A tree trunk is approximated by a circular cylinder of height 100 meters and diameter 4 meters. The tree is growing taller at a rate of 2 meters per year and the diameter is increasing at a rate of 5 cm per year. The density of the wood is 2000 Kg per cubic meter. How quickly is the mass of the tree increasing?

Step-by-step explanation:

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facundo3141592 facundo3141592 · Answer 2 · 2021-11-05T14:28:42+01:00

We want to find at which rate the mass of the tree increases given that we know how the volume increases.

The equation that says how fast is the mass increasing as a function of time is:

M'(t) = (3.14*2000kg/m^3)*[3*0.00125m^3*t^2 + 2*0.262m^3*t + 260m^3]

First, we know that for a cylinder of radius R and height H, the volume is:

V = 3.14*R^2*H

Here we know that:

The diameter is of 4 meters (so the radius is 2 meters) and it increases 5 cm per year (so the radius increases 2.5 cm per year).

knowing that:

100cm = 1m

we can write:

2.5cm = (2.5/100) m = 0.025m

Then the radius is given by.

R(t) = 2m + 0.025m*t

Where t is time in years.

For the height, we know that initially is of 100 meters and increases 2 meters per year, then the height equation is:

H(t) = 100m + 2m*t

Then the volume of the tree as a function of time is given by:

V(t) = 3.14*R(t)^2*H(t)

Replacing the actual functions we get:

V(t) = 3.14*( 2m + 0.025m*t)^2*(100m + 2m*t)

And the mass is given as the density times the volume, we know that:

density = 2000kg/m^3

Then the mass as a function of time is given as:

M(t) = (3.14*2000kg/m^3)*[( 2m + 0.025m*t)^2*(100m + 2m*t)]

M(t) = (3.14*2000kg/m^3)*[0.00125m^3*t^3 + 0.262m^3*t^2 + 260m^3*t + 400m^3]

To get how fast the mass increases, we derive the mass with respect to the time to get:

M'(t) = (3.14*2000kg/m^3)*[3*0.00125m^3*t^2 + 2*0.262m^3*t + 260m^3]

This equation gives how fast is the mass increasing as a function of time.

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