Find the length of the curve r(t) = 9t, 3 cos(t), 3 sin(t) , -3 ≤ t ≤ 3?

Solution:

r(t) = 9t, 3 cos(t), 3 sin(t) [Given]

We know that

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

So we get

x = 9t

y = 3 cos(t)

z = 3 sin(t)

The arc length formula can be used

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

We know that the derivatives are

x’ = 9

y’ = -3 sin (t)

z’ = 3 cos (t)

Now substitute these values

\( \\L =\int_{-3}^{3}\sqrt{(9)^{2}+(-3 \sin t)^{2}+(3 \cos t)^{2}}dt \\=\int_{-3}^{3}\sqrt{9^{2}+9}dt \\ =\int_{-3}^{3}3\sqrt{10}dt \\ =3\sqrt{10}t|^{3}_{-3}\\L=18\sqrt{10} \)

Therefore, the length of the curve is 18√10.

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Find the length of the curve r(t) = 9t, 3 cos(t), 3 sin(t) , -3 ≤ t ≤ 3?

Summary:

The length of the curve r(t) = 9t, 3 cos(t), 3 sin(t) , -3 ≤ t ≤ 3 is 18√10.