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?

Question

106377 views around the world

You can reuse this answer
Creative Commons License

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