# (a) Find a parametrization (functions for x and y , each as a function of parameter t ) for...

## Question:

(a) Find a parametrization (functions for {eq}x {/eq} and {eq}y {/eq}, each as a function of parameter {eq}t {/eq}) for the line segment between the points {eq}(3,3) {/eq} and {eq}(-1, -5) {/eq}. Include the domain of {eq}t {/eq}.

(b) Write down, but do not evaluate, the integral for the arc length of the curve defined by:

{eq}\begin{array}{} x(t) = 3t^2 + 2, & y(t) = \sin (t), & 0 \leq t \leq 2 \pi \end{array} {/eq}

## Parametric Equations:

To parameterize a line segment, we can always do so on {eq}t \in [0,1] {/eq} by treating the parameter like an off/on switch, so that is what we will do for the first problem. To find the arc length of a function on an interval given a parameterization, we can use the following integral formula:

{eq}\begin{align*} L &= \int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \ dt \end{align*} {/eq}

We will need it for the second problem.

Part A

We start with the domain of the parameter; we want {eq}t \in [0,1] {/eq}. Then the parameter is really just an off/on switch. We need to subtract 4 from {eq}x {/eq} to get it from 3 to -1, and we need to subtract 8 from {eq}y {/eq} to get it from 3 to -5. So a parameter is

{eq}\begin{align*} x &= 3 - 4t \\ y &= 3 - 8t \end{align*} {/eq}

Part B

We have

{eq}\begin{align*} \frac{dx}{dt} &= \frac{d}{dt} \left( 3t^2+2 \right) \\ &= 6t \end{align*} {/eq}

and

{eq}\begin{align*} \frac{dy}{dt} &= \frac{d}{dt} \left( \sin t \right) \\ &= \cos t \end{align*} {/eq}

Then we have

{eq}\begin{align*} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } &= \sqrt{ \left( 6t \right)^2 + \left(\cos t \right)^2 } \\ &= \sqrt{36t^2 + \cos^2 t} \end{align*} {/eq}

And so the integral for the arc length on {eq}t \in [0,2\pi] {/eq} is

{eq}\begin{align*} L &= \int_0^{2\pi} \sqrt{36t^2 + \cos^2 t}\ dt \end{align*} {/eq}