Arc Length with Vector Functions — Question 7

Question 7

For $$\mathbf r(t)=\left\langle t^2,\ \frac 16(4t+1)^{3/2},\ 0\right\rangle,\qquad t\ge 0,$$ find the parameter value $T>0$ at which the distance traveled from $t=0$ first equals $35/4$.

Tasks

1. Derive the arc-length function $s(T)$.

2. Solve the resulting equation exactly.

3. Explain why the solution is unique.

Question 7 – Solution

Strategy. Differentiate carefully; the squared speed is a perfect square on the stated domain.

Step 1: Speed $$\mathbf r'(t)=\left\langle 2t,\sqrt{4t+1},0\right\rangle.$$ For $t\ge 0$, $$\begin{align*} \|\mathbf r'(t)\| &=\sqrt{4t^2+4t+1}\\ &=\sqrt{(2t+1)^2}=|2t+1|=2t+1. \end{align*}$$

Step 2: Arc-length function $$\begin{align*} s(T)&=\int_0^T(2t+1)\,dt\\ &=\left[t^2+t\right]_0^T=T^2+T. \end{align*}$$ Set this equal to the prescribed distance: $$T^2+T=\frac{35}{4}.$$ Multiplying by $4$ and factoring, $$4T^2+4T-35=(2T-5)(2T+7)=0.$$ The candidates are $T=5/2$ and $T=-7/2$; only the first lies in the domain. Hence $$\boxed{T=\frac 52}.$$

Step 3: Uniqueness For $T>0$, $s'(T)=2T+1>0$, so $s$ is strictly increasing. Therefore the displayed equation has at most one positive solution. Since $s(0)=0$ and $s(T)\to\infty$, it has exactly one. Direct substitution gives $s(5/2)=25/4+10/4=35/4$, verifying the result.