If a snowball melts so its surface area decreases at a rate of 1 cm^2 /min , find the rate at which the diameter decreases when the diameter is 6 cm?

2 Answers
Dec 7, 2017

#d'(t_0)=-1/(12π) # #"cm/min"#

Explanation:

  • #Area _((snowball))=4πr^2#
  • #A(t)=4πr^2(t)#
  • #A'(t)=2*4πr(t)r'(t)# #=# #4π*2r(t)r'(t)#

For #color(purple)(t=t_0)# , #A'(t_0)=-1#

, #d(t)=6# #<=># #2*r(t)=6#

, #d'(t)=2*r'(t)#

                          ----
  • #A'(t_0)=4π*2r(t_0)r'(t_0)# #<=>#

#<=># #-1=4π*d(t_0)*r'(t_0)# #<=>#

#<=># #-1=4π*6*r'(t_0)# #<=>#

#<=># #-1=2π*6*2r'(t_0)# #<=>#

#<=># #-1=2π*6*d'(t_0)# #<=>#

#<=># #-1=12π*d'(t_0)# #<=>#

#<=># #d'(t_0)=-1/(12π)# #"cm/min"#

Dec 8, 2017

The diameter is decreasing at a rate of #1/(12pi) \ "cm " "min"^(-1)#

Explanation:

Let us setup the following variables:

# { (D, "Diameter of snowball at time t","(cm)"), (S, "Surface area of snowball at time t", "(cm"^3")"), (t, "time", "(min)") :} #

The standard formula for Surface Area of a sphere with radius #r# is :

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

Differentiating wrt #D#, we have:

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

We are given that the constant rate at which surface area decreases is 1, Thus:

# (dS)/(dt) = -1 #

And, we seek the value of #(dD)/dt# when #D=6#, which we expect to be negative (as the volume is decreasing). Applying the chain rule, we have:

# (dD)/(dt) = (dD)/(dS) * (dS)/(dt) #
# \ \ \ \ \ \ = 1/(2piD) * (-1) #
# \ \ \ \ \ \ = -1/(2piD) #

So then:

# [ (dD)/(dt) ]_(D=5) = -1/(2pi * 6) = -1/(12pi)#

Thus the diameter is decreasing at a rate of #1/(12pi) \ "cm " "min"^(-1)#