material needed to make ball, air held, % of box not filled,

[Jackie12]

I'm taking an online Geometry course to get ahead for school, and there is a gnarly extra credit question on one of the projects that would totally knock my grade up a notch. Sorry about all the slang. I've got two-thirds of it I'm pretty sure, here it is, maybe you guys could help me out with the rest.

A basketball has a radius of approximately 4.75 inches when filled. How much material is needed to make one? How much air will it hold? If the basketball is stored in a cubic box whose edges are 9.5 inches long. what percent of the box is not filled by the basketball? % (Round your answers to the nearest hundredth.)

I've got the surface area(material needed), it's 283.53 in. squared. I've also got the volume, it's 448.92 in. cubed. I've got as far as to subtract the volume of the basketball from the volume of the box, but I'm stuck on the whole percentage left. I'm pretty sure the box's volume is 857.375. Please help!

There's another problem too, I don't have a clue where to start on it. Give it a shot, if you feel like it. Here it is,

EC: A roll of paper used in printing has a diameter of 3 feet, a hollow core of 4 inches, and a width of 5 feet. If the thickness of the paper is 0.0015 inch, what is the length of the paper on the roll? ______ ft


Thanks in advance.

 

[galactus]

Re: Kind of lame question,

A roll of paper used in printing has a diameter of 3 feet, a hollow core of 4 inches, and a width of 5 feet. If the thickness of the paper is 0.0015 inch, what is the length of the paper on the roll? ______ ft

This problem can be tackled in many ways, but here is how I do it. This is a model of the Spiral of Archimedes.

By width, I assume that means the length of the roll. We do not need that to find the length of the paper.

The core has radius 1/6 feet and outer radius of 3/2.

To create a model for this, assume that as the paper is wound around the core its distance r from the center increases linearly at the rate of .0015 inches per revolution.

\(\displaystyle (.0015)\frac{\theta}{2{\pi}}=\frac{3\theta}{4000\pi}, \;\ \frac{2000\pi}{9}<{\theta}<2000{\pi}, \;\ why?\). Where theta is in radians.

To find the coordinates of the point (x,y) corresponding to a given radius is \(\displaystyle x=rcos{\theta}\)

and \(\displaystyle y=rsin{\theta}\).


Subbing for r we get the parametric equations:

\(\displaystyle x=\left(\frac{3{\theta}}{4000\pi}\right)cos{\theta}\)

and

\(\displaystyle y=\left(\frac{3{\theta}}{4000\pi}\right)sin{\theta}\)

We can use the arc length formula to find the total length of the paper.

\(\displaystyle \int_{\frac{2000\pi}{9}}^{2000\pi}\sqrt{\left(\frac{dx}{d{\theta}}\right)^{2}+\left(\frac{dy}{d{\theta}}\right)^{2}}d{\theta}\)

\(\displaystyle =\boxed{4654.2 \;\ feet \;\ or \;\ 55851 \;\ inches}\)

 

[Deleted member 4993]

Re: Kind of lame question,

EC: A roll of paper used in printing has a diameter of 3 feet, a hollow core of 4 inches, and a width of 5 feet. If the thickness of the paper is 0.0015 inch, what is the length of the paper on the roll? ______ ft


Thanks in advance.

\(\displaystyle Volume \, of \, paper\, = \, \pi\cdot ({r_o}^2 - {r_i}^2)\cdot w\) <<<< edit - corrected

\(\displaystyle Length \, of \, paper\, =\, \frac{Volume \, of \, paper}{Thickness \, of \, paper\, \cdot w}\)

Galactus's method is more rigorous than shown above - however for thin paper this method is good enough for "government work".

 

[Jackie12]

Re: Kind of lame question,

A roll of paper used in printing has a diameter of 3 feet, a hollow core of 4 inches, and a width of 5 feet. If the thickness of the paper is 0.0015 inch, what is the length of the paper on the roll? ______ ft

\(\displaystyle \int_{\frac{2000\pi}{9}}^{2000\pi}\sqrt{\left(\frac{dx}{d{\theta}}\right)^{2}+\left(\frac{dy}{d{\theta}}\right)^{2}}d{\theta}\)

There, solve that you've earned your extra credit.

Thanks lol.

