Question

Q. Find the arc length of the curve
γ(t)=(t,lnt,tlnt), 1≤t≤2

Answer 1

Answer on Question #72686, Math / Geometry

Find the arc length of the curve

$$\gamma(t) = (t, \ln t, t \ln t), 1 \leq t \leq 2$$

Solution

The length of a curve from $t = a$ to $t = b$ is given by

$$\int_{a}^{b} \sqrt{(x'(t))^2 + (y'(t))^2 + (z'(t))^2} \, dt$$

We call this arc length.

Then

$$L = \int_{1}^{2} \sqrt{((t)')^2 + ((\ln t)')^2 + ((t \ln t)')^2} \, dt$$ $$L = \int_{1}^{2} \sqrt{1 + \frac{1}{t^2} + (\ln t)^2 + 2 \ln t + 1} \, dt \approx 1.858054696497771$$

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