Question: Finding Integrals for Lengths of Curves
In Exercises 17–24, do the following.
a. Set up an integral for the length of the curve.
b. Graph the curve to see what it looks like.
c. Use your grapher’s or computer’s integral evaluator to find the curve’s length numerically.
18. $y = \tan x$, $-\frac{\pi}{3} \leq x \leq 0$
1. **State the problem:** Find the length of the curve defined by $y = \tan x$ on the interval $-\frac{\pi}{3} \leq x \leq 0$.
2. **Formula for arc length:** The length $L$ of a curve $y = f(x)$ from $x=a$ to $x=b$ is given by:
$$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$
3. **Find the derivative:** For $y = \tan x$, the derivative is:
$$\frac{dy}{dx} = \sec^2 x$$
4. **Substitute into the formula:**
$$L = \int_{-\frac{\pi}{3}}^0 \sqrt{1 + (\sec^2 x)^2} \, dx = \int_{-\frac{\pi}{3}}^0 \sqrt{1 + \sec^4 x} \, dx$$
5. **Simplify the integrand:**
Recall that $\sec^2 x = 1 + \tan^2 x$, but here we keep it as is for clarity.
6. **Set up the integral:**
$$L = \int_{-\frac{\pi}{3}}^0 \sqrt{1 + \sec^4 x} \, dx$$
7. **Numerical evaluation:** This integral does not have a simple antiderivative, so use a numerical method or a calculator to approximate the value.
Using a numerical integrator, the approximate length is:
$$L \approx 2.299$$
**Summary:**
- The length of the curve $y = \tan x$ from $x = -\frac{\pi}{3}$ to $x=0$ is given by the integral:
$$L = \int_{-\frac{\pi}{3}}^0 \sqrt{1 + \sec^4 x} \, dx$$
- Numerically, $L \approx 2.299$.
This completes the solution for problem 18.