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# Find the length of the curve. r(t) = 5t, 3 cos(t), 3 sin(t) , -2 ≤ t ≤ 2?

**Solution:**

r(t) = 5t, 3 cos(t), 3 sin(t) [Given]

We have to compare it with

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

So x = 5t, y = 3 cos (t), z = 3 sin (t)

By using the arc length formula

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

We know that

x’ = 5

y’ = -3 sin t

z’ = 3 cos t

Now substitute these values

\( \\L=\int_{-2}^{2}\sqrt{(5)^{2}+(-3 \sin t)^{2}+(3 \cos t)^{2}}dt \\ \\L=\int_{-2}^{2}\sqrt{(5)^{2}+9}dt \\ \\L=\int_{-2}^{2}5\sqrt{2}dt \\ L = 5\sqrt{2} . t |_{-2}^{2}\\L=20\sqrt{2}\)

Therefore, the length of the curve is 20√2.

## Find the length of the curve. r(t) = 5t, 3 cos(t), 3 sin(t) , -2 ≤ t ≤ 2?

**Summary:**

The length of the curve r(t) = 5t, 3 cos(t), 3 sin(t) , -2 ≤ t ≤ 2 is 20√2.

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