NOTE: You may view the original PDFs of the notes written in class by clicking on the link for the 1:30 Section or 2:30 Section, respectively.

Lecture 7 - Instantenous Rate of Change

Goal Interpret the role of the derivative in various contexts

In Precalculus, you were asked to find the average rate of change over some interval, say $ [x, x+h] $ for some $h>0$. In this case, the average rate of change of a function $f$ over $[x,x+h]$ is given by $\displaystyle \frac{f(x+h) - f(x)}{h}$, the slope of the line segment connecting the end points (See Figure 1).

Hand-drawn graph showing a secant line from $x$ to $x+h$ lying above a dipping curve. — Figure 1

A hand-drawn coordinate graph on a whiteboard. The vertical and horizontal axes are drawn as thin black lines intersecting near the lower left. A black curve rises steeply from the origin, reaches a red dot marked at the point above $x$, then dips downward into a concave valley before rising again to a second red dot marked above $x+h$, after which the curve continues upward past the second point. A straight red line, the secant line, connects the two red dots directly, passing above the dip in the black curve, illustrating that the chord lies above the function between $x$ and $x+h$. Along the horizontal axis, two bracket marks labeled $x$ and $x+h$ indicate the interval endpoints corresponding to the two red points. The letter $h$ appears to the right of the graph, denoting the distance between $x$ and $x+h$. This illustration likely depicts the concept of a secant line versus a curve's concavity, relevant to discussions of convexity, average rate of change, or the definition of the derivative.

Now, in application problems, the derivative is interpreted as "change at a point." In other words, the instantenous change at a point $x$ in the limit of the average rate of change of $f$ on $[x, x+h]$ as $h \to 0$ (See Figure 2).

Hand-drawn graph with axes labeled $a t$ and $x$, showing a curve rising, dipping, then rising again with a tangent line at one point. — Figure 2

A hand-drawn sketch on a white background depicts a coordinate system with a vertical axis labeled $a t$ (upper left) and a horizontal axis labeled $x$ (lower middle), with the letter $x$ also appearing at the top right of the image as a separate label. The curve starts at the origin where the axes meet, rises steeply and roughly linearly toward the upper right, reaches a small black dot marking a point on the curve at a moderate height, then curves downward slightly to form a shallow dip, and afterward rises again to reach a second black dot at a higher position further to the right. A red straight line is drawn tangent to the curve near the first black dot, illustrating the slope or rate of change of the curve at that point. The overall shape suggests a function with an initial steep increase, a local minimum or plateau, and then continued increase, with two marked points of interest along the path and a tangent line indicating instantaneous slope at the first point.

Physics

(Instantenous) Velocity:

Recall the definition of velocity. \[ \text{Velocity} = \frac{\text{Change in Position}}{\text{Time}} = \frac{\Delta s(t)}{\Delta t} \]

If $ s(t) $ represents the position of an object at time $ t $, then the (instanenous) velocity $ v(t) $ is given by

\[ v(t) \overset{\text{def}}{=\!=} \frac{ds}{dt} \]

NOTE:

When interpreting what the derivative, it is important to know the units of the variables in play. Hence in this lecture, unless otherwise stated, $t$ represents time in seconds.

Example 1

The position of a particle (in meters) moving in a straight line can be modeled via

\[ s(t) = 7\cos t + t^2 \]

What's the velocity?

\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}\left( 7\cos t + t^2 \right) = \left( 2t - 7\sin t \right) \frac{m}{s} \]

NOTE:

The unit of the derivative is given by: \[ \text{Unit of Derivative}= \dfrac{\text{Unit of Dependent Variable}}{\text{Unit of the Independent Variable}}\] $s$ is measured in meters ($m$) and $t$ is measured in seconds $(s)$; therefore, the unit is meters per second $m/s$.

Biology

Example 2

A ball is tossed straight into the air. The height from the ground (in feet) can be modeled via

\[ s(t) = 80t - 16t^2 \]

Before starting the example, we note that this is the usual kinematic equation
$s(t) = [s(0)] + [v(0)]t + \frac{1}{2}g^2$ you may have seen in a physics class. Here, $ g \approx -9.81 \frac{m}{s^2} \approx 32\ ft/s^2 $ is the acceleration due to gravity.

