Math Problem Statement

Find the exact length of the curve. y = ln(sec(x)), 0 ≤ x ≤ 𝜋/6

Solution

To find the exact length of the curve y=ln(sec(x))y = \ln(\sec(x)) over the interval 0xπ60 \leq x \leq \frac{\pi}{6}, we use the formula for the arc length of a function y=f(x)y = f(x), which is given by:

L=ab1+(dydx)2dxL = \int_a^b \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx

Step 1: Find dydx\frac{dy}{dx}

Given y=ln(sec(x))y = \ln(\sec(x)), we need to differentiate this with respect to xx:

dydx=ddx(ln(sec(x)))\frac{dy}{dx} = \frac{d}{dx} \left( \ln(\sec(x)) \right)

Using the chain rule, we know that the derivative of ln(u)\ln(u) is 1ududx\frac{1}{u} \cdot \frac{du}{dx}. Here, u=sec(x)u = \sec(x), and the derivative of sec(x)\sec(x) is sec(x)tan(x)\sec(x)\tan(x). So:

dydx=1sec(x)sec(x)tan(x)=tan(x)\frac{dy}{dx} = \frac{1}{\sec(x)} \cdot \sec(x)\tan(x) = \tan(x)

Step 2: Set up the arc length formula

Now, substitute dydx=tan(x)\frac{dy}{dx} = \tan(x) into the arc length formula:

L=0π61+tan2(x)dxL = \int_0^{\frac{\pi}{6}} \sqrt{1 + \tan^2(x)} \, dx

Recall the trigonometric identity 1+tan2(x)=sec2(x)1 + \tan^2(x) = \sec^2(x), so:

L=0π6sec2(x)dxL = \int_0^{\frac{\pi}{6}} \sqrt{\sec^2(x)} \, dx

L=0π6sec(x)dxL = \int_0^{\frac{\pi}{6}} \sec(x) \, dx

Step 3: Evaluate the integral

The integral of sec(x)\sec(x) is well-known:

sec(x)dx=lnsec(x)+tan(x)+C\int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C

Thus, the arc length is:

L=[lnsec(x)+tan(x)]0π6L = \left[ \ln|\sec(x) + \tan(x)| \right]_0^{\frac{\pi}{6}}

Step 4: Compute the values at the limits

At x=π6x = \frac{\pi}{6}:

sec(π6)=23,tan(π6)=13\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}, \quad \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}

So:

ln(sec(π6)+tan(π6))=ln(23+13)=ln(33)=ln(3)\ln\left( \sec\left(\frac{\pi}{6}\right) + \tan\left(\frac{\pi}{6}\right) \right) = \ln\left( \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} \right) = \ln\left( \frac{3}{\sqrt{3}} \right) = \ln(\sqrt{3})

At x=0x = 0:

sec(0)=1,tan(0)=0\sec(0) = 1, \quad \tan(0) = 0

So:

ln(sec(0)+tan(0))=ln(1)=0\ln\left( \sec(0) + \tan(0) \right) = \ln(1) = 0

Step 5: Final result

The exact length of the curve is:

L=ln(3)0=ln(3)=12ln(3)L = \ln(\sqrt{3}) - 0 = \ln(\sqrt{3}) = \frac{1}{2} \ln(3)

Thus, the exact length of the curve is:

L=12ln(3)L = \frac{1}{2} \ln(3)

Would you like further clarification or details on any step? Here are some related questions:

  1. How do you differentiate logarithmic functions?
  2. What is the trigonometric identity 1+tan2(x)1 + \tan^2(x)?
  3. How do you integrate sec(x)\sec(x)?
  4. What are the properties of the natural logarithm function?
  5. How does the chain rule apply to logarithmic differentiation?

Tip: Knowing common integrals, such as the integral of sec(x)\sec(x), can greatly speed up solving problems involving arc lengths.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Arc Length
Differentiation
Trigonometric Functions

Formulas

Arc length formula: L = ∫_a^b √(1 + (dy/dx)^2) dx
Derivative of ln(sec(x)): d/dx [ln(sec(x))] = tan(x)
Trigonometric identity: 1 + tan^2(x) = sec^2(x)

Theorems

Chain Rule
Trigonometric Identity for sec(x)
Integral of sec(x)

Suitable Grade Level

Undergraduate Calculus (Grades 11-12 and higher)