# Population Max/Min & Cone Question

• May 25th 2007, 01:07 PM
SportfreundeKeaneKent
Population Max/Min & Cone Question
Most population questions ask about when the population will reach etc. This one's different though. The second one is with cones but I'm not sure if the volume is increasing per unit of time or height.

http://img525.imageshack.us/img525/2494/96lq4.png
• May 25th 2007, 01:32 PM
ecMathGeek
Quote:

Originally Posted by janvdl
But wait, its RATE, so it must be time. :)

I think the units should be $\displaystyle m^3/h$ because given the situation, the volume would have to be increasing on the second problem.
• May 25th 2007, 01:41 PM
Jhevon
Quote:

Originally Posted by SportfreundeKeaneKent
Most population questions ask about when the population will reach etc. This one's different though. The second one is with cones but I'm not sure if the volume is increasing per unit of time or height.

http://img525.imageshack.us/img525/2494/96lq4.png

Here's the second.

For a cone, the volume, V, is given by:

$\displaystyle V = \frac {1}{3} \pi r^2 h$
where $\displaystyle r$ is the radius and $\displaystyle h$ is the height

we want $\displaystyle \frac {dh}{dt}$ when $\displaystyle r = 9$

we are told that $\displaystyle h = \frac {2}{3} r \Rightarrow r = \frac {3}{2}h$

So our formula becomes:

$\displaystyle V = \frac {1}{3} \pi \left( \frac {3}{2} h \right)^2 h$

$\displaystyle \Rightarrow V = \frac {3}{4} \pi h^3$

Differentiating implicitly, we obtain:

$\displaystyle \frac {dV}{dt} = \frac {9}{4} \pi h^2 \frac {dh}{dt}$

Now, when $\displaystyle r = 9$, $\displaystyle h = \frac {2}{3} (9) = 6$

$\displaystyle \Rightarrow 243 = \frac {9}{4} \pi (6)^2 \frac {dh}{dt}$

$\displaystyle \Rightarrow 243 = 81 \pi \frac {dh}{dt}$

$\displaystyle \Rightarrow \frac {dh}{dt} = \frac {243}{81 \pi } = \frac {3}{ \pi } \mbox { m/h }$
• May 25th 2007, 01:44 PM
ecMathGeek
Quote:

Originally Posted by SportfreundeKeaneKent
Most population questions ask about when the population will reach etc. This one's different though. The second one is with cones but I'm not sure if the volume is increasing per unit of time or height.

http://img525.imageshack.us/img525/2494/96lq4.png

I'll give you some hints, but I'm sure janvdl will give you the exact process.

For problem 1:

$\displaystyle L(p)=3\sqrt{p^3+10p^2+599}$
$\displaystyle p(t)=32-\frac{6}{t+1}$

We want: $\displaystyle L'(t=5)$, but as you see, $\displaystyle L$ is not a function of t.

Here's how we take care of that.
Find: $\displaystyle \frac{dL}{dp}\frac{dp}{dt}$ at $\displaystyle t=5$. I hope that makes sense to you.

For problem 2:

$\displaystyle \frac{dV}{dt}=243m^3/h$
$\displaystyle h=\frac{2}{3}r$

We want: $\displaystyle \frac{dh}{dt}$ when $\displaystyle r=9$.

Use the fact that:
$\displaystyle V=\frac{1}{3}\pi r^2h$
Thus:
$\displaystyle \frac{dV}{dt}=\frac{2}{3}\pi rh\frac{dr}{dt} + \frac{1}{3}\pi r^2\frac{dh}{dt}$, where $\displaystyle \frac{dr}{dt}=\frac{3}{2}\frac{dh}{dt}$ to solve this.

Good luck! :D
• May 25th 2007, 01:45 PM
ecMathGeek
Yours looks prettier than mine. ;) (I should have substituted for r as well.)
• May 25th 2007, 02:09 PM
Jhevon
Quote:

Originally Posted by SportfreundeKeaneKent
Most population questions ask about when the population will reach etc. This one's different though. The second one is with cones but I'm not sure if the volume is increasing per unit of time or height.

http://img525.imageshack.us/img525/2494/96lq4.png

is L the cuberoot of p^3 + 10p^2 + 599 or is it 3 times the squareroot of that?
• May 25th 2007, 02:28 PM
Jhevon
Quote:

Originally Posted by SportfreundeKeaneKent
Most population questions ask about when the population will reach etc. This one's different though. The second one is with cones but I'm not sure if the volume is increasing per unit of time or height.

http://img525.imageshack.us/img525/2494/96lq4.png

Let's use the hint of the great ecMathGeek

Someone please check my computation, I caught myself making a lot of mistakes while doing this, I don't know if i caught all my mistakes.

We want $\displaystyle \frac {dL}{dt}$, which is the rate of change of $\displaystyle L$ with respect to time $\displaystyle t$. We will use the chain rule to obtain this. we can find $\displaystyle \frac {dL}{dt}$ by the following operation:

$\displaystyle \frac {dL}{dt} = \frac {dL}{dp} \cdot \frac {dp}{dt}$

since derivative notations can function as fractions, the $\displaystyle dp's$ would cancel, leaving $\displaystyle \frac {dL}{dt}$, which is what we want to find.

So let's find $\displaystyle \frac {dp}{dt}$

$\displaystyle p = 32 - \frac {6}{t + 1} = 32 - 6(t + 1)^{-1}$

$\displaystyle \Rightarrow \frac {dp}{dt} = 6(t + 1)^{-2}$

when $\displaystyle t = 5$, $\displaystyle \frac {dp}{dt} = 6(6)^{-2} = \frac {1}{6}$

Now let's find $\displaystyle \frac {dL}{dp}$:

$\displaystyle L = 3 \sqrt {p^3 + 10p^2 + 599}$

$\displaystyle \Rightarrow \frac {dL}{dp} = \frac {3}{2} (p^3 + 10p^2 + 599)^{- \frac {1}{2}} \cdot (3p^2 + 20p)$ ......by the Chain Rule

Now, when $\displaystyle t = 5$, $\displaystyle p = 32 - \frac {6}{5 + 1} = 31$

So when $\displaystyle t = 5$:

$\displaystyle \frac {dL}{dp} = \frac {3}{2} ((31)^3 + 10(31)^2 + 599)^{- \frac {1}{2}} \cdot (3(31)^2 + 20(31))$

$\displaystyle \Rightarrow \frac {dL}{dp} = \frac {10509}{400}$

So now, finally, we find $\displaystyle \frac {dL}{dt}$:

When $\displaystyle t = 5$:

$\displaystyle \frac {dL}{dt} = \frac {dL}{dp} \cdot \frac {dp}{dt} = \frac {1}{6} \cdot \frac {10509}{400} = \frac {3503}{800} \mbox { ppm/year }$ ...weird number:D