Hello,
I'm having many doubts while doing this exercise on spiral length and surface area, and I'd like to know if I'm on the right path on what I'm doing. My doubts are written as comments along my attempt at solving it. They're in bold. I had to search what an archimedean spiral was, so this is also based on what I found about it.
So, in this exercise, an archimedean spiral starts at (5,0) and ends at (-31/2,0) after 7.5 loops. It's asked two things:
i) to write an adequate parametrization and then find it's length;
ii) consider a spring which at first approximation has the geometry of the given spiral, but has 0.2 units of thickness by 5 of height. I'm asked to write a new parametrization and the integral expression which gives the total surface of the spring.
This is the reference image regarding the two questions:
I'm not sure if the way I've solved i) is correct, but my main doubts are in ii).
i)
The spiral has each loop at a constant distance given by 2πb. The general expression is ρ = a + bθ; a=5.
(15.5-5)/7.5= 1.4 (ρ increments 1.4 units each loop); 2πb = 1.4 , b = 14/20π.
Therefore, ρ = 5 + (14/20π)θ.
A parametrization γ in polar coordinates,
x = ρcos(θ) and y = ρsin(θ) , x = (5 + 14/20πθ)cos(θ) and y = (5 + 14/20πθ)sin(θ)
where θ varies from zero to 7.5*π = 15π.
γ(θ) = ((5 + (14/20π)θ)cos(θ) ; (5 + (14/20π)θ)sin(θ))
γ'(θ) = (14/20π*cos(θ) - (5 + (14/20π)θ)sin(θ); 14/20π*sin(θ) + (5 + (14/20π)θ)cos(θ))
||γ'(θ)|| = sqrt ((14/20π*cos(θ) - (5 + (14/20π)θ)sin(θ))^2 + (14/20π*sin(θ) + (5 + (14/20π)θ)cos(θ))^2)
Aux. calc:
(14/20π*cos(θ) - (5 + (14/20π)θ)sin(θ))^2 = (14/20π*cos(θ))^2 - 2 * (14/20π*cos(θ) * (5 + (14/20π)θ)sin(θ)) + ((5 + (14/20π)θ)sin(θ))^2
(14/20π*sin(θ) + (5 + (14/20π)θ)cos(θ))^2 = (14/20π*sin(θ))^2 + 2 * (14/20π*sin(θ) * (5 + (14/20π)θ)cos(θ)) + ((5 + (14/20π)θ)cos(θ))^2
The factors multiplied by 2 cancel out, and (14/20π*cos(θ))^2 + (14/20π*sin(θ))^2 + ((5 + (14/20π)θ)sin(θ))^2 * ((5 + (14/20π)θ)cos(θ))^2 is equal to (14/20π)^2 * 1 + (5 + (14/20π)θ)^2 * 1
And so the length of the spiral is the integral from 0 to 15π, of sqrt((14/20π)^2 + (5 + (14/20π)θ)^2). The result I got from sagemath, approximately 406 units.
ii)
I'm not sure if I'm meant to consider the additional spacing in between loops, so that the spiral ends at 15.5 or not, since it's said that in first approximation the geometry is similar to the one given in i). Because of that, I didn't count with the spacing,
Instead of an increment of 1.4 units as in i), I'll consider 0.2 units, corresponding to the thickness of the spring
So, 2πb = 0.2 , b = 2/20π , ρ = 5 + (2/20π)θ.
Using the cylindrical parametrization,
x = (5 + (2/20π)θ)cos(θ) , y = (5 + (2/20π)θ)sin(θ) , z = z ; θ varies from 0 to 15π, z from 0 to 5.
Therefore, γ(θ,z) = ( (5 + (2/20π)θ)cos(θ) , (5 + (2/20π)θ)sin(θ) , z )
I then calculated the Jacobian matrix, its transpose, and then multiplied it by the first one. Applying the determinant, it gave a sum of squares similar to the one I got in i):
(2/20π*cos(θ) - (5 + (2/20π)θ)sin(θ))^2 = (2/20π*cos(θ))^2 - 2 * (2/20π*cos(θ) * (5 + (2/20π)θ)sin(θ)) + ((5 + (2/20π)θ)sin(θ))^2
+
(2/20π*sin(θ) + (5 + (2/20π)θ)cos(θ))^2 = (2/20π*sin(θ))^2 + 2 * (2/20π*sin(θ) * (5 + (2/20π)θ)cos(θ)) + ((5 + (2/20π)θ)cos(θ))^2
=
(2/20π)^2 * 1 + (5 + (2/20π)θ)^2 * 1
The surface area is then given by:
integral from 0 to 5 in z * integral from 0 to 15π of sqrt ((2/20π)^2 + (5 + (2/20π)θ)^2). This latter I solved through sagemath. The result is approximately 5 * 585.
This is where I'm a bit confused... The area I calculated here, if it's correct, is just one side, right? I think I still need to multiply this by 2, and calculate the remaining area on the top and bottom of the spring, which would be through polar coordinates. Am I on the right path?
Thank you very much for helping me.
