Word problem using rate of change and integrals

Question

A giant spherical snowball is melting such that its radius decreases at a rate of 5 mm per hour. How fast is the surface area changing when the diameter is two meters? How fast is the volume changing at that time?

 

Answer 1

Let r = radius (in meters) at time t

 

      A = surface area (in square meters) at time t

 

      V = volume (in cubic meters) at time t

 

A = 4πr²      V = (4/3)πr³

 

Given:  dr/dt =  0.005m/hr

 

Find:  dA/dt and dV/dt when r = 1 meter 

 

To find dA/dt, differentiate the surface area equation implicitly with respect to t:

 

      A = 4πr²    dA/dt = 8πr(dr/dt) = 8π(1)(0.005) = (0.04π) m²/hr

 

To find dV/dt, differentiate the volume equation implicitly with respect to t:

 

     V = (4/3)πr³     dV/dt = 4πr²(dr/dt) = 4π(1)²(0.005)

 

                                                                     = (π/50) m³/hr

 

 

Answer 2

When we have rates, we are dealing with derivatives.   Be sure to convert units here too.  You have millimeters and meters.

 

 

Here is what we know:

 

r = radius = 1m    ------>    diameter is 2m

A = 4πr²          ------->    area of sphere

 

dr / dt = 5 mm = 0.005 m      ----->   this is the rate of radius

 

 

What we need to find:

 

dA / dt

 

 

We can write a formula to obtain dA / dt.

 

 

dA / dt = (dA / dr) * (dr / dt)

 

 

We need to first find dA/dr.   This is the derivative of the area in terms of r.

 

dA / dr = 8πr

 

 

plugging in r=1 into dA/dr, we obtain

 

dA / dr = 8π

 

 

dA / dt = 8π * 0.05

           = 160π

 

The surface area is decreasing (2π / 5) (m² / hr).