2. A metal rod has the shape of a right circular cylinder. As it is being heated, its length
is increasing at a rate of 0.005 cm/min and its diameter is increasing at 0.002
cm/min. At what rate is the volume changing when the rod has a length of 40 cm
and a diameter of 3 cm?
Let's denote the length of the cylinder as L and its diameter as D. The radius of the cylinder, r, is half of the diameter, so r = D/2.
The volume V of a right circular cylinder is given by the formula:
V = πr^2L
We are given that dL/dt = 0.005 cm/min and dD/dt = 0.002 cm/min. We want to find dV/dt, the rate at which the volume is changing.
First, we'll express the volume in terms of D and L to make differentiation straightforward:
V = π(D/2)^2L
V = (π/4)D^2L
Now, to find dV/dt, we take the derivative of V with respect to time t using the product rule since V is a product of the functions of L and D:
dV/dt = d/dt [(π/4)D^2L]
Using the product rule, we have:
dV/dt = (π/4) [2D(dD/dt)L + D^2(dL/dt)]
Plugging in the values for dD/dt and dL/dt:
dV/dt = (π/4) [2D(0.002)L + D^2(0.005)]
Substituting D = 3 cm and L = 40 cm:
dV/dt = (π/4) [2(3)(0.002)(40) + (3)^2(0.005)]
dV/dt = (π/4) [0.24 + 0.45]
dV/dt = (π/4) [0.69]
Simplifying this further:
dV/dt = (π/4) * 0.69
dV/dt = (π/4) * 69/100
dV/dt = (69π)/400
And calculating the value:
dV/dt ≈ (69π)/400 cm³/min
dV/dt ≈ 0.1725π cm³/min
dV/dt ≈ 0.5415 cm³/min (with π ≈ 3.14159)
So, the volume of the cylinder is increasing at a rate of approximately 0.5415 cm³/min when the length is 40 cm and the diameter is 3 cm.