[Guest]

A reel is 2 inches in diameter. A thin tape (1/50 of an inch in thickness) is wrapped round the reel until the end of the tape is reached. The diameter of the reel and tape combined is now 4 inches. How long is the tape?

[Guest]

[spoiler='Or we could take the physicist's approach

'], and just say, "how much area consists of tape, how long a rectangle would the tape be if the thickness were 1/50?"

The area of the spool is pi.

The area of the spool with tape on it is 4*pi.

The area of the tape alone is 4*pi - pi = 3*pi

If that area were a long skinny rectangle of thickness 1/50, its length would be 3*pi / (1/50) = 150*pi ~= 471.xxx inches

Bravo!!! ....got'ta love those physicists! KISS Keep It Simple...........

[Guest]

But isn't the length of the tape influenced by the slow change in diameter? On the first wrap the diameter is 2, with a radius of 1 and the circumferance 3.141592654, the length of the tape at this point. The second wrap the diameter is now 2.02, or 1.01 radius and the tape would be 3.204738666 inches. Add the two together and the tape is 6.34633132 inches long at this point. Keep going until you get to the diameter of 4 or radius of 2 and you get 740.8975035 inches.

Consider the first wrap. If it wrapped all the way around and touched the beginning of the tape, it would be a relatively perfect circle. You must raise the tape up 1/50th of an inch to reach the top of the surface of the tape. Assuming you climbed at a 45 degree angle, you would need to add (sqrt(2)-1)/50 inches of tape to reach the same radius one tape-width higher. A low estimate of 50 times that amount would add more than 0.41 inches by the 50th wrap.

Edit: Did I really spell higher, "hier"?

Edited May 15, 2009 by Phatfingers

[Guest]

Using a calculus approach

If you use the arc length formula of calculus, Integral[sqrt[1 + f(x)^2]dx] (integrate from x=-1-(i/50) to x=1+(i/50)) where f(x) = (1 + i/50)Cos[x]) and i increases from 0-50 in increments of one and summing all of the integrals, the answer I got was 222.809. this gives the top half of the graph so you need to multiply by 2, = 445.6222. As phatfingers pointed out, this still doesnt account for the steps up, this equation would assume that after each time around the spool you would cut the tape and wrapped it once more.

BTW, for whoever doesnt know, mathematica is a program that runs math equations. matlab could do this too and I know for sure you can write a simple program to do all of the iterations for you, its just a matter of syntax.

[Guest]

I approached this a little differently; in case anyone is interested, I think this is a simple and intuitive way to think of it:

As others have mentioned, the tape has to be wrapped around the spool 50 times (each wrap around increases the diameter by 2/50, so after 50 wraps the diameter is 2 inches + 50*(2/50) = 4 inches, as required).

Now add the lengths of the 1st and 50th wraps. Clearly, this must be the same distance as the combined length of the 2nd and 49th wraps (because we are simultaneously adding and subtracting 2*pi*1/50 ). This is also the same length as the 3rd plus 48th, 4th plus 47th, etc. Continue this pairing until you reach the 25th and 26th wraps. Now all the wraps are accounted for: 25 pairs, each of which has a combined length the same as that of the 1st plus the 50th: (2 inches)*pi + (4 inches)*pi = 6*pi inches. Thus, the total length of the tape is 25*6*pi = 150*pi ~= 471.2 inches.

Of course, as already mentioned, this is an average based on an imprecise definition of length. Other variations can be found using summations:

The total length is given by the summation of pi*(2+2*n/50) from n=0 to n=49, which is 149*pi ~= 468.1 inches.

The total length is given by the summation of pi*(2+2*n/50) from n=1 to n=50, which is 151*pi ~= 474.4 inches.

The total length is given by the summation of pi*(2+1/50+2*n/50) from n=0 to n=49, which is 150*pi ~= 471.2 inches. Note that this is the same answer obtained through the intuitive method, and is also simply the average of the above two lengths calculated from the inside and the outside.

