Arc Length with Vector Functions — Question 1

Question 1

Find the exact length of $$\mathbf r(t)=\left\langle 3t,\ 4t,\ \frac 12t^2\right\rangle,\qquad 0\le t\le 2.$$ Tasks

1. Compute the speed and simplify it completely.

2. Set up and evaluate the arc-length integral.

3. Check that the result exceeds the straight-line distance between the endpoints.

Question 1 – Solution

Strategy. Arc length is the integral of speed: $L=\int_a^b\|\mathbf r'(t)\|\,dt$.

Step 1: Velocity and speed $$\mathbf r'(t)=\left\langle 3,4,t\right\rangle.$$ Therefore $$\|\mathbf r'(t)\|=\sqrt{3^2+4^2+t^2}=\sqrt{25+t^2}.$$

Step 2: Integrate Use $$\int\sqrt{t^2+a^2}\,dt =\frac 12\left(t\sqrt{t^2+a^2}+a^2\ln\left|t+\sqrt{t^2+a^2}\right|\right).$$ With $a=5$, $$\begin{align*} L&=\int_0^2\sqrt{25+t^2}\,dt\\ &=\frac 12\left[t\sqrt{25+t^2}+25\ln\left(t+\sqrt{25+t^2}\right)\right]_0^2\\ &=\boxed{\sqrt{29}+\frac{25}{2}\ln\left(\frac{2+\sqrt{29}}5\right)}. \end{align*}$$

Step 3: Endpoint check The displacement is $$\mathbf r(2)-\mathbf r(0)=\left\langle 6,8,2\right\rangle,$$ whose length is $\sqrt{36+64+4}=2\sqrt{26}\approx 10.198$. The arc length is approximately $10.261$, which is slightly larger, as a nonstraight path requires.