Math Problem Statement

Decide whether the following problem can be solved using precalculus, or whether calculus is required. If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, use a graphical or numerical approach to estimate the solution. A cyclist is riding on a path whose elevation is modeled by the function f(x)=0.08(16x−x 2 ) where x and are measured in miles. Find the rate of change of elevation when x = 4.

Solution

To determine the rate of change of elevation at x=4x = 4, we need to find the derivative of the given elevation function f(x)f(x). Calculus is required here because the "rate of change" refers to the derivative, which measures how the function's output (elevation) changes with respect to its input (distance).

The given function is: f(x)=0.08(16xx2)f(x) = 0.08(16x - x^2)

Step 1: Differentiate the function

We use the power rule to differentiate the function with respect to xx.

First, expand the expression: f(x)=0.08(16xx2)=1.28x0.08x2f(x) = 0.08(16x - x^2) = 1.28x - 0.08x^2

Now differentiate term by term: f(x)=ddx(1.28x)ddx(0.08x2)f'(x) = \frac{d}{dx}(1.28x) - \frac{d}{dx}(0.08x^2)

f(x)=1.280.16xf'(x) = 1.28 - 0.16x

Step 2: Evaluate the derivative at x=4x = 4

Substitute x=4x = 4 into the derivative: f(4)=1.280.16(4)f'(4) = 1.28 - 0.16(4) f(4)=1.280.64f'(4) = 1.28 - 0.64 f(4)=0.64f'(4) = 0.64

Conclusion:

The rate of change of elevation when x=4x = 4 miles is 0.640.64 miles per mile (which can also be interpreted as a slope of 0.64).

Would you like further details or have any questions?

Here are some related questions:

  1. What is the physical interpretation of a derivative in this context?
  2. How does the rate of change vary with respect to distance for this function?
  3. What happens to the elevation function as xx increases beyond 4 miles?
  4. Can we find where the maximum elevation occurs using calculus?
  5. How would you approximate the derivative numerically if you did not know calculus?

Tip: The derivative of a function at a point gives the slope of the tangent line to the graph at that point, which is the rate of change of the function at that specific value.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Rate of Change

Formulas

Power Rule for Derivatives
f'(x) = d/dx [0.08(16x - x^2)]
f'(x) = 1.28 - 0.16x

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12 (Advanced Calculus)