1) A Ferris wheel with a radius of 10 m is rotating at a rate of one revolution

Question

1) A Ferris wheel with a radius of 10 m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when his seat is 16 m above ground level?
??? m/min

2) If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm.
?? cm/min

3) A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 7 ft/s along a straight path. How fast is the tip of his shadow moving when he is 30 ft from the pole?
??? ft/s

Explanation / Answer

1. we ve

h = 10 sin theta

dh/dt = 10 cos theta d theta/dt

d theta / dt = 2 pie/2 = pie

when the seat is 16m above ground level we ve

h = 16-10 = 6m

cos theta = sqrt (16^2 - 6^2)/10 = 8/10

dh/dt = 10*8/10 d theta /dt

= 80/10 * pie = 8 pie m/min...

2. let S denote the surface area

and x be diameter of sphere...

since

S = 4 pie r^2 = 4pie (x/2)^2

= pie x^2

ds/dt = 2piex * dx/dt

also ds/dt = -3

and x = 10

hence

dx/dt = -3/20 pie

3 . From similarity of triangle we ve

6/15 = (A - 30) / A

AT = 50 ft

as we know dat

dv = ds/dt

hence

S = v* t

the total time guy spared t = S/v = 30/7 = 4.29 seconds

same time spend by the tip of the shadow, but the space (") is bigger

50 = v * t = v* 4.29

hence

v = 50/4.29 = 11.66 ft/s

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