Answer:
the rate of change in volume is dV/dt = 4π mm³/s = 12.56 mm³/s
Step-by-step explanation:
since the volume V of a cylinder is related with the height H and the radius R through:
V = πR²*H
then the change in time is given by the derivative with respect to time t
dV/dt = (∂V/∂R)*(dR/dt) + (∂V/∂H)*(dH/dt)
the change in volume with radius at constant height is
(∂V/∂R) = 2*πR*H
the change in volume with height at constant radius is
(∂V/∂H) = πR²
then
dV/dt = 2π*R*H *(dR/dt) + πR²*(dH/dt)
replacing values
dV/dt = 2π* 2 mm * 20 mm * (-0.1 mm/s) + π (2 mm) ²* 3 mm/s = 4π mm³/s
dV/dt = 4π mm³/s = 12.56 mm³/s
Rate of change is simply how much a quantity changes, over another quantity.
The rate of change of the volume of the tube is 12.57 cubic inches per second
The volume of a cylinder is:
[tex]V = \pi r^2h[/tex]
Differentiate with respect to time
[tex]\frac{dV}{dt} = 2\pi rh \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}[/tex]
From the question, we have:
[tex]r = 2 mm[/tex] --- the radius
[tex]h = 20mm[/tex] -- the height
[tex]\frac{dr}{dt} = -0.1mms^{-1}[/tex] --- the rate at which the radius decreases
[tex]\frac{dh}{dt} = 3mms^{-1}[/tex] --- the rate at which the height increases
Recall that:
[tex]\frac{dV}{dt} = 2\pi rh \frac{dr}{dt} + \pi r^2 \frac{dh}{dt}[/tex]
The equation becomes
[tex]\frac{dV}{dt} = 2 \times 3.142 \times 2 \times 20 \times -0.1 + 3.142 \times 2^2 \times 3[/tex]
Using a calculator:
[tex]\frac{dV}{dt} = 12.568[/tex]
Approximate
[tex]\frac{dV}{dt} = 12.57[/tex]
Hence, the rate of change of the volume of the tube is 12.57 cubic inches per second
Read more about rates of change at:
https://brainly.com/question/15278071
