Find the length of the curve r(t) = 7t, 3 cos(t), 3 sin(t) , -2 ≤ t ≤ 2

Solution:

It is given that,

r(t) = 7t, 3 cos(t), 3 sin(t)

Let x = 7t, y = 3cos(t), z = 3 sin (t)

Now, we are going to use arc length formula,

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

By finding the derivative,

x’= 7, y’ = -3 sin(t), z’ = 3 cos (t)

By substituting the values we get,

\( L = \int_{-2}^{2}\sqrt{(7)^{2}+(-3sin(t)^{2})+(3cos(t)^{2})}dt \)

Then,

\( L = \int_{-2}^{2}\sqrt{(7)^{2}+9}dt \)

L = √58 × 2 - (-2√58)

L = 4√58

Therefore, the length of the curve is 4√58.

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Find the length of the curve r(t) = 7t, 3 cos(t), 3 sin(t) , -2 ≤ t ≤ 2

Summary:

The length of the curve r(t) = 7t, 3 cos(t), 3 sin(t), -2 ≤ t ≤ 2 is 4√58.