The hurdle and runner form a right triangle (see attached picture) such that
sin(30°) = <em>h</em> / (5 ft)
and
cos(30°) = <em>x</em> / (5 ft)
where <em>h</em> is the height of the hurdle and <em>x</em> is the horizontal distance from where the runner jumps to the hurdle. So
<em>h</em> = (5 ft) sin(30°) = 5/2 ft = 2.5 ft
<em>x</em> = (5 ft) cos(30°) = (5√3)/2 ft ≈ 4.33 ft
Cone details:
Sphere details:
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From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.
<u>Using Pythagoras Theorem</u>
(a)
TO² + TU² = OU²
(h-10)² + r² = 10² [insert values]
r² = 10² - (h-10)² [change sides]
r² = 100 - (h² -20h + 100) [expand]
r² = 100 - h² + 20h -100 [simplify]
r² = 20h - h² [shown]
r = √20h - h² ["r" in terms of "h"]
(b)
volume of cone = 1/3 * π * r² * h
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To find maximum/minimum, we have to find first derivative.
(c)
<u>First derivative</u>
<u>apply chain rule</u>
<u>Equate the first derivative to zero, that is V'(x) = 0</u>
<u />
<u>maximum volume:</u> <u>when h = 40/3</u>
<u>minimum volume:</u> <u>when h = 0</u>
