Find the length of the curve. r(t) = 3t, 3 cos(t), 3 sin(t) , -4 ≤ t ≤ 4?

Solution:

Given: r(t) = 3t, 3 cos(t), 3 sin(t)

We know that

r(t) = (x, y, z)

So we get

x = 3t, y = 3 cos(t), z = 3 sin(t)

The length formula can be used

\(L = \int_{b}^{a}\sqrt{(x')^{2}+(y')^{2}+(z')^{2}}dt\)

We know that

x = 3t, y = 3 cos(t), z = 3 sin(t)

Differentiate with respect to 't'

x’ = 3

y’ = -3 sin (t)

z’ = 3 cos (t)

Now substitute these values, we get

\(L=\int_{-4}^{4}\sqrt{(3)^{2}+(-3\sin t)^{2}+(3\cos t)^{2}}dt \\ =\int_{-4}^{4}\sqrt{(3)^{2}+9 \sin^2{t}+9 \cos^2{t}}dt\\ =\int_{-4}^{4}\sqrt{(3)^{2}+9(\sin^2{t}+\cos^2{t})}dt\\ =\int_{-4}^{4}\sqrt{(3)^{2}+9}dt \\ =\int_{-4}^{4}3\sqrt{2}dt \\ =3\sqrt{2} t|_{-4}^{4}\\ L=(4-(-4)×3\sqrt{2}\\ \\L=8× 3\sqrt{2}\\ \\L=24\sqrt{2}\)

Therefore, the length of the curve is 24√2.

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Find the length of the curve. r(t) = 3t, 3 cos(t), 3 sin(t) , -4 ≤ t ≤ 4?

Summary:

The length of the curve r(t) = 3t, 3 cos(t), 3 sin(t) , -4 ≤ t ≤ 4 is 24√2.