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

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

[tex]x=6t+5\implies\dfrac{\mathrm dx}{\mathrm dt}=6[/tex]
[tex]y=7-7t\implies\dfrac{\mathrm dy}{\mathrm dt}=-7[/tex]

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

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jainveenamrata jainveenamrata · Answer 2 · 2022-04-27T09:07:05+02:00

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

[tex]\rm \int _{-1}^3 \sqrt{(\dfrac{dx}{dt})^2 + (\dfrac{dy}{dt})^2} \ dt[/tex]

Then we have

[tex]\rm \dfrac{dx}{dt} = 6\\\\\dfrac{dy}{dt} = -7[/tex]

Then the arc length will be

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

More about the integration link is given below.

https://brainly.com/question/18651211