# Growing tree

• Mar 24th 2008, 11:28 AM
a.a
Growing tree
The trunk of a tree is approximately cylindrical in shape and has a diameter of 1 m when the height is 15m. If the radius is inreasing at 3mm/a and the height is increasing at 0.4 m/a, find the rate of increase of the volume of the trunk.

using related triangles i got r = h/30 and h=30r
and using the volume of a cylinder i got V = ( pi h^3)/900
then i said that dV/dh = (pi h^2)/ 300
also dV/dt= dV/dh times dh/dt and i got dV/dt = (pi h^2)/ 750

then just to check i used subed in h = 30r in the volume formula and did all the same steps and got a different answer... so i kno im doing something wrong.. :S

• Mar 24th 2008, 12:28 PM
Soroban
Hello, a.a!

Quote:

The trunk of a tree is approximately cylindrical in shape
and has a diameter of 1 m when the height is 15 m.
If the radius is increasing at 3mm/year (0.003 m/year)
and the height is increasing at 0.4 m/year,
find the rate of increase of the volume of the trunk. . . . . when?

The volume of a cylinder is: .$\displaystyle V \;=\;\pi r^2h$

At $\displaystyle t = 0\:\;\;r = 0.5,\:h = 15$

After $\displaystyle t$ years: .$\displaystyle \begin{array}{ccc}r &=&0.5 + 0.003t \\ h &=& 15 + 0.4t\end{array}$

Then: . $\displaystyle V \;=\;\pi(0.5+0.003t)^2(15 + 0.4t) \;=\;\pi(0.000036t^3 + 0.00255t^2 + 0.55t + 3.75)$

Therefore: . $\displaystyle \frac{dV}{dt} \;=\;\pi(0.000108t^2 + 0.0051t + 0.55) \;\text{ m}^3\text{/year}$
• Mar 25th 2008, 10:03 PM
a.a
Quote:

Originally Posted by Soroban
Hello, a.a!

The volume of a cylinder is: .$\displaystyle V \;=\;\pi r^2h$

At $\displaystyle t = 0\:\;\;r = 0.5,\:h = 15$

After $\displaystyle t$ years: .$\displaystyle \begin{array}{ccc}r &=&0.5 + 0.003t \\ h &=& 15 + 0.4t\end{array}$

Then: . $\displaystyle V \;=\;\pi(0.5+0.003t)^2(15 + 0.4t) \;=\;\pi(0.000036t^3 + 0.00255t^2 + 0.55t + 3.75)$

Therefore: . $\displaystyle \frac{dV}{dt} \;=\;\pi(0.000108t^2 + 0.0051t + 0.55) \;\text{ m}^3\text{/year}$

can we just differentiate V = pi r^2 w.r.t t and use product rule?