Question

A snowball is melting at a rate of 3 cm3 per second. If it remains spherical, how fast is the surface area decreasing when the diametre is 8 cm?

Answer 1

Explanation:

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