The math help thread

I'M IN DIS THREAD, NECRO-POSTING UR PROBLEMS!!

There's one problem that's been bothering me for a while. No picture this time xD

A 1'' steel plate is taken, and a 3/4'' hole is drilled out. Then a countersink drill bit (90 degrees at the point) replaces the regular drill bit and bores out even more steel. The area it would normally have drilled out is in the shape of a simple cone, and the base diameter is 1 1/4''. Both drills were used at the exact same spot. How much steel was bored out by the countersink drill bit?

My first try at it was integration and revolutions, but for some reason it wouldn't work out. I managed to find the answer with the Pappus-Guldinus theory. But I want to know what is the proper integral that defines the answer?

I'm afraid I don't follow. Would you please mind to draw a sketch of what we're actually going to calculate?

1" wide, 1" long?
3/4" deep, wide, or the radius of the hole?

Giving us those data would help us quite a lot.

Noah

I'm afraid I don't follow. Would you please mind to draw a sketch of what we're actually going to calculate?

1" wide, 1" long?
3/4" deep, wide, or the radius of the hole?

Giving us those data would help us quite a lot.

Noah

The hole drilled out looks like this. In carpentry at least, any mention of measurement concerning drill bits are always diameters. You're looking for how much material the countersink bit took out.

Code:

\       /
\     /
  \   /
  |   |
  |   |
  |   |
  |   |

(also in the original post I accidentally said 1 1/4" radius instead of diameter, sorry)

Call the line through the center of the hole the x axis (straight up and down in the above figure), and a line across the base of the hole (top of the figure) the y axis. Then the straight hole can be represented as the line y = 3/4 rotated about the x axis, and the countersink is y = 5/4 - x rotated about the x axis.

Then it should be just an integral of outer radius squared - inner radius squared from x = 0 to the intersection of the two.

It would be nice if you said what number you got for an answer. Also, I said those numbers were diameters if you didn't understand what was going on. No one describes drill bits in terms of radius.

In the ways I've interpreted your equation so far, I've not come up with the correct answer.

EDIT: Nvm got it! Oh those tricky integrals >_> Now I am a little bit wiser.

ᴨ * [[SIZE="5"]ʃ[/SIZE]((5/8 - x)^2 - (3/8)^2)dx ,0 , 1/4] = .0900 in.^3

I didn't crunch it.

What's the correct answer?

ᴨ * [[SIZE="5"]ʃ[/SIZE]((5/8 - x)^2 - (3/8)^2)dx ,0 , 1/4] = .0900 in.^3

I love using triple integrals for volume

So I'd do theta = 0 to 2pi, r = 3/8 to 5/8, x = 0 to 5/8 - r

ʃʃʃ r dx dr dtheta
ʃʃ (5/8 - r)r dr dtheta
ʃ [5/16r^2 - 1/3r^3]_{r=3/8}^{5/8} dtheta
ʃ 0.0143229167 dtheta
0.0899935396

Obviously your method has rounding issues, it's not actually 0.09 but 11/384 pi, which is a little lower.


edit: heh, stole your integral sign.

edit: heh, stole your integral sign.

It's actually the consonant ʃ in "washing", clever though. xD

It's actually the consonant ʃ in "washing", clever though. xD

Technically it is the small letter Esh xD
THIS is the real integral sign (I just found it): ∫
Two-part integral sign:
⌠
⌡
This is being pulled from the Character Map in case it wasn't obvious enough.



@ Stereo, the Pappus-Guldinus method also turned up just shy of .09, but who cares, it rounds to .09 anyways.

Yup, that's what I would've got, except I mixed up radius and diameter so I doubled both the two radiuses and the upper limit of the integral. And I neglected the pi in front.

And I should know drill bits are diameter too >_>

XYZ function I think, maybe linear programming but I doubt it.
ih8wordproblems.

Algebra 2 is so much easier when words aren't involved.

Got a little problem, but i don't know if it will make sense D:

e is the same as 10^x?

[NOPARSE][/NOPARSE]

XYZ function I think, maybe linear programming but I doubt it.
ih8wordproblems.

Algebra 2 is so much easier when words aren't involved.

Typical 3 variable 3 equation problem.

Code:

x + y + z = 50,000
2y = z
1.12x + 1.08y + 1.05z = 54,500

Solve.

Got a little problem, but i don't know if it will make sense D:

e is the same as 10^x?

Finding x I take it? I don't remember how to do this on paper x_X

Got a little problem, but i don't know if it will make sense D:

e is the same as 10^x?

log(e)=0.434294482=x

10^0.434294482 = e
About... lolrounding.

I have a physics assignment due in 6 hours and this 1 question is just freaking impossible, i mean it doesn't even give enough information to solve it!

A cyclotron accelerates protons from rest to a speed of 2.6×10^7 m/s.
How much work does it do on each proton?

I don't even know where to start for this...

So that number squared, times the mass of the proton (iunno what units it would be... kg? holycrapsmallnumbertime) divided by 2?

So that number squared, times the mass of the proton (iunno what units it would be... kg? holycrapsmallnumbertime) divided by 2?

Wow thx, holycrapyouaresosmart!!!

Wow thx, holycrapyouaresosmart!!!

Not really. :|
http://en.wikipedia.org/wiki/Work_%28physics%29

Work is force over a distance. Force is mass times acceleration. So to find work, you multiply the acceleration, the mass of a proton, and the distance traveled by each proton. Assume it takes 1 second to accelerate it (because I think it doesn't matter), and from your given velocity you get that your acceleration is 2.6*10^7 m/s/s (it takes 1 second to accelerate from 0 to that velocity). You also know, then, that v = 2.6*10^7*t, which you can integrate to get the distance traveled, which ends up being 2.6e7*(1/2 (1)^2) = 1.3e7 meters.

2.6*10^7 m/s/s * whatever g * 1.3*10^7 m = something J

At least, that would be my calc-bashing approach to this problem, with my limited knowledge of physics formulas, having not taken physics yet. E = 1/2 mv^2 as Hazzy said would probably be easier