Slide around a Cylinder

To solve this question we are going to "unroll" the wire or whatever it was from the cylinder. The wire is sloping downwards at a constant rate, so when we "unroll" it it forms a straight line; the hypotenuse of a right angled triangle. The legs of this triangle will be the height (40pi) and the base, 5 times the circumference ( because the wire has gone around the cylinder exactly 5 times): 30pi. This is a lovely little pythagorean triple with a factor of ten, so we know the hypotenuse will have a length of 50pi. Therefore, our answer is 50.

Since we know the Radius is 3m the circumference of the base is $6\pi$ m. The slide wraps around the cylinder 5 times so the length of the base is $30\pi$ m. The height is $40\pi$ m. So by the Pythagorean Theorem we get that the length of the slide is $\boxed{50}\pi$ m

Step 1: The cylinder is wrapped with 5 times then the height of cylinder is divided in five equal part is (40 pie)/5= 8 pie meter, it is the height of one complete roll.

Step 2. When unroll the cylinder then it form a rectangle and its Width is =2 pie r= 2 pie 3=6*pie meter.

Step 3. The height of one complete roll (8 pie) and the width of unroll cylinder (6 pie) is formed a rectangle and wire wrapped on its diagonal.

Step 4. The length of diagonal (length of one complete roll of wire) is= pie [sqrt{(8 8)+(6 6)}]= 10 pie.

Step 5. The length of total wire wrapped on cylinder= 5* length of one complete wrapped wire= 5 10 pie=50*pie

Step 6. x pie=50 pie then x=50.

i was actually rolling and unrolling the thread on a cylinder and finally drawn a correct inference.

Here is another solution by solving in cylindrical coordinates (a more general problem can be solved by this approach, if it was a cone, sphere, ellipsoid, paraboloid, hyperboloid rather than a cylinder with a slight modification to the arc length formula)

In cylindrical co-ordinates arc-length is given by dL = sqrt(R^2 d theta^2 + dz^2),

Where L = differential lenght of the arc R = Radius of Cylinder z = vertical distance

If H is the height of the cylinder and the arc wraps around N times over the height H, then

d theta / dz = (N* 2pi)/H
d theta^2 = (N*2pi/H)^2 * dz^2

integrating the arc lenght dL from L--> 0 to L and z --> 0 to H

integral (dL) = integral [ sqrt(R^2 * N^2 * 4pi^2/H^2 + 1) * dz], for L--> 0 to L and z --> 0 to H

L = sqrt(R^2 * N^2 * 4pi^2/H^2 + 1) * H

Given R = 3, N = 5, H = 40pi

L = sqrt(9 * 25 * 4pi^2/(1600pi^2) +1) * 40pi
= sqrt(900/1600 +1) * 40pi
= sqrt(2500/1600) * 40pi
= 5/4 * 40pi
= 50pi, hence x = 50

I did this in a different way i guess. I divided the cylinder in 5 small cylinders of height $8\pi$ also, of course, dividing the wire in 5 parts. Analysing only this part, viewing from the front, the wire will go from the top, cross half of the circumference, and stop at a height $4\pi$ (remembering he was initially at $8\pi$ , and then, the opposite side, it'll go from $4\pi$ cross the other half of the circumference and stop at height $0$ . Then i "flattened" the cylinder and got a rectangle with 2 straight lines forming two triangles (each one, the part of the wire that crosses the front, and the part crossing the back side). Knowing that the base of the rectangle will be half of the circumference, $3\pi$ , the height $8\pi$ , and we'll have two triangles with the same base and height $4\pi$ , we just need to do the Pythagorean Theorem and discover that each half way around the cylinder is $5\pi$ , then, since we got the wire crossing 10 times half of the cylinder, we have a length of $50\pi$

If you unroll the wire, you'll get a rectangle where the:

Length = 5 times the circumference of the base, and the

Width = Height

Length $= 5(2\pi r) = 10\pi r = 10\pi (3) =- 30\pi$

Width $=40\pi$

Using the Pythagorean Theorem, which is $c=\sqrt{a^2+b^2}$

The length of the wire is $\sqrt {(30\pi)^2+(40\pi)^2} = \sqrt {900\pi^2+1600\pi^2} = \sqrt {2500\pi^2} = 50\pi$

Therefore, the final answer is 50 .

we must draw the picture in 6 right angle triangle each of equal height =40*pi /6 and each have height of diameter that is 6 double of 3.. there fore each hypotenuse is multiplied to 6 and ans is 50