Help? Functions

**Slipery** · February 13th 2009, 10:35 PM

Hey guys.. on my last question for the 2nd last lesson for the course.. meaning I only have 1 more lesson after this. I would like to mail this in tomorrow morning, so some help would be nice..

Here it is:

A gas tank on a dock has a small puncture and is leaking gas at the rate of 1 cm^3/min into the lake. It forms a circular slick that is 1mm thick on the surface of the water.

a) Find the amount of gas leaked as a function of time

I believe this will be f(x) = 1ml(x) with x being minutes (Should I change this to seconds?)

b) State the radius of the slick as a function of it's volume

Not sure what to do here? Is this a sphere with a radius of 1mm?

c) State the radius of the slick as a function of time

Again.. Since this slick is 1mm thick, does this make it a sphere?

d) What is the radius of the slick after 40 minutes?

**Reckoner** · February 13th 2009, 10:50 PM

Originally Posted by Slipery

a) Find the amount of gas leaked as a function of time

I believe this will be f(x) = 1ml(x) with x being minutes (Should I change this to seconds?)

Good. Minutes should be fine if you were not told otherwise.

b) State the radius of the slick as a function of it's volume

Not sure what to do here? Is this a sphere with a radius of 1mm?

The slick is a right circular cylinder with a height of 1 mm, not a sphere. The volume of a cylinder of height $h$ and radius $r$ is

$V = \pi r^2h.$

Solve for $r$ in terms of $V$ .

c) State the radius of the slick as a function of time

Again.. Since this slick is 1mm thick, does this make it a sphere?

No. See above.

This function should be the composition of the function from part (b) with that from part (a).

d) What is the radius of the slick after 40 minutes?

Once you find part (c), this is just substitution.

**Slipery** · February 13th 2009, 10:57 PM

Thanks! How would I switch V=pi r^2(h) to a function? I am trying to switched all values to the opposite side and its not working so well.. State the radius as a function of its volume.. I am having trouble picturing that word statement in a formula :/

I have r= sqrt of v/h/pi

Just doesn't look right :/

**Reckoner** · February 13th 2009, 11:02 PM

Originally Posted by Slipery

Thanks! How would I switch V=pi r^2(h) to a function? I am trying to switched all values to the opposite side and its not working so well.. State the radius as a function of its volume.. I am having trouble picturing that word statement in a formula :/

$V = \pi r^2h = \pi r^2\cdot1 = \pi r^2$

$\Rightarrow r^2 = \frac V\pi\Rightarrow r=\pm\sqrt{\frac V\pi}$

We can ignore the negative root since $r > 0,$ so we may write $r = \sqrt{\frac V\pi}.$ $r$ is now written as a function of $V$ (if you like, you can call it $r(V)$ , or you can give it another name like $g(x)$ ).

Edit:

Originally Posted by Slipery

I have r= sqrt of v/h/pi

Just doesn't look right :/

This is correct:

$r = \sqrt{\frac{\left(\frac Vh\right)}\pi} = \sqrt{\frac V{h\pi}}.$ Now substitute for $h$ .

**Slipery** · February 13th 2009, 11:07 PM

Ahh =) Thank you so much! Heh, I promised my girlfriend that I would have the course done before she comes for valentines, so hopefully I can get some sleep in as well!

Appreciate it

So... I came up with a radius of 3.57 mm, but I am unsure if this was right..

My brother, home from university helped me, and he got a radius of 1.2732 metres.

He did this:

Instead of changing my original 'volume' function to 1ml (min) he kept it as 1 cm^3/min..

So I would get v= sqrt of (40cm^3) / (0.314cm)
=127.38cm

Is this correct?

**Reckoner** · February 13th 2009, 11:56 PM

Originally Posted by Slipery

So... I came up with a radius of 3.57 mm, but I am unsure if this was right..

My brother, home from university helped me, and he got a radius of 1.2732 metres.

He did this:

Instead of changing my original 'volume' function to 1ml (min) he kept it as 1 cm^3/min..

So I would get v= sqrt of (40cm^3) / (0.314cm)
=127.38cm

Is this correct?

Sorry, I neglected to explain the units in my post.

In the functions I gave, $V$ was in ml, which is equivalent to cm^3, and $h$ was in mm. I probably should have converted $h$ .

You should have

$r\left(f(40\text{ min.})\right) = r\left(40\text{ cm}^3\right)$

$= \sqrt{\frac{40\text{ cm}^3}{(0.1\text{ cm})\pi}} = \sqrt{\frac{400}\pi\;\text{cm}^2} = \sqrt{\frac{400}\pi}\text{ cm}\approx127.32\text{ cm.}$

Sorry for the confusion.

**Slipery** · February 14th 2009, 12:15 AM

Thank you very much =)

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