What is the velocity?
\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}\left( 80t - 16t^2 \right) = \left( 80 - 32t \right) \frac{ft}{s} \]
When $ t = 3 $, is the ball going up or down ?
\[ \begin{gathered} v(3) = 80 - 32(3) = 80 - 96 = -16\ \frac{ft}{s} \\[0.5em] v(3) < 0 \implies \text{height decreasing} \implies \text{Ball is going Down} \end{gathered} \]

Figure 3

A hand-drawn sketch with a vertical axis labeled s and a horizontal axis labeled t. A curve rises steeply from the origin, arcs over a rounded peak, and falls back down. A short horizontal tangent line is drawn touching the curve at the top of the arc, where a small circle marks the peak point. A dashed vertical line drops from the peak to the t-axis. Two additional points marked with small circles and x marks lie on the curve, one on the rising left side and one on the falling right side, illustrating positive and negative velocity.
When will the ball reach its peak?
\[ \begin{gathered} v(t) = 80 - 32t \overset{\text{set}}{=\!=} 0 \\[0.5em] 80 = 32t \\[0.5em] t = \frac{80}{32} = 2.5 \text{ seconds} \end{gathered} \]

Biology

Population Growth

Sketch of population versus time with a step function of half-open intervals and a smooth increasing curve labeled P of t — Figure 4

A hand-drawn graph with a vertical axis and a horizontal axis. A sequence of short horizontal segments forms a rising staircase; each segment begins with a filled dot on the left and ends with an open circle on the right, representing discrete population counts over successive time intervals. A smooth red S-shaped increasing curve is drawn through the staircase and is labeled P of t, showing the continuous model approximating the discrete population data.

$ P(t) $ represents the population at time $ t $. Any function $P(t)$ that is non-negative for any $t$ would work. But, here is an example population function:

\[ P(t) = P_0 e^{rt} \]

$ r $: Some Constant (controls how quickly the population grows/decays)

$ P_0 $: Initial Population $P(0)$

$ \frac{dP}{dt} $ is called the population growth; measures how fast the population grows or shrinks

Example 3

The population of a city since 2000 can be modeled via the equation

\[ P(t) = 100t^2 - 600t + 10000 \]

What is the rate at which the population is changing ?
\[ P'(t) = \left( 200t - 600 \right) \frac{\text{people}}{\text{year}} \]
What year is the population increasing at a rate of 1000 people/yr.
\[ \begin{gathered} 200t - 600 \overset{\text{set}}{=\!=} 1000 \\[0.5em] 200t = 1600 \\[0.5em] t = 8 \\[0.5em] \implies \text{In the year 2008.} \end{gathered} \]

Blood Flow

This equation works for any kind of pipe. Though, we are putting it in the context of blood flowing through a vein or artery. The velocity of the blood is modeled via

Sketch of a cylindrical blood vessel of length l and radius R with red flow arrows and distance x from the center axis — Figure 5

A hand-drawn diagram of a blood vessel shown as a long horizontal cylinder viewed from the side, with elliptical openings at the left and right ends. A dashed horizontal line runs down the middle of the cylinder representing the center axis. Many red arrows point rightward from the left opening toward the middle, representing blood flowing through the vessel; the arrows are longer near the center axis and shorter near the walls. Near the right end, a short label $ x $ marks a small distance from the center axis, and a vertical arrow labeled $ R $ marks the radius from the axis to the vessel wall. A blue horizontal line beneath the cylinder spans its full length and is labeled $ l $ for the length of the vessel.

\[ V(x) = \frac{P}{4\eta l}\left( R^2 - x^2 \right) \]

$ P $ $\leftarrow$ Difference in pressure between the start and the end

$ \eta $ $\leftarrow$ Viscocity of the blood

$ l $ $\leftarrow$ Length of the vein/artery

$ R $ $\leftarrow$ Radius of the vein/artery

$ x $ $\leftarrow$ Distance from the center axis. This is our variable; in this lecture, everything else is a constant

$ \frac{dV}{dx} $ is called the velocity gradient.

