500=1200\). Also,
As \(h\to 0\), this approaches \(16\), as calculated by differentiating.
Worked Example: Area as a Function of Radius
The area of a circle with radius \(r\) is \(A(r)=\pi r^2\). Find the instantaneous rate of change of area with respect to radius at \(r=3\) centimeters.
Differentiate with respect to \(r\):
Evaluate at \(r=3\):
The rate is \(6\pi\) square centimeters per centimeter. The units follow from area units divided by radius units. For a check, the difference quotient at \(r=3\) is:
As \(h\to 0\), the quotient approaches \(6\pi\), agreeing with \(A'(3)\).
Worked Example: A Composite Quantity Formula
A sensor’s output is modeled by \(S(x)=(2x+1)^3\) output units, where \(x\) is measured in input units. Find the derivative at \(x=2\).
Apply the chain rule: the derivative of the outside cube is \(3(2x+1)^2\), then multiply by the derivative of the inside expression, which is \(2\):
Substitute \(x=2\):
The derivative is \(150\) output units per input unit. To check the differentiation, expand the original formula: \(S(x)=8x^3+12x^2+6x+1\). Differentiating the expanded expression gives \(24x^2+24x+6\). Meanwhile, \(6(2x+1)^2=6(4x^2+4x+1)=24x^2+24x+6\), so both forms agree. At \(x=2\), this is \(24(4)+24(2)+6=96+48+6=150\).
A Common Substitution Error
A frequent mistake is to substitute the specified input into the original quantity formula and report that result as the rate. In the water example, \(V(2)=24\) liters. The incorrect calculation “the rate at \(t=2\) is \(24\) liters per minute” confuses the amount of water with how fast that amount is changing.
A related error is to substitute too early and treat the input as a constant before differentiating. Once \(t=2\) has been substituted into \(V(t)\), the result \(V(2)=24\) is a number; differentiating that number cannot recover the rate at \(t=2\). Keep the input variable until after differentiation.
Check the Formula and the Units
The derivative calculation is meaningful only if the given formula is appropriate at the requested input and differentiable there. For the polynomial and power formulas in the examples, the derivatives exist at every input in their domains. In other problems, first make sure the requested input is in the formula’s domain; a formula that is undefined at that input cannot supply a derivative there.
Units provide a useful final check. A derivative of volume with respect to time should have volume-per-time units; a derivative of cost with respect to number of items should have cost-per-item units. If the written units instead match the original output quantity alone, you may have calculated the quantity value rather than its rate.
Check Your Understanding
For each question, differentiate the formula before substituting the specified input, and include units when they are given.
- If \(P(t)=4t^2+3t+7\) is measured in grams and \(t\) in seconds, find \(P'(2)\).
- A quantity is given by \(F(x)=\pi x^2\). Find \(F'(4)\).
- If \(C(q)=0.5q^2+8q+100\) dollars, where \(q\) is a number of items, find \(C'(10)\) and its units.
- For \(G(x)=(3x-1)^2\), find \(G'(2)\) using the chain rule.
- Explain why substituting \(t=3\) into \(Q(t)\) does not, by itself, calculate the rate of change at \(t=3\).
