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

Solution:

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

We have to compare it with

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

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

By using the length formula

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

We know that,

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

Differentiate with respect to 't'

x’ = 3

y’ = -3 sint

z’ = 3 cost

Now substitute these values

\(L=\int_{-3}^{3}\sqrt{(3)^{2}+(-3 sint)^{2}+(3cost)^{2}}dt \\ \\L=\int_{-3}^{3}\sqrt{(3)^{2}+9sin^2{t}+9cos^2{t}}dt\\ \\L=\int_{-3}^{3}\sqrt{(3)^{2}+9(sin^2{t}+cos^2{t})}dt\\ \\L=\int_{-3}^{3}\sqrt{(3)^{2}+9}dt \\ \\L=\int_{-3}^{3}3\sqrt{2}dt \\ \\L=(3-(-3)×3\sqrt{2}\\ \\L=6× 3\sqrt{2}\\ \\L=18\sqrt{2}\)

Therefore, the length of the curve is 18√2.

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

Summary:

The length of the curve r(t) = 3t, 3 cos(t), 3 sin(t) , -3 ≤ t ≤ 3 is 18√2.