Question

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?

Answer 1

`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"`

Answer 2

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)`