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How rise over run can be used to estimate the rate of change and the derivative when the original function is not known.

In real world applications, measurements of certain quantities such as position, temperature, number of individuals, and concentrations are taken at specific points in time. The rate of change of these quantities can be estimated using the rise over run formula for slope. This may also be used to estimate the derivative of a function f that is not known.
slope = rise run = Δ f Δ x = f ( x + Δ x ) − f ( x ) x + Δ x − x

derivative = f ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x

Professionals from various fields must estimate rates of change such as velocities and growth rates when the functions describing these quantities are not known. Rates of change may be approximated using estimates of the derivative. Here are some examples.

Veterinarians use video analysis to understand the movement of horses and other animals over surfaces. Video analysis can also be used to diagnose lameness and disease. White dots are placed on the horse to track the position of the knee and other joints. The velocity of these parts can be calculated using the rise over run to estimate the derivative of the position function. Such estimates are needed when the function describing the position over time is not known.
Epidemiologists at the Center for Disease Control track the number of clinical cases of a disease over time. From this data, they can estimate the growth rate of a disease. This is done by calculating the rise over run of the number of cases as a function of time. The rise over run gives an estimate of the derivative, and the derivative describes the growth rate of the disease. This growth rate may be used to make policy decisions, such as whether or not to restrict travel.
Civil engineers often track the flow rate of water through a drainage system. The rate of change of the volume of water is used to determine the flow rate. The flow rate is found by taking the derivative of the function describing the volume over time. The derivative can be estimated using the rise over the run. Such estimates are needed since the function describing the volume may not be known. Monitoring the flow rate is important for proper drainage during floods.

GRAPHICAL DEFINITION OF DERIVATIVES
Show tangent line
Show rise/run lines
Function 1  Tt=50e-0.5t
Function 2  Ht=20t-4.9t2
Function 2  Nt=100000e-(t-10)28
Coordinates of Fixed Point = ( ,  ) Slope of secant line  rise/run = 9139.39
Coordinates of Adjustable Point = ( ,  ) Slope of tangent line  T′0 = 15353.33
A chemist has placed a solution at 50°C in an ice bath and recorded the temperature over time using a digital thermometer. The temperature recordings over time for the first 10 minutes are shown in the corresponding graph. The derivative of the function gives the instantaneous rate of cooling at any point in time within the first ten minutes.
1 2 3 4
1. What is the temperature after 1 minute? Choose the t-coordinate of the fixed point to be equal to 1. What is the corresponding y-value?
2. Choose a coordinate of the adjustable point. By moving it closer the fixed point, estimate the value of the slope of the tangent line to the graph at the fixed point. Since the slope of the tangent gives the derivative of the function, what is the cooling rate (the derivative) after 1 minute has passed?
3. Choose the t-coordinate of the fixed point to be 2 and drag the adjustable point so that its t-coordinate is 4. Which of the following represents the rise over the run?
4. Assuming that the temperature was measured only after 2 and 4 minutes, what would be the estimate of the cooling rate using the slope of the secant line?
Many plants use projectile motion to disperse seeds far from the plant. In this example, a seed is projected from the ground (height = 0) up into the air and eventually returns to the ground. A scientist tracks the position of the seed over time using a video camera. The actual vertical position of the seed is shown in the accompanying graph. The derivative of this function is the vertical velocity of the seed.
1 2 3 4
1. Assume that the exact moment of launch was captured and that the camera films only one frame every 2 seconds. Choose the t-coordinate of the fixed point to be equal to 0. Drag the adjustable point so that its t-coordinate is 2. What is the rise over the run?
2. Using the slope of the secant line, what would be a scientist's best estimate of the initial vertical velocity of the seed using the data at times 0 and 2 seconds?
3. Now assume that the camera can film one frame every second. Drag the adjustable point so that its t-coordinate is 1. Using the slope of the secant line, what would be a scientist's best estimate of the initial velocity of the seed?
4. Continue to drag the adjustable point near the fixed point. Zoom in to the fixed point, and compare the slopes of the secant and tangent lines. Using the slope of the tangent line, what is the seed's initial velocity?
An epidemiologist tracks the number of reported cases of the flu in New York City. Reports of the total number of flu cases are sent in daily. The accompanying graph shows the number cases as a function of time. The derivative of the function gives the growth rate of the disease.
1 2 3 4
1. The epidemiologist receives her report only once a day. Any estimates of the growth rate must use current and older data because she does not have future reports. What is the best estimate of the growth rate of the disease that she can make on day 6? Choose the t-coordinate of the fixed point to be equal to 6. Drag the adjustable point such that the t-coordinate is 5. What is the slope of this secant line?
2. On day 10, the epidemiologist has data for only days 0 through 10. What is the best estimate of the growth rate of the disease she can make on day 10? Remember that she will need to use data from day 9 for this calculation.
3. Move the adjustable point toward the fixed point at day 10. Zoom into the fixed point, and compare the slopes of the secant and tangent lines. What is the actual growth rate of the disease on day 10?
4. The epidemiologist can estimate the growth rate of the disease on a particular day by setting the x-coordinate of the fixed point to that day and setting the x-coordinate of the adjustable point to the previous day. What is the first day that she can see the growth rate of the disease is negative and the number of cases is declining?
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