[Guest]

A reel is 2 inches in diameter. A thin tape (1/50 of an inch in thickness) is wrapped round the reel until the end of the tape is reached. The diameter of the reel and tape combined is now 4 inches. How long is the tape?

after wrapping tape radius will become=2 inches

radius increase=(2-1)=1 inch

1/50 inch increase in 1 round.

in first round tape used=2*3.14*1 inches

second round tape used= 2*3.14*(1+1/50)

third round tape used=2*3.14*(1+2/50)

n'th round tape used=2*3.14*(1+(n-1)/50)

last round tape used=2*3.14*(2-(1/50))

suppose n round we wrap the tape.

2*3.14*(2-(1/50))=2*3.14*(1+(n-1)/50)

n=50

so total length of tape is=2*3.14*1+2*3.14*(1+1/50)+2*3.14*(1+2/50)+......+2*3.14(1+49/50)

=2*3.14(1+1/50+2/50+...49/50)

=467.86inches

i think i made the solution

[Guest]

I need to correct my math, I was mixing polar and rectangular coordinates.

The f(x) I listed is actually supposed to be sqrt((1+(i/50))^2+x^2), not 1+(i/50)cos(x). With this correction the answer I get is 609.587, seems high compared to what everyone else is getting with geometry.

[Guest]

A reel is 2 inches in diameter. A thin tape (1/50 of an inch in thickness) is wrapped round the reel until the end of the tape is reached. The diameter of the reel and tape combined is now 4 inches. How long is the tape?

148 Pi = 464.72 inches

Explaination :-

1st round circumference:- Pi * 2

2nd round circumference:- Pi (2+2/50)

.

.

50th round circumference:- Pi(2+98/50)

Adding all we get 148*Pi=464.72

Edited May 15, 2009 by suryakiran

[Guest]

ans - 493.42

sol-

pi(2) + pi(2+2/50) + pi(2+4/50) + pi(2+6/50) ...........pi(2+100/50) -------------- add + 2/50 as increase in dia after every circuit

=3.14*153

[Guest]

148 Pi = 464.72 inches

Explaination :-

1st round circumference:- Pi * 2

2nd round circumference:- Pi (2+2/50)

.

.

50th round circumference:- Pi(2+98/50)

Adding all we get 148*Pi=464.72

surya, our approach similar......but for the last circumference dia = 4, so need to add another round - pi(2+100/50) -

[Guest]

The tape is 303 inch long.

This can be solved by airthmetic progression

[Guest]

This is very easy , the problem can be easily solved by airthmetic progression.

The series will be as follows-

2.00+2.02+2.04+2.06+2.08+. . . . . . . . . . . . . . +4.00

so the sum of the series will be 303.

And hence the length of the tape will be 303 inches

[Guest]

This can very easy, the problem can be easily solved by airthmetic progression.

The series will be as follows-

2.00+2.02+2.04+2.06+2.08+. . . . . . . . . . . . . . +4.00

so the sum of the series will be 303.

And hence the length of the tape will be 303 inches

[Guest]

...

wanted to delete because it was too close to another post

Edited May 15, 2009 by FunkyM0nk

[Guest]

A reel is 2 inches in diameter. A thin tape (1/50 of an inch in thickness) is wrapped round the reel until the end of the tape is reached. The diameter of the reel and tape combined is now 4 inches. How long is the tape?

import java.math.*;

public class Puzzle {
public static void main(String[] args) {
double filmWidth = 2/50F;
double reelDiameter = 2F;
double currentFilmDiameter = reelDiameter + filmWidth;
double filmLength = 0F;
for (int i = 1; i <= 50; ++i) {
filmLength += Math.PI * currentFilmDiameter ; // circumference
currentFilmDiameter += filmWidth;
}
System.out.printf("Film length %.2f inches", filmLength);
}
}

Output: Film length 474.38 inches

[Guest]

ans - 493.42

sol-

pi(2) + pi(2+2/50) + pi(2+4/50) + pi(2+6/50) ...........pi(2+100/50) -------------- add + 2/50 as increase in dia after every circuit

=3.14*153

don we've to take the increase in circumference (2*pi*r ) in terms of radius.. so 1/50 ll be added each time i believe

[Guest]

don we've to take the increase in circumference (2*pi*r ) in terms of radius.. so 1/50 ll be added each time i believe

yes Dev, even if v take increase in radius, "2r" (or simply D) wud make it 2*1/50. it will be much clear if make a diagram for d puzzle

[Guest]

Sorry guys cal mistake

153 * 3.14 = 480.42

[Guest]

But isn't the length of the tape influenced by the slow change in diameter? On the first wrap the diameter is 2, with a radius of 1 and the circumferance 3.141592654, the length of the tape at this point. The second wrap the diameter is now 2.02, or 1.01 radius and the tape would be 3.204738666 inches. Add the two together and the tape is 6.34633132 inches long at this point. Keep going until you get to the diameter of 4 or radius of 2 and you get 740.8975035 inches.

