Rates of Change (mind blank) (1 Viewer)

Aquawhite · Jun 23, 2009

I think I may have seen numerous question like this before but I need help with this one:

Sand is falling from a funnel forms a conical pile such that the height of the pile is one and a half times the radius.

a) Show that the volume of the pile is given by V = (pi.r^2)/2 [I can do this, you can prove if you want to ^_^]

b) If the sand is falling at the rate of pi/10 m^3 per minute, find the rate at which the height is increasing when the pile of sand is 3m high.

Thanks in advance.

Edit: If it's also possible, this too:

A water trough is 200cm long and has a cross-section of a right-angled isosceles triangle. Show that when the depth of the water is x cm, the volume of water in the tank is 200x^2cm^3. Water is poured at a constant rate of 5 litres per minute. Find the rate at which the water level is rising when the depth is 30cm.

Pwnage101 · Jun 23, 2009

im pretty sure 'related rates' is 3Unit work, not 2 unit...

Aquawhite · Jun 23, 2009

Pwnage101 said:
im pretty sure 'related rates' is 3Unit work, not 2 unit...

I was informed that parts of the Rates of Change are part of the 2unit course. Such as this question... but you may be right. I'm sure a 2unit student could do it ^_^

shady145 · Jun 23, 2009

Aquawhite said:
I was informed that parts of the Rates of Change are part of the 2unit course. Such as this question... but you may be right. I'm sure a 2unit student could do it ^_^

ooo i notice ur from kempsey... which school do u go to, melville high or kempsey high, or a private one???
i did a question very similar to this in the 3u thread a few days ago, u can check it out its the same method id say

Aquawhite · Jun 23, 2009

shady145 said:
ooo i notice ur from kempsey... which school do u go to, melville high or kempsey high, or a private one???
i did a question very similar to this in the 3u thread a few days ago, u can check it out its the same method id say

How do you know which school are in Kempsey?

(answer this one

)

I attend St. Paul's College (it's a private catholic college, hence the SOR

)

I shall check it out.

EDIT: Yea, it was very similar... thanks for that. I was only one small bit off.

just need to shove some things around...

Drongoski · Jun 23, 2009

Here's my attempt. (looks more like a 3U problem)

$\80dpi Vol\ of\ cone\ \ v\ =\ \frac{1}{3}\times base\ area\ \times height\ =\ \frac{1}{3}\times \pi r^{2}\times \frac{3r}{2}\ =\ \frac{\pi r^{3}}{2}\\ \\ Now\ \ h\ =\ \frac{3r}{2}\ \ \Rightarrow \ \ v\ =\ \frac{4\pi h^{3}}{27}\ \ \Rightarrow \ \ \frac{\mathrm{d} v}{\mathrm{d} t}\ =\ \frac{\pi}{10} \ \ =\ \frac{\mathrm{d} v}{\mathrm{d} h}\times \frac{\mathrm{d} h}{\mathrm{d} t}\ = \ \frac{4\pi h^{2}}{9 }\times \frac{\mathrm{d} h}{\mathrm{d} t}\\ \\ \Rightarrow \ \ \frac{\mathrm{d} h}{\mathrm{d} t} \ =\ \frac{9}{40 h^{2}}\ \ =\ \ \frac{9}{40 \times 3^{2}}\ \ =\ \ \frac{1}{40 }\ \ m/min$

Edit

In my original posting I used dv/dt = 10 instead of pi/10; hopefully amended solution is now OK.

Aquawhite · Jun 24, 2009

Drongoski said:
Here's my attempt. (looks more like a 3U problem)

$\80dpi Vol\ of\ cone\ \ v\ =\ \frac{1}{3}\times base\ area\ \times height\ =\ \frac{1}{3}\times \pi r^{2}\times \frac{3r}{2}\ =\ \frac{\pi r^{3}}{2}\\ \\ Now\ \ h\ =\ \frac{3r}{2}\ \ \Rightarrow \ \ v\ =\ \frac{4\pi h^{3}}{27}\ \ \Rightarrow \ \ \frac{\mathrm{d} v}{\mathrm{d} t}\ =\ \frac{\pi}{10} \ \ =\ \frac{\mathrm{d} v}{\mathrm{d} h}\times \frac{\mathrm{d} h}{\mathrm{d} t}\ = \ \frac{4\pi h^{2}}{9 }\times \frac{\mathrm{d} h}{\mathrm{d} t}\\ \\ \Rightarrow \ \ \frac{\mathrm{d} h}{\mathrm{d} t} \ =\ \frac{9}{40 h^{2}}\ \ =\ \ \frac{9}{40 \times 3^{2}}\ \ =\ \ \frac{1}{40 }\ \ m/min$

Edit

In my original posting I used dv/dt = 10 instead of pi/10; hopefully amended solution is now OK.

Yea, thanks. It was a 3U question... apologies for posting it in the wrong section.

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