Checking My Answer (water dripping from filter to cup)

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Spoon-

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Nov 26, 2007
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Here's the word problem I got:

Water in a paper conical filter drips into a cylindrical cup. The radius of the filer is 6cm and the height is 10cm. Suppose 15cm[sup:12uewqxe]3[/sup:12uewqxe] of water is poured into the filter. At time t, the height of the water filter is x cm and the height of the water in the cup is y cm.

a) Express the radius r of the filter as a function of x
b) Express y as a function of x

My answers:
a) 15 = (1/3)(2*pi*r)(x)

r = 45 / (2*pi*x)

b) 15 = (1/3)(2*pi*r)(x) 15 = (2*pi*r)(y)

(2*pi*r)(y) = (1/3)(2*pi*r)(x)
y = x/3

Did I do the question right or did I misread the question? I have a feeling I did it wrong... can I even equate them for part b?

Thanks.
 
Re: Checking My Answer

Your problem statement seems a little confusing. You have the 'height of the water filter' is x. That remains constant. It's 10 cm.

You can use similar triangles to get what you need, perhaps.

\(\displaystyle \frac{r}{h}=\frac{6}{10}\), where r is the radius of the water and h is the height of the water.

Now, you can solve for a variable and plug it into the cone volume formula. Hope this helps a little.
 
Yeah I was confuse by that too but I'm sure the x cm is referring to the height of the water in the cone.
 
Yeah I did do part b) wrong but I have one simple question.

When it asks for "Express y as a function of x" does my function HAVE to have only x as a variable? Or am I allowed to put in other variables?
 
Spoon- said:
Here's the word problem I got:

Water in a paper conical filter drips into a cylindrical cup. The radius of the filer is 6cm and the height is 10cm. Suppose 15cm[sup:ufupg7s9]3[/sup:ufupg7s9] of water is poured into the filter. At time t, the height of the water filter is x cm and the height of the water in the cup is y cm.

a) Express the radius r of the filter as a function of x
b) Express y as a function of x

My answers:
a) 15 = (1/3)(2*pi*r)(x) <--- That equation is incorrect

Volume of a cone = 1/3 * ? * r^2 * h

r = 45 / (2*pi*x)

b) 15 = (1/3)(2*pi*r)(x) 15 = (2*pi*r)(y)

(2*pi*r)(y) = (1/3)(2*pi*r)(x)
y = x/3

Did I do the question right or did I misread the question? I have a feeling I did it wrong... can I even equate them for part b?

Thanks.
Click to expand...
 
Spoon- said:
Yeah I did do part b) wrong but I have one simple question.

When it asks for "Express y as a function of x" does my function HAVE to have only x as a variable? Or am I allowed to put in other variables?
Click to expand...

Generally it means that - 'y' is on the left-hand-side - and there is no 'y' on the right-hand-side. So you can have whole bunch of other stuff on the right-hand-side - but no 'y'.
 
One of the way to do this problem (part b) should be following:

15 = Volume of water in filter + volume of water in cup

15 = 1/3 * ? * \(\displaystyle r_f^{2}\) * x + ? * \(\displaystyle r_c^{2}\) * y

where

\(\displaystyle r_f^{2}\) = radius of water cone in the filter = 6/10 * x = 0.6x

\(\displaystyle r_c^{2}\) = radius of the cup (unknown) = r

15 = 1/3 * ? * \(\displaystyle r_f^{2}\) * x + ? * \(\displaystyle r_c^{2}\) * y

? * \(\displaystyle r^{2}\) * y = 15 - 1/3 * ? * \(\displaystyle {0.6x}^{2}\) * x

? * \(\displaystyle r^{2}\) * y = 15 - 0.12 * ? * \(\displaystyle x^{3}\)

\(\displaystyle y = \frac{15 - 0.12 \pi x^{3}}{ \pi r^{2}}\)
 
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