Calculate the arc length of the curve x = 3 (cos t + ln

Calculate the arc length of the curve $x = 3\left(\cos t + \ln\tan\frac{t}{2}\right)$, $y = 3\sin t$ between the points corresponding to parameter values $t_1 = \frac{\pi}{2}$ and $t_2 = \frac{2\pi}{3}$.

We differentiate carefully (the logarithmic term simplifies via a standard half-angle identity), then integrate $\cot t$ over the given range.

Step 1: Differentiate the logarithmic term

\[\frac{d}{dt}\ln\tan\frac t2 = \frac{1}{\tan\frac t2}\cdot\sec^2\frac t2\cdot\frac12 = \frac{1}{2\sin\frac t2\cos\frac t2} = \frac{1}{\sin t}\]

Step 2: Compute both derivatives

\[\frac{dx}{dt} = 3\left(-\sin t+\frac{1}{\sin t}\right) = 3\cdot\frac{1-\sin^2t}{\sin t} = \frac{3\cos^2t}{\sin t}, \qquad \frac{dy}{dt}=3\cos t\]

Step 3: Simplify the arc-length integrand

\[\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 = 9\cos^2t\left(\frac{\cos^2t}{\sin^2t}+1\right) = 9\cos^2t\cdot\frac{1}{\sin^2t} = \frac{9\cos^2t}{\sin^2t}\]

So the integrand is $\left|\dfrac{3\cos t}{\sin t}\right|$. For $t\in\left[\frac\pi2,\frac{2\pi}3\right]$: $\sin t>0$ but $\cos t<0$, so this equals $-\dfrac{3\cos t}{\sin t}$.

Step 4: Set up and evaluate the integral

\[L = -3\int_{\pi/2}^{2\pi/3}\cot t\,dt = -3\left[\ln|\sin t|\right]_{\pi/2}^{2\pi/3} = -3\left[\ln\frac{\sqrt3}{2}-\ln1\right] = -3\ln\frac{\sqrt3}{2}\]

Step 5: Simplify

\[-3\ln\frac{\sqrt3}{2} = -\frac32\ln3+3\ln2 = \frac32\left(2\ln2-\ln3\right) = \frac32\ln\frac43\]

\[\boxed{L = \frac32\ln\frac43 \text{ units}}\]

Step 6: Numerical value

\[L = \frac32\ln\frac43 \approx 0.432 \text{ units}\]

Summary

\[\boxed{L = \frac32\ln\frac43 \approx 0.432 \text{ units}}\]

Key insight: The identity $\frac{d}{dt}\ln\tan\frac t2=\frac1{\sin t}$ is exactly what causes the arc-length integrand to collapse to a bare $|\cot t|$ — without it, this curve's arc length would look far less tractable.