Thread: Cone Calculus Problem

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

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

Originally Posted by sitho

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

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

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

Originally Posted by Punch

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

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.

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

Did you mean negative 2 or negative 3?

Originally Posted by Punch

Did you mean negative 2 or negative 3?

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

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.

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

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

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

I guess I will require a detailed working...

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 $\displaystyle 2cm^3/s$