 

[Jackie12]

Re: Kind of lame question,

I'm so close to solving the first problem...can anyone help with that?

 

[stapel]

I'm so close to solving the first problem...can anyone help with that?

What have you got so far? Where are you stuck?

Please be complete. Thank you!

Eliz.

 

[galactus]

Re: Kind of lame question,

You have the volume of the box and the volume of the sphere. Subtract the two volumes and divide that result by the volume of the box.

 

[Jackie12]

Re:

I'm so close to solving the first problem...can anyone help with that?

What have you got so far? Where are you stuck?

Please be complete. Thank you!

Eliz.

All of my progress I put in the first post. I'm stuck on the last bit.

 

[Jackie12]

Re: Kind of lame question,

You have the volume of the box and the volume of the sphere. Subtract the two volumes and divide that result by the volume of the box.

So I subtract the basketball volume from the volume of the box, and then divide that number by the volume of the box? That doesn't really sound right...

 

[galactus]

Re: Kind of lame question,

Why does that not seem right?.

The volume of the ball is \(\displaystyle \frac{4}{3}{\pi}(4.75)^{3}=448.92\)

The volume of the box is \(\displaystyle (9.5)^{3}=857.375\)

It is rather intuitive that the volume of the remaining box not taken up by the ball is 857.375-448.92=408.45

What % is this of the entire volume of the box?. \(\displaystyle \frac{408.45}{857.375}=.476\)

About 47.6%

 

[Jackie12]

Re: Kind of lame question,

Why does that not seem right?.

The volume of the ball is \(\displaystyle \frac{4}{3}{\pi}(4.75)^{3}=448.92\)

The volume of the box is \(\displaystyle (9.5)^{3}=857.375\)

It is rather intuitive that the volume of the remaining box not taken up by the ball is 857.375-448.92=408.45

What % is this of the entire volume of the box?. \(\displaystyle \frac{408.45}{857.375}=.476\)

About 47.6%

I guess that is right, I just get confused a lot when it comes to dividing with fractions you know? I don't know whether or not the numerator or the denominator are the dividend or the divisor...

But thanks for the help, I really appreciate it.

 

[Jackie12]

Re: Kind of lame question,

EC: A roll of paper used in printing has a diameter of 3 feet, a hollow core of 4 inches, and a width of 5 feet. If the thickness of the paper is 0.0015 inch, what is the length of the paper on the roll? ______ ft


Thanks in advance.

\(\displaystyle Volume \, of \, paper\, = \, \pi\cdot ({r_o}^2 - {r_i}^2)\cdot w\) <<<< edit - corrected

\(\displaystyle Length \, of \, paper\, =\, \frac{Volume \, of \, paper}{Thickness \, of \, paper\, \cdot w}\)

Galactus's method is more rigorous than shown above - however for thin paper this method is good enough for "government work".

I got 55841.27778 as an answer for the total length of the roll by using your method right there. Does that sound about right?

 

[galactus]

Re: Kind of lame question,

I don't think so. I get 4654.21

 

[Deleted member 4993]

Re: Kind of lame question,

I don't think so. I get 4654.21

Galactus,

Are you getting that by integration?

 

[galactus]

Re: Kind of lame question,

EDIT: My answer is in feet and yours in in inches.

I get 4654.1 feet, which is 55851 inches.

 

[Jackie12]

Re: Kind of lame question,

Yes, I am. I could be wrong. What do you get?.

I finally submitted it and the answer is 55,850.

 

[LK101]

Re: Kind of lame question,

Yes, I am. I could be wrong. What do you get?.

I finally submitted it and the answer is 55,850.

I have the exact same questions too. One is 20 extra credit and the other is 30.

which question is 55,850 the answer too?

and whats the answer to the other one?

thanks

 

[Deleted member 4993]

Re: Kind of lame question,

I finally submitted it and the answer is 55,850.

I have the exact same questions too. One is 20 extra credit and the other is 30.

which question is 55,850 the answer too?

and whats the answer to the other one?

You want to get extra credit without doing one lick of work! What are you planning to learn - why are you taking such courses where you have no intention of learning?

Your abhorrence to learning is quite obvious from your questions - you even refuse to read the answers discussed - you want the answers spoon-fed.