The length of the path of the ant from the top of the cylinder down the
point Q is a helical path.
Correct responses:
- The shortest path for PQ when R = 4'', H = 8' and n = 3 is 40.69 inches
- [tex]The \ shortest \ path \ for \ PQ \ is \ \underline{ \sqrt{\left(2\cdot n \cdot R \right)^2 + H^2 }}[/tex]
Steps by which the above values are found
The given description are;
The radius of the cylinder = R
Height of the cylinder = H
Point at which the ant starts = P at the top edge
Point at which the the ant finishes = Q at the bottom edge
a. Please find attached the drawing of the cylinder showing the points P
and Q, the height of the cylinder, H, and the radius R, created with MS
Visio.
b. The length of the radius, R, the height of the cylinder, H, the and a turn
of the ant, n, are labelled.
c. When R = 4 inches; H = 8 feet; n = 3, we have;
The shortest possible path is a helical line, given by the following equation;
[tex]Length \ of \ helix = \mathbf{ n \cdot \sqrt{C^2 + p^2}}[/tex]
Where;
C = Circumference of the circle = 2×π×R
[tex]\displaystyle p = The \ pitch = \mathbf{\frac{H}{n}}[/tex]
Therefore;
C = 2 × π × 4 inches = 8·π inches
[tex]\displaystyle p = \frac{8 \, ft.}{3} = \frac{96 \ inches}{3} = \mathbf{ 32 \ inches}[/tex]
Therefore;
Length of the helix = [tex]3 \times \sqrt{(8 \cdot \pi)^2 + 32^2} \approx 40.69[/tex]
- The length of the helical path of the ant ≈ 40.69 inches
d. The rule for the cylinder is as follows;
[tex]\displaystyle Length \ of \ ant \ path = n \times \sqrt{\left(2 \cdot R\right)^2 + \left(\frac{H}{n} \right)^2} = \sqrt{\left(2 \cdot R \cdot n\right)^2 + H^2 }[/tex]
The rule for the cylinder is therefore;
- [tex]\displaystyle Length \ of \ ant \ path \ from \ P \ to \ \underline{Q = \sqrt{\left(2 \cdot R \cdot n\right)^2 + H^2 }}[/tex]
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