Find the exact mass of a thin wire in the shape of the helix $x = 2 \sin ( t ) , y = 2 \cos ( t ) , z = 3 t , 0 \leq t \leq 2 \pi$ if the density is 5 .
A) $4 \sqrt { 13 }$
B) $8 \sqrt { 10 } \pi ^ { 2 }$
C) $10 \sqrt { 13 }$
D) $10 \sqrt { 13 } \pi$
E) $\sqrt { 13 } \pi$

Question

Quizplus · Accepted Answer

First, we need a parametric equation for the helix in terms of the variable t:
$$\vec{r}(t) = \langle 2\sin(t), 2\cos(t), 3tangle$$

Next, we need to find the arc length formula for a helix. Since the helix is a three-dimensional curve, we actually need the three-dimensional arc length formula:

$$L = \int_a^b \sqrt{\left(\frac{dx}{dt}ight)^2 + \left(\frac{dy}{dt}ight)^2 + \left(\frac{dz}{dt}ight)^2} dt$$

Plugging in our parametric equations for x, y, and z, we get:

$$L = \int_0^{2\pi} \sqrt{4\cos^2(t) + 4\sin^2(t) + 9} dt$$

Simplifying under the square root, we get:

$$L = \int_0^{2\pi} \sqrt{13} dt$$

$$L = \sqrt{13} \int_0^{2\pi} 1 dt$$

$$L = \sqrt{13} \cdot 2\pi$$

Finally, we can use the formula for the mass of a wire:

$$M = 	ext{density} \cdot 	ext{length}$$

Plugging in our values, we get:

$$M = 5 \cdot \sqrt{13} \cdot 2\pi$$

$$M = 10\sqrt{13}\pi$$

Therefore, the answer is D.

Find the Exact Mass of a Thin Wire in the Shape