Question

A family of exponential functionsb. Verify that the arc length of the curve y=f(x) on the interval [0, ln 2] is A(2^a-1) - 1/4a²A (2^-a - 1).

Accepted Answer

Identify the given function and the interval. The function is of the form $$y = f(x)$$, where $$f(x) is an $$exponential function involving a parameter $$a. $$The interval is $$[0, \ln 2]. $$Recall the formula for the arc length $$L of a $$curve $$y = f(x) on $$the interval $$[a, b]$$:

$$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ Compute the derivative $$\frac{dy}{dx} of $$the function $$y = f(x). $$Since $$f(x) is $$exponential, use the chain rule carefully to express $$\frac{dy}{dx} in $$terms of $$x$$ and $$a. $$Substitute $$\frac{dy}{dx}$$ into the arc length formula and simplify the expression inside the square root. This may involve algebraic manipulation and properties of exponents to rewrite the integrand in a more manageable form. Evaluate the definite integral from $$0 to$$ $$\ln 2. $$Use substitution if necessary to integrate the expression, and simplify the result to verify that it matches the given formula:

$$A(2^a - 1) - \frac{1}{4a^2} A (2^{-a} - 1)$$

A family of exponential functions

b. Verify that the arc length of the curve y=f(x) on the interval [0, ln 2] is A(2^a-1) - 1/4a²A (2^-a - 1).

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Key Concepts

Arc Length Formula

Derivative of Exponential Functions

Integration Techniques for Exponentials