I'm having many doubts while doing this exercise on spiral length and surface area, and I'd like to know if I'm on the right path on what I'm doing. My doubts are written as comments along my attempt at solving it. They're in bold. I had to search what an archimedean spiral was, so this is also based on what I found about it.
So, in this exercise, an archimedean spiral starts at (5,0) and ends at (-31/2,0) after 7.5 loops. It's asked two things:
i) to write an adequate parametrization and then find it's length;
ii) consider a spring which at first approximation has the geometry of the given spiral, but has 0.2 units of thickness by 5 of height. I'm asked to write a new parametrization and the integral expression which gives the total surface of the spring.
This is the reference image regarding the two questions:
I'm not sure if the way I've solved i) is correct, but my main doubts are in ii).
i)
The spiral has each loop at a constant distance given by 2πb. The general expression is ρ = a + bθ; a=5.
(15.5-5)/7.5= 1.4 (ρ increments 1.4 units each loop); 2πb = 1.4 , b = 14/20π.
Therefore, ρ = 5 + (14/20π)θ.
A parametrization γ in polar coordinates,
x = ρcos(θ) and y = ρsin(θ) , x = (5 + 14/20πθ)cos(θ) and y = (5 + 14/20πθ)sin(θ)
where θ varies from zero to 7.5*π = 15π.
γ(θ) = ((5 + (14/20π)θ)cos(θ) ; (5 + (14/20π)θ)sin(θ))
γ'(θ) = (14/20π*cos(θ) - (5 + (14/20π)θ)sin(θ); 14/20π*sin(θ) + (5 + (14/20π)θ)cos(θ))
||γ'(θ)|| = sqrt ((14/20π*cos(θ) - (5 + (14/20π)θ)sin(θ))^2 + (14/20π*sin(θ) + (5 + (14/20π)θ)cos(θ))^2)
Aux. calc:
(14/20π*cos(θ) - (5 + (14/20π)θ)sin(θ))^2 = (14/20π*cos(θ))^2 - 2 * (14/20π*cos(θ) * (5 + (14/20π)θ)sin(θ)) + ((5 + (14/20π)θ)sin(θ))^2
(14/20π*sin(θ) + (5 + (14/20π)θ)cos(θ))^2 = (14/20π*sin(θ))^2 + 2 * (14/20π*sin(θ) * (5 + (14/20π)θ)cos(θ)) + ((5 + (14/20π)θ)cos(θ))^2
The factors multiplied by 2 cancel out, and (14/20π*cos(θ))^2 + (14/20π*sin(θ))^2 + ((5 + (14/20π)θ)sin(θ))^2 * ((5 + (14/20π)θ)cos(θ))^2 is equal to (14/20π)^2 * 1 + (5 + (14/20π)θ)^2 * 1
And so the length of the spiral is the integral from 0 to 15π, of sqrt((14/20π)^2 + (5 + (14/20π)θ)^2). The result I got from sagemath, approximately 406 units.
ii)
I'm not sure if I'm meant to consider the additional spacing in between loops, so that the spiral ends at 15.5 or not, since it's said that in first approximation the geometry is similar to the one given in i). Because of that, I didn't count with the spacing,
Instead of an increment of 1.4 units as in i), I'll consider 0.2 units, corresponding to the thickness of the spring
So, 2πb = 0.2 , b = 2/20π , ρ = 5 + (2/20π)θ.
Using the cylindrical parametrization,
x = (5 + (2/20π)θ)cos(θ) , y = (5 + (2/20π)θ)sin(θ) , z = z ; θ varies from 0 to 15π, z from 0 to 5.
Therefore, γ(θ,z) = ( (5 + (2/20π)θ)cos(θ) , (5 + (2/20π)θ)sin(θ) , z )
I then calculated the Jacobian matrix, its transpose, and then multiplied it by the first one. Applying the determinant, it gave a sum of squares similar to the one I got in i):
(2/20π*cos(θ) - (5 + (2/20π)θ)sin(θ))^2 = (2/20π*cos(θ))^2 - 2 * (2/20π*cos(θ) * (5 + (2/20π)θ)sin(θ)) + ((5 + (2/20π)θ)sin(θ))^2
+
(2/20π*sin(θ) + (5 + (2/20π)θ)cos(θ))^2 = (2/20π*sin(θ))^2 + 2 * (2/20π*sin(θ) * (5 + (2/20π)θ)cos(θ)) + ((5 + (2/20π)θ)cos(θ))^2
=
(2/20π)^2 * 1 + (5 + (2/20π)θ)^2 * 1
The surface area is then given by:
integral from 0 to 5 in z * integral from 0 to 15π of sqrt ((2/20π)^2 + (5 + (2/20π)θ)^2). This latter I solved through sagemath. The result is approximately 5 * 585.
This is where I'm a bit confused... The area I calculated here, if it's correct, is just one side, right? I think I still need to multiply this by 2, and calculate the remaining area on the top and bottom of the spring, which would be through polar coordinates. Am I on the right path?
Thank you very much for helping me.