Example 4

In a small artery, the velocity of the blood is modeled via

\[ V(x) = \left( 1.85 \times 10^{4} \right)\left( 6.4 \times 10^{-5} - x^2 \right) \]

where $ x $ is the distance from the center axis in centimeters.

What is the velocity gradient ?
\[ \begin{gathered} V'(x) = \left( 1.85 \times 10^{4} \right) \cdot \frac{d}{dx}\left( 6.4 \times 10^{-5} - x^2 \right) \\[0.5em] = -2\left( 1.85 \times 10^{4} \right) x \frac{(\text{cm/s})}{\text{cm}} \end{gathered} \]
Evaluate at $ x = 0.002 \text{ cm} = 2 \times 10^{-3} \text{ cm} $
\[ V'(0.002) = -2\left( 1.85 \times 10^{4} \right)\left( 2 \times 10^{-3} \right) = -74 \frac{(\text{cm/s})}{\text{cm}} \]

NOTE:

$ 1 \text{ cm} = 10\,000\,\mu m $

\[ \frac{-74 \frac{\text{cm}}{\text{s}}}{1 \text{ cm}} = \frac{-74 \cdot 10000 \frac{\mu m}{\text{s}}}{1 \cdot 10000\, \mu m} = -74 \frac{\mu m/\text{s}}{\mu m} \]

I.e., when we move $ 1\, \mu m $ away from the center of the artery, the velocity decreases by $ 74\, \mu m/\text{s} $.

Economics

Cost Functions

Sketch of axes with P vertical and q horizontal showing a wiggly increasing curve labeled C of q starting above the origin — Figure 6

A hand-drawn graph with a vertical axis labeled P and a horizontal axis labeled q. A curve labeled $ C(q) $ begins at a marked dot on the vertical axis above the origin, rises steeply, dips down into a local minimum, and then rises again steeply toward the upper right, where an arrowhead points up and to the right beside the label $ C(q) $. The overall shape resembles a cubic cost curve.

The cost function $ P = C(q) $ represents the cost to produce $ q $ items.

$ C(0) $ is called the overhead costs.

$ \frac{dP}{dq} = \frac{dC}{dq} $ is called the marginal cost.

NOTE:

It is common to model cost functions as cubics.

\[ C(q) = a + bq - cq^2 + dq^3 \]

$ a = C(0) = $ Overhead costs

$ b = C'(0) = $ Initial Marginal Cost

$ c > 0 $ and $ d > 0 $ determine the point of diminishing returns.

Example 5

In USD, a company estimates the cost of producing $ q $ items as

\[ C(q) = 10000 + 5q + 0.01 q^3 \]

What is the marginal cost at 50 items?

\[ \begin{gathered} C'(q) = 5 + 0.03 q^2 \\[0.5em] C'(50) = 5 + 0.03(50)^2 = \$80/\text{item} \end{gathered} \]

NOTE:

\[ C'(50) \approx \frac{C(51) - C(50)}{51 - 50} = \$81.51/\text{item} \]

So, if we have already made 50 items, $ C'(50) $ is roughly how much money we need to produce the 51^st item.

Profit

The Revenue Function $ R(q) $ is the amount of money made (without considering costs) after producing $ q $ items.

Profit $ \overset{\text{def}}{=} $ Revenue $ - $ Cost

So, the Profit function $ P(q) $ is defined by

\[ P(q) \overset{\text{def}}{=} R(q) - C(q) \]

Example 6

In USD, if the revenue function is $ R(q) = q $ and $ C(q) = 50 - 6q^2 + 4q^3 $, what is the marginal profit.

\[ \begin{array}{rl} P'(q) &= R'(q) - C'(q) \\ &= [q]' - [50 - 6q^2 + 4q^3]' \\ &= 1 - [- 12q + 12q^2] \\ &= \$(-12q^2 + 12q + 1)/\text{item} \end{array} \] If you were to graph $P$ and $P'$ you will see that... you're not making any profit.