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?

Question

46735 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-07T13:30:09

#d'(t_0)=-1/(12π) # #"cm/min"#

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

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

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

                          ----

#<=># #-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"#

Answer 2 · 2017-12-08T01:24:46

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

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)#