ab85c4…: “Down a proud fact”

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Suppose that the curve $C$ in the $xy$ -plane is traced out by the vector-valued function

$\vec {r}(t) = \vector {2t^{3/2},4t+1}, 0 \leq t \leq 4.$

In order to determine if this continuously differentiable curve is parameterized by arclength, we could check either if

$\int _0^t \left |\vec {r}'(\tau )\right | \d \tau = t$
$\left |\vec {r}'(t)\right |=1$ for all $t$ .

If either of these holds, then the curve uses arclength as a parameter. Furthermore, in order to establish the first result, we would have to compute $\left |\vec {r}'(t)\right |$ anyways, so let’s take this approach.

We calculate that $\vec {r}'(t) = \vector {\answer {3t^{1/2}},\answer {4}}$ , and hence $\left |\vec {r}'(t)\right | = \answer {\sqrt {9t+16}}$ .

Since $\left |\vec {r}'(t)\right | \neq 1$ for all $t$ , the curve doesdoes not use arclength as a parameter.

In order to find a description $\vec {p}(s)$ that does use arclength as a parameter, we can take the following steps.

Find $s$ in terms of $t$ by computing $s = \int _0^t \left |\vec {r}'(\tau )\right | \d \tau$ .
Since $\left |\vec {r}'(t)\right | = \sqrt {9t+16}$ , $\left |\vec {r}'(\tau )\right | = \sqrt {9\tau +16}$ , and thus
$\begin{align*} s &= \int_0^t \left|\vec{r}'(\tau)\right| \d \tau \\ &= \int_0^t \sqrt{9\tau+16} \d \tau \end{align*}$
Note that we need to find the antiderivative $\int \sqrt {9x+16} \d x$ to proceed, and
$\int \sqrt {9x+16} \d x = \answer {\frac {2}{27} (9x+16)^{3/2}} +C.$
Thus, we find that $s = \answer {\frac {2}{27} (9t+16)^{3/2} - \frac {128}{27}}$ .

We can calculate $\int \sqrt {9x+16} \d x$ with the substitution $u=9x+16$ . Don’t forget to evaluate the antiderivative at $t=0$ when you find $s$ .
Solve for $t$ in terms of $s$ .
This requires some careful algebra, after which we find
$t = \answer {\frac {\left (\frac {27}{2}s+64\right )^{2/3}-16}{9}}$

Let’s work through the start of the algebra one step at a time. $\begin{align*} s &= \frac{2}{27} (9t+16)^{3/2} - \frac{128}{27} \\ s +\frac{128}{27} &= \frac{2}{27} (9t+16)^{3/2} \\ \frac{27}{2}s +64 &= (9t+16)^{3/2}& (\textrm{ multiply both sides by } \frac{27}{2} \textrm{ and simplify. } ) \\ \left(\frac{27}{2}s +64\right)^{\answer{2/3}} &= 9t+16 \\ \end{align*}$
From here, there’s not too much more work necessary to solve for $t$ .
We can now find the parameterization $\vec {p}(s)$ by substituting the expression above for $t$ in the original parameterization. $\begin{align*} \vec{r}(t) &= \vector{2t^{3/2},4t+1} , 0 \leq t \leq 4 \\ \vec{p}(s) &= \vector{2\left(\answer{\frac{\left(\frac{27}{2}s+64\right)^{2/3}-16}{9} } \right)^{3/2},4\left( \answer{\frac{\left(\frac{27}{2}s+64\right)^{2/3}-16}{9}} \right)+1} \end{align*}$
Note that we also need to transform the domain as well. Since the original domain is $0 \leq t \leq 4$ , we have $0 \leq \frac {\left (\frac {27}{2}s+64\right )^{2/3}-16}{9} \leq 4$ , or
$\textrm { Domain in } s: ~ 0 \leq s \leq \answer {\frac {2}{27}\left ((52)^{3/2}-64\right )}$

Note that the arclength parameterization here is quite painful. In practice, the best we can hope for is that this parameterization is painful, but possible. Since finding $s$ in terms of $t$ in general requires that we evaluate an integral that has a square root, we often will not be able to compute the necessary antiderivative explicitly. Even if we can, we still have to solve for $t$ in terms of $s$ .

In general, arclength is not generally a useful parameter to use explicitly, but it is a fundamentally important one to use when formulating definitions that have physical meaning, as a later exercise will explore. A common theme in results that require this is to take the definition in terms of arclength and establish a computational result in terms of the parameter you want to use. This will be the subject of a later problem in this assignment.