Using the method above to get the correct answer:

471 inches

2*pi*(1.01+1.03+1.05+....+1.99), since you have to take the radius from the center of the tape.

[Guest]

Using the method above to get the correct answer:

471 inches

2*pi*(1.01+1.03+1.05+....+1.99), since you have to take the radius from the center of the tape.

A person can quickly add the series of 50 numbers (1.01+1.03+...+1.97+1.99) without a calculator by pairing the numbers from the outside inward because each pair adds up to 3. 1.01+1.99=3, 1.03+1.97=3, ..., 1.49+1.51=3, so the sum of that series can be calculated as 3 * 25, or 75.

[Guest]

My friends,

The perimeter of a circle is ∏*d not ∏*(r²).

The other thing you need to consider is every time you roll the tape the diameter increases in 2/50 (not 1/50) because increases in both sides... 2/50 = 1/25

Then, my answer is; The length of the tape (L) is the sum of all the "perimeters" around the real from the diameter of 2 to 4 inches.

L= ∏*2 + ∏*(2+1/25) + ∏*(2+2/25) + ... + ∏*(2+50/25) = ∏*(2*50 + (1/25+2/25+ ... + 50/25)) = ∏*(100+((50*51/2)/25)) =

L = ∏*(100+51) = 474.38 in

Hope I was clear!

[Guest]

I think there might be a small mistake in some calculations. Most people are calculating one too many revolutions.

i used C=2*pi*r

d = 2*pi*(1 + 1.02 + 1.04 + ... + 1.96 + 1.98)

if we did it till r = 2 then the diameter would be 4.04"

so the legnth would be 149pi" or approx. 468.1"

[Guest]

I take back my last post. Corp_Carp has it right.

150pi" or approx. 471.24"

[Guest]

As I see it, you guys are missing the boat.

The tape has a thickness of 1/50, therefore with each wrap the diameter of the spool increases by .04

The radius increases by .02

You need 50 wraps of the spool.

Using 2*pi*r

I come up with a tape length of approx. 468.0973 inches.

[Guest]

The area that the tape makes in a cross section is

pi (2)² - pi (1)² = 3 pi

The length of the tape is then 3pi*50 = 150 pi = 471 in

The tape must be wrapped 50 times to have a diameter of 4 inches:

tapeThickness=1/50;

tapeLength=0;

diameter=2;

for i=1:50

diameter=diameter+2*tapeThickness;

tapeLength=tapeLength+diameter*pi;

end

diameter = 4.0000

tapeLength = 474.38

[Guest]

Well the formula would be:

Initial diameter is 2 inches so initial radius is 1 inch.

The increment in radius is 0,02 inches.

Since the total increase in diameter is 2 inches this means for the radius the increase was 1 inch

Dividing radius increase by the thickness of the tape to find how many times the tape can go around the roll in order to reach that specific diameter.

1/(1/50)=50

So the tape went round the roll 50 times

Length = 2.pi.1 + 2.pi.(1+0,02) + 2.pi.(1+0,02+ 0,02) + ... + 2.pi.(1+0,02+0,02+...+0,02) <last parethesis has 49 additions of 0,02

=2pi.(50) + 2.pi. (0,02 + 2.0,02 + 3.0,02 + ... + 49.0,02)=

=100pi+2.pi.0,02( 1 + 2 + 3 + ... + 49)=

For the parenthesis we can use this formula

(1+N)*N/2 where N is the last number (49 in our case)

=100pi+2.pi.0,02.[(1+49)*49/2]=100pi+2.pi.0,02*1225=100pi+49pi=149pi

~=468 inches

Edited May 15, 2009 by Lemeshianos