Section 3.4 Derivatives as Rates of Change
Reference.
Now that we know how to evaluate the derivatives of some common functions efficiently, let us look at some of the places that this is useful in scientific problems. We will look at velocity and acceleration, population growth, and marginal cost.
Rate of Change of One Quantity Relative to Another.
The basic idea of a rate of change is that there is a relationship between changes in one quantity \(x\) and another, \(y\text{,}\) due to there being a function connecting the two quantities, \(y=f(x)\text{.}\) (Note: you might not have a formula for this function!) The pairs might be
- time and position,
- pressure and volume in a quantity of gas,
- horizontal and vertical position of a person traveling over hilly terrain,
or many other combinations.
Average Rate of Change of One Quantity Relative to Another.
Suppose that the first quantity changes in value from \(x_1\) to \(x_2\text{,}\) the second changes from \(y_1\) to \(y_2\text{;}\) in terms of the function, from \(f(x_1)\) to \(f(x_2)\text{.}\)
The change in the first quantity is \(\Delta x = x_2 - x_1\) (spoken “delta x”), the corresponding change in the second quantity is \(\Delta y = y_2 - y_1 = f(x_2)-f(x_1)\) (“delta y”).
The average rate of change of \(\mathbf y\) with respect to \(\mathbf x\) over the interval from \(x_1\) to \(x_2\) is given by the difference quotient
\begin{equation*}
\frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{f(x_2)-f(x_1)}{x_2 - x_1}.
\end{equation*}
This is the slope of the secant line between the two points \(P(x_1,y_1)\) and \(Q(x_2,y_2)\) on the graph of \(y\) vs. \(x\text{.}\)
Instantaneous Rate of Change of One Quantity Relative to Another.
In the limit of \(x_2 \to x_1\text{,}\) with \(x_1\) unchanging, so \(\Delta x \to 0\text{,}\) we get the instantaneous rate of change of \(y\) with respect to \(x\) at \(x_1\text{,}\)
\begin{equation*}
\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx}(x_1) = f'(x_1).
\end{equation*}
Thus, whenever a derivative arises like this in a physical context, it has an interpretation as a rate of change.
Physics: Velocity and Acceleration.
We have already seen one example: when position \(s\) (along a line) is a function of time \(t\text{,}\) \(s=f(t)\) the the rate of change of position with respect to time is the velocity,
\begin{equation*}
v(t) = s'(t) = f'(t) = \frac{ds}{dt}.
\end{equation*}
Many basic laws of physics describe the way that a force causes the velocity of an object to change at a certain rate, and the rate of change of velocity is acceleration,
\begin{equation*}
a(t) = v'(t) = \frac{dv}{dt} = s''(t) = f''(t) = \frac{d^2s}{dt^2}.
\end{equation*}
An object moving back and forth along a line is at position \(s=f(t)=t^3-6t^2+9t\) meters to the right of its starting point at time \(t\) seconds (a negative \(s\) value means that it is to the left.)
- Find its velocity at time \(t\text{.}\)
- What is the velocity after 2 seconds? After 4 seconds?
- When is the object at rest?
- When is it moving to the right?
- Draw a diagram to represent the motion of the object.
- Find the total distance traveled by the object during the first five seconds.
Bacterial Population Growth.
A population of bacteria often grows by cells dividing at a roughly fixed time period, \(T\text{.}\) If the initial number of bacteria is \(P_0\text{,}\) then at a time \(nT\) the number of bacteria is \(P = f(nT)=P_0 2^n\text{.}\) To write this as a function of \(t\text{,}\) use \(t=nT\) so \(n=t/T\text{,}\) so
\begin{equation*}
P = f(t)=P_0 2^{t/T}.
\end{equation*}
If this exponential patterns hold at all times, not just multiples of \(T\) (the bacteria are probably not synchronized to divide at the same time), we can try to compute the rate of change of the population,
\begin{equation*}
\frac{dP}{dt} = \frac{d (P_0 2^{t/T})}{dt}= P_0 \ln 2 \cdot 2^{t/T} \frac 1 T = kP, \quad k=\frac{\ln 2}{T}.
\end{equation*}
The exponential function fits the expected pattern of population growth rate being proportional to current population size. As always, the exponential can be rewritten using the natural exponential:
\begin{equation*}
P(t) = P_0 e^{\ln 2 \cdot t/T} = P_0 e^{kt}, \quad k=\frac{\ln 2}{T}.
\end{equation*}
Example 3.4.2.
If the initial size of a bacterial population is \(P_0=100\) and it doubles every five hours:
- How fast is it growing initially?
- How fast is it growing after two days?
- Compare the second answer to the average rate of change between days 1 and 3.
The population as a function of time is \(100 \cdot 2^{t/5}\text{,}\) measuring time in hours. Thus the growth rate is \(100 k 2^{t/5}\text{,}\) \(k=(\ln 2)/5\text{,}\) approximately \(13.86 2^{t/5}\text{.}\) This gives the initial growth rate as \(100 k \approx 13.86\) bacteria per hour.
The growth rate two days [48 hours] later as \(100 k 2^{48/5} \approx 10,758\) bacteria per hour.
The average growth rate “around” day two, time \(t=48\text{,}\) from \(t=24\) to \(t=72\text{,}\) is \([f(72)-f(24)]/48 = 44,981\text{.}\) The average is distinctly higher, because the growth rate increases so much between days 2 and 3.
Economics: Cost Functions and Marginal Cost.
If the cost \(C\) of producing \(x\) units of a product depends only on the number produced, it is given by some function \(C(x)\text{.}\) Increasing production level from \(x_1\) units to \(x_2\) incurs a change (increase) in cost of \(\Delta C = C(x_2)-C(x_1)\text{,}\) for a change (increase) in production of \(\Delta x = x_2-x_1\text{.}\) The average rate of increase of cost with respect to increase in production is
\begin{equation*}
\frac{\Delta C}{\Delta x} = \frac{C(x_2)-C(x_1)}{x_2-x_1} = \frac{C(x_1+ \Delta x)-C(x_1)}{\Delta x}
\end{equation*}
Even though the production level is an integer, it can be extremely large, so that \(x\) might be in units of thousands or millions, and then increments \(\Delta x\) in production level can be small, so the average rate of change of cost is well approximated by the derivative at \(x_1\text{:}\)
\begin{equation*}
\frac{\Delta C}{\Delta x} \approx \frac{d C}{d x}(x_1) = C'(x_1).
\end{equation*}
This is called the marginal cost, also well approximated by \({\Delta C}/{\Delta x}\) for an increase in production by a single unit (which might be \(\Delta x\) far smaller than 1, depending on the units used.) A decision to increase production might be made by comparing this marginal cost to the price at which extra units could be sold.
Checkpoint 3.4.3.
Consider the cost function \(C(x) = 10,000 + 5x + 0.01x^2\)
- Find the marginal cost where the production level is 500 units.
- Compare this to the actual added cost of making one more unit (501 instead of 500).
- At what production level does the marginal cost reach $20 per item? (Important to know if that is the selling price!)
Exercises Exercises
Study Calculus Volume 1, Section 3.4 2 ; in particular Examples 34 to 36, Checkpoint item 22, and Exercises 151, 159 and 165.
openstax.org/books/calculus-volume-1/pages/3-4-derivatives-as-rates-of-changeopenstax.org/books/calculus-volume-1/pages/3-4-derivatives-as-rates-of-change
