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NNADVOKAT [17]
3 years ago
15

PLEASE PLEASE HELP I AM DESPERATE 20 POINTS!!!

Mathematics
1 answer:
lina2011 [118]3 years ago
3 0
Https://www.khanacademy.org/math/algebra/quadratics/solving-quadratics-using-the-quadratic-formula/v...
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if a snowball rolls down a hill and its surface area increaces at a rate of 18cm^2/min, find the rate at which the radius increa
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Answer: dr/dt = 9/(24pi)  cm per minute

9/(24pi) is approximately equal to 0.119366

=============================================

Work Shown:

Given info

dS/dt = 18 cm^2/min is the rate of change of the surface area

r = 6 cm is the radius, from the fact that the diameter is 12 cm

--------

Use the surface area equation given, apply the derivative, plug in the given values and then isolate dr/dt which represents the rate of change for the radius

S = 4*pi*r^2

dS/dt = 2*4*pi*r*dr/dt

dS/dt = 8*pi*r*dr/dt

18 = 8*pi*6*dr/dt

18 = 48*pi*dr/dt

48pi*dr/dt = 18

dr/dt = 18/(48pi)

dr/dt = (9*2)/(24*2pi)

dr/dt = 9/(24pi)

The units are cm per minute, which can be written as cm/min.

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