Question

114. {Use of Tech} Arc length of the natural logarithm Consider the curve y = ln(x).c. As a increases, L(a) increases as what power of a?

Accepted Answer

Recall the formula for the arc length $$L of a $$curve $$y = f(x)$$ from $$x = 1 to$$ $$x = a$$:  
$$L(a) = \int_1^a \sqrt{1 + \left(\frac{dy}{dx}ight)^2} \, dx$$ For the curve $$y = \ln(x)$$, compute the derivative:  
$$\frac{dy}{dx} = \frac{1}{x}$$ Substitute the derivative into the arc length formula:  
$$L(a) = \int_1^a \sqrt{1 + \left(\frac{1}{x}ight)^2} \, dx = \int_1^a \sqrt{1 + \frac{1}{x^2}} \, dx = \int_1^a \sqrt{\frac{x^2 + 1}{x^2}} \, dx = \int_1^a \frac{\sqrt{x^2 + 1}}{x} \, dx To $$understand how $$L(a)$$ grows as $$a$$ becomes large, analyze the behavior of the integrand $$\frac{\sqrt{x^2$ + 1}}{x}$$ for large $$x. $$Simplify the integrand for large $$x$$:  
$$\frac{\sqrt{x^2 + 1}}{x} = \frac{x \sqrt{1 + \frac{1}{x^2}}}{x} = \sqrt{1 + \frac{1}{x^2}}$$ which approaches 1 as $$x 	o \infty. $$Since the integrand approaches 1 for large $$x$$, the arc length integral behaves roughly like $$\$$int_1$$^a 1 \, dx = a - 1$$, which grows linearly with $$a. $$Therefore, $$L(a)$$ increases approximately as a power of $$a$$ equal to 1.

114. {Use of Tech} Arc length of the natural logarithm Consider the curve y = ln(x).
c. As a increases, L(a) increases as what power of a?

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Key Concepts

Arc Length Formula

Derivative of the Natural Logarithm

Asymptotic Behavior and Power Law Growth