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Question

You decide to warm your house by lighting your wood fire place. The smoke is coming out the chimney in the shape of an inverted right circular cone, the diameter of which is equal to its height. If the rate of smoke emission is 100m^3/min, how fast is the height increasing when the height is 30m?

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Answer 1 · 2017-08-15T12:34:03

Please see below.

Variables

Let #h# = the height of the cone

let #r# = the radius of the base of the cone

let #V# = the volume of the cone

(We also have #t# = time in minutes)

Rates of change

#(dV)/dt = 100# #m^3#/#min#

We need to find #(dh)/dt# when #h = 30# #m#

Additional information

#h = 2r#

Equation relating the variables

#V = 1/3 pir^2h#

Using the additional information, we can write #v# as a function of the variable we are most interested in, which is #h#.

#V = 1/12pih^3#

Finish the problem

Differentiate with respect to #t# (Use implicit differentiation / the chain rule.)
Substitute what we know and solve for what we don't know.

#d/dt(V) = d/dt(1/12pih^3)#

#(dV)/dt = 1/4pih^2 (dh)/dt#

#100 = 1/4pi(30)^2 (dh)/dt#

#100 = 15^2 pi (dh)/dt#

#(dh)/dt = 100/(15^2pi) = 4/(9pi)# #m# /#min#