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)