2 answers

A snowball, in the shape of a sphere, is melting so that the radius is decreasing at a uniform rate

Question:

A snowball, in the shape of a sphere, is melting so that the radius is decreasing at a uniform rate of 1cm/s. How fast is the volume decreasing when the radius equals 6cm?

Answers

We can write the first relationship (in #"cm/s"#) as:

#(dr)/(dt) = 1#

We know that the formula for the volume of a sphere is:

#V(r) = 4/3 pi(r(t))^3#

Also, the derivative of #V# with respect to #t#, #(dV)/(dt)#, implies that we also multiply by #(dr)/(dt)#, since #r = r(t)#, whose slope is #1#, and #(dV(r))/(dt) = (dV(r))/(dr(t))*(dr(t))/(dt) = (dV)/(dr)*(dr)/(dt)#.

Thus, if we differentiate this with respect to #t#, we can find a use for #(dr)/(dt)# through the Chain Rule.

#(dV)/(dt) = 4/3pi d/(dt)[(r(t))^3]#

#= 4/3pi d/(dr)[r^3] ((dr)/(dt))#

#= 4/3pi*3r^2((dr)/(dt))#

Finally, we can plug it in:

#= 4/3 pi (3r^2)(1) = color(green)(4 pir^2)#

Therefore, when #r = 6#, we get the rate of decrease in the volume while the radius is currently #"6 cm"#:

#(dV(6))/(dt) = V'(6) = 4 pi (36)#

# = color(blue)(144pi "cm"^3"/s")#

.

The volume is decreasing at a rate of #144picm^3.s^-1#

Explanation:

We are told that #(dr)/(dt)=-1"cm/s"#

We need to find #(dV)/(dt)# when #r=6"cm"#

We know:

#V=(4)/(3).pir^3#

So:

#(dV)/(dr)=(12)/(3).pir^2=4pir^2#

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

#(dV)/(dt)=4pir^(2)xx(-1)=-4pir^2#

So when the radius #r=6"cm"# we can substitute this in :

#(dV)/(dt)=-4pi.6xx6=-144picm^3.s^-1#

.

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