# What is the arclength of (t-3,t+4) on t in [2,4]?

Apr 2, 2018

$A = 2 \sqrt{2}$

#### Explanation:

The formula for parametric arc length is:
$A = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$

We begin by finding the two derivatives:

$\frac{\mathrm{dx}}{\mathrm{dt}} = 1$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$

This gives that the arc length is:
$A = {\int}_{2}^{4} \sqrt{{1}^{2} + {1}^{2}} \setminus \mathrm{dt} = {\int}_{2}^{4} \sqrt{2} \setminus \mathrm{dt} = {\left[\sqrt{2} t\right]}_{2}^{4} = 4 \sqrt{2} - 2 \sqrt{2} = 2 \sqrt{2}$

In fact, since the parametric function is so simple (it is a straight line), we don't even need the integral formula. If we plot the function in a graph, we can just use the regular distance formula:

$A = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = \sqrt{4 + 4} = \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$

This gives us the same result as the integral, showing that either method works, although in this case, I'd recommend the graphical method because it is simpler.
The text contains a mathematical question about finding the arclength of a parametric function, and it provides a detailed solution to the problem.
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# What is the arclength of (t-3,t+4) on t in [2,4]?

Apr 2, 2018

$A = 2 \sqrt{2}$

#### Explanation:

The formula for parametric arc length is:
$A = {\int}_{a}^{b} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$

We begin by finding the two derivatives:

$\frac{\mathrm{dx}}{\mathrm{dt}} = 1$ and $\frac{\mathrm{dy}}{\mathrm{dt}} = 1$

This gives that the arc length is:
$A = {\int}_{2}^{4} \sqrt{{1}^{2} + {1}^{2}} \setminus \mathrm{dt} = {\int}_{2}^{4} \sqrt{2} \setminus \mathrm{dt} = {\left[\sqrt{2} t\right]}_{2}^{4} = 4 \sqrt{2} - 2 \sqrt{2} = 2 \sqrt{2}$

In fact, since the parametric function is so simple (it is a straight line), we don't even need the integral formula. If we plot the function in a graph, we can just use the regular distance formula:

$A = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}} = \sqrt{4 + 4} = \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}$

This gives us the same result as the integral, showing that either method works, although in this case, I'd recommend the graphical method because it is simpler.
user
The text contains a mathematical question about finding the arclength of a parametric function, and it provides a detailed solution to the problem.

TYPE: SINGLE_QUESTION_WITH_ANSWER