How fast is the radius changing when diameter of the snowball is 10 cm given a spherical snowball with an outer layer of ice melts so that the volume of the snowball decreases at a rate of 2cm per 3min?

Question

5357 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-02-28T20:09:41

The radius is decreasing at a rate of #0.0021"cm/min"#

We need to find #(dr)/(dt)#

We know #(dV)/(dt)=-2/3=-0.66cm^(3)permin#

So lets set up:

#(dr)/(dt).(dV)/(dr)=(dV)/(dt)#

#V=4/3pir^3#

So #(dV)/(dr)=4pir^2#

So:

# (dr)/(dt).4pir^2=-0.66#

#(dr)/(dt)=-(0.66)/(4xx3.142xx5^2)#

#(dr)/(dt)=-0.0021"cm/min"#

Answer 2 · 2015-02-28T20:26:10

You can try this:
enter image source here

(check my maths anyway!!! Hope it helps)