Question

Find the arc length of the curve on the given interval. (round your answer to three decimal places.) parametric equations interval x = 6t + 5, y = 7 − 7t −1 ≤ t ≤ 3

Answer 1

$\displaystyle\int_{-1}^3\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

$x=6t+5\implies\dfrac{\mathrm dx}{\mathrm dt}=6$
$y=7-7t\implies\dfrac{\mathrm dy}{\mathrm dt}=-7$

$\displaystyle\int_{-1}^3\sqrt{36+49}\,\mathrm dt=\sqrt{85}(3-(-1))=4\sqrt{85}$

Answer 2

The arc length of the curve on the given interval −1 ≤ t ≤ 3 with parametric equation x = 6t + 5 and y = 7 − 7t −1 is 4√85.

What is integration?

It is the reverse of differentiation.

The arc length of the curve on the given interval.

Parametric equations interval

x = 6t + 5,  −1 ≤ t ≤ 3

y = 7 − 7t,  −1 ≤ t ≤ 3

We know that the parametric form of the arc length will be given as

$\rm \int _{-1}^3 \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2} \ dt$

Then we have

$\rm \dfrac{dx}{dt} = 6\\\\\dfrac{dy}{dt} = -7$

Then the arc length will be

$\rightarrow \rm \int _{-1}^3 \sqrt{(3)^2 + (-7)^2} \ dt\\\\\rightarrow \sqrt{85} [t]_{-1}^3 \\\\\rightarrow 4 \sqrt{85}$

More about the integration link is given below.

https://brainly.com/question/18651211