Math Problem Statement

Estimate the second point on the tangent line and calculate the rate of change and percentage rate of change for the function represented by a graph showing thousand employees over time, with a tangent line at point (4.8, 9368).

Solution

The image shows a calculus problem (LCALCCON5 2.2.011) where you are asked to analyze a graph with a tangent line drawn at a specific labeled point (4.8, 9368). The task involves:

Part (a):

You have already estimated the second point on the tangent line as: (t,p)=(5,9400)(t, p) = (5, 9400) This seems correct based on the graph, as the point (5, 9400) lies on the tangent line drawn near the labeled point (4.8, 9368).

Part (b):

Here, you need to calculate the rate of change of the function at the labeled point, which corresponds to the slope of the tangent line.

To find the slope of the tangent line (rate of change), use the two points on the tangent line: (4.8,9368) and (5,9400)(4.8, 9368) \text{ and } (5, 9400) The slope mm is calculated as: m=y2y1x2x1=9400936854.8=320.2=160m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{9400 - 9368}{5 - 4.8} = \frac{32}{0.2} = 160 So, the rate of change at the labeled point is approximately 160 thousand employees per year.

Part (c):

You are asked to calculate the percentage rate of change of the function at the labeled point.

The formula for the percentage rate of change is: Percentage rate of change=Rate of changeValue of the function at the point×100\text{Percentage rate of change} = \frac{\text{Rate of change}}{\text{Value of the function at the point}} \times 100 Using the rate of change found earlier and the value of the function at t=4.8t = 4.8 (which is 9368), we calculate: Percentage rate of change=1609368×100=1.708%\text{Percentage rate of change} = \frac{160}{9368} \times 100 = 1.708\% Thus, the percentage rate of change is approximately 1.708% per year.

Let me know if you'd like further explanation on any of the steps!

Relative questions:

  1. What is the significance of the tangent line in determining the rate of change?
  2. How is the percentage rate of change different from the slope of the tangent line?
  3. Why do we need two points to calculate the slope of the tangent line?
  4. How would the interpretation of the problem change if the slope of the tangent was negative?
  5. How can the concept of the derivative be extended to non-linear graphs?

Tip:

When estimating the slope from a graph, it's helpful to pick points that are easy to read to minimize calculation errors.

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Math Problem Analysis

Mathematical Concepts

Calculus
Rate of Change
Tangent Lines

Formulas

Slope of a line: m = (y2 - y1) / (x2 - x1)
Percentage rate of change = (Rate of change / Function value) × 100

Theorems

Slope of the tangent line as the derivative of the function at a point

Suitable Grade Level

Grades 11-12, Calculus I