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

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 the derivatives of x, y and z are

x’ = 5

y’ = -3 sin t

z’ = 3 cos t

Now substitute these values

\(L= \int_{-5}^{5}\sqrt{(5)^{2}+(-3 sint)^{2}+(3cost)^{2}}dt \\\\ =\int_{-5}^{5}\sqrt{25+9(cos^2t+sin^2 t)} \\\\=\int_{-5}^{5}\sqrt{25 +9}. dt \\\\ =\sqrt{34}\int_{-5}^{5} dt \\\\ =\sqrt{34}. t |_{-5}^{5}\\\\=\sqrt{34} [5-(-5)]\\\\= \sqrt{34}\times 10\\\\=58.3\)

Therefore, the length of the curve is 58.3

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

Summary:

The length of the curve r(t) = 5t, 3 cos(t), 3 sin(t) , -5 ≤ t ≤ 5 is 58.3