Cone Calculus Problem

**Punch** · Mar 4th 2010, 01:26 AM

Is there no reply because none can see the image I posted?

**sitho** · Mar 4th 2010, 01:44 AM

After 3 minutes the volume is $390 - 3 \cdot 60~cm^3 = 210~cm^3$ .
The volume of a cone is $V = \frac{\pi}{3}r^2 h$ .
The 45 degree angle gives $r = h~\Rightarrow~V = \frac{\pi}{3}h^3~\Leftrightarrow~h = \left(\frac{3 V}{\pi}\right)^{1/3}$ .
$\frac{dh}{dt} = \frac{dh}{dV}\cdot\frac{dV}{dt} = \frac{1}{\left(9\pi V^2\right)^{1/3}}\frac{dV}{dt} \approx 0.028~cm/s$ .

**Punch** · Mar 4th 2010, 01:46 AM

Originally Posted by sitho

After 3 minutes the volume is $390 - 3 \cdot 60~cm^3 = 210~cm^3$ .
The volume of a cone is $V = \frac{\pi}{3}r^2 h$ .
The 45 degree angle gives $r = h~\Rightarrow~V = \frac{\pi}{3}h^3~\Leftrightarrow~h = \left(\frac{3 V}{\pi}\right)^{1/3}$ .
$\frac{dh}{dt} = \frac{dh}{dV}\cdot\frac{dV}{dt} = \frac{1}{\left(9\pi V^2\right)^{1/3}}\frac{dV}{dt} \approx 0.028~cm/s$ .

Correction: After 3mins, the volume is $390-(3\cdot60\cdot3)$

**Punch** · Mar 4th 2010, 01:48 AM

Also the answer is $-0.0680 cm^3/s$

**sitho** · Mar 4th 2010, 01:59 AM

Originally Posted by Punch

Correction: After 3mins, the volume is $390-(3\cdot60\cdot3)$

Ah, that's true. My mistake. But that means the cone is emptied after 3 mins, or more precisely after 2 mins and 10 secs.

**sitho** · Mar 4th 2010, 02:10 AM

I guess there was a typo in the question. It should be $\frac{dV}{dt} = -2~cm^3/s$ . Then after 3 mins the volume is $390 - 2 \cdot 180~cm^3 = 30~cm^3~\Rightarrow$
$\frac{dh}{dt} = -0.0680~cm/s$

**Punch** · Mar 4th 2010, 02:17 AM

Did you mean negative 2 or negative 3?

**sitho** · Mar 4th 2010, 02:22 AM

Originally Posted by Punch

Did you mean negative 2 or negative 3?

Negative 2. Hence the volume decreases $2~cm^3/s$ , because if it would decrease $3~cm^3/s$ as it says in the question it would give a negative volume after 3 mins. Replacing 3 with 2 also gives the correct answer.

**Punch** · Mar 4th 2010, 02:27 AM

However, I really doubt that there is any mistake with the question.... This is because my teacher said she have already done this question in the past years with the past batches and said this question is relatively common.

**sitho** · Mar 4th 2010, 02:43 AM

Originally Posted by Punch

However, I really doubt that there is any mistake with the question.... This is because my teacher said she have already done this question in the past years with the past batches and said this question is relatively common.

Yeah, that's a bit strange. Try showing her the expression

$\frac{dh}{dt} = \frac{1}{\left(9\pi\left(V_0 + t\cdot \frac{dV}{dt}\right)^2\right)^{1/3}}\cdot\frac{dV}{dt}$

where $V_0$ is the initial volume in $cm^3$ , $t$ is the time in seconds and $\frac{dV}{dt}$ is the change in volume w.r.t. time $[cm^3/s]$ , and ask her if she gets the same. Because if it's true, then you get the correct answer when inserting 3 mins and $2~cm^3/s$ .

**Punch** · Mar 4th 2010, 02:45 AM

I guess I will require a detailed working...

**Punch** · Mar 5th 2010, 01:40 AM

Although my teacher have not confirmed this, I conclude that the question has a typo mistake in it and the rate of volume flowing out should instead be $2cm^3/s$

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