If a snowball melts so that its surface area decreases at a rate of 3 cm2/min, how do you find the rate at which the diameter decreases when the diameter is 12 cm?

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Answer 1 · 2016-12-10T22:29:29

diameter is decreasing at rate of #3/(24pi)# (#0.040# (2sf)) #"cm"^2"/min"#

Assuming the snowball is a perfect sphere, then if #A# denotes the surface area and #D# the diameter then:

# A = 4pir^2 = 4pi(D/2)^2 = piD^2 #

Differentiating wrt #r# we have:

# (dA)/(dD) = 2piD #

We are told that # (dA)/dt=-3# and we want to find #(dD)/dt#

By the chain rule we have:

# (dA)/(dD) = (dA)/(dt)*(dt)/(dD) = ((dA)/(dt)) / ((dD)/dt)#
# :. 2piD = -3/ ((dD)/dt) #
# :. (dD)/dt = -3/ (2piD) #

When #D=12 => (dD)/dt = -3/ (24pi) = -1/(8pi) # (#~~0.039788...#)
The sign confirms that #D# is decreasing