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

Paisley Johnson · Accepted Answer

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

Elijah Abram · Answer

undefined To find the rate at which the diameter decreases when the diameter is 12 cm, you can use the formula for the surface area of a sphere:

$$ A = 4\pi r^2 $$

Differentiating both sides with respect to time $ t $, you get:

$$ \frac{dA}{dt} = 8\pi r \frac{dr}{dt} $$

Given that $ \frac{dA}{dt} = -3 \, \text{cm}^2/\text{min} $ (negative because the surface area is decreasing), and when $ r = 6 \, \text{cm} $, you can substitute these values into the equation and solve for $ \frac{dr}{dt} $.

$$ -3 = 8\pi (6) \frac{dr}{dt} $$

Solving for $ \frac{dr}{dt} $:

$$ \frac{dr}{dt} = -\frac{1}{8\pi} \, \text{cm/min} $$

So, the rate at which the diameter decreases when the diameter is 12 cm is $ -\frac{1}{8\pi} \, \text{cm/min} $.