What is the arc length parametrization of α(t) and why is the s so tiny?

In summary: Maybe because the itex tag is for inline tex and it's trying to conserve space. If you put it on its own line with tex tags instead it looks like this$$t(s) = sinh^{-1}\frac{s}{2}$$Then again, maybe not. It still looks small compared to the ##2##. Something to do with the default fonts I suppose.
  • #1
Shackleford
1,656
2
3.b Find the arc length of the given curve

α(t) = (t, 1, (1/6)t^3 + (1/2)t^-1) from t = 1 to t = 3.

Of course, I need to find the first derivative and integrate its norm.

α'(t) = (1, 0, (1/2)t^2 - (1/2)t^-2)

∫ [1 + (1/4)t^4 + (1/4)t^-4]^(1/2) dt, t = 1 to t = 3.

Have I simply forgotten useful integrals?

5. Let α(t) = (e^t, e^-t, root2*t). Calculate first derivative, norm, and re-parametrize alpha by its arc length, starting at t = 0.

α'(t) = (e^t, -e^-t, root2)

∫ [e^2u + e^-2u + 2]^(1/2) du, u = 0 to u = t.
 
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  • #2
With over 1500 posts to this group, I would say it is time you learned to type simple equations like yours in Latex to make them more readable for everyone.

Shackleford said:
Of course, I need to find the first derivative and integrate its norm.

α'(t) = (1, 0, (1/2)t^2 - (1/2)t^-2)

∫ [1 + (1/4)t^4 + (1/4)t^-4]^(1/2) dt, t = 1 to t = 3.

Have I simply forgotten useful integrals?

Apparently you have forgotten algebra too. ##(a-b)^2\ne a^2+b^2##. Once you calculate the norm correctly, use algebra to make a perfect square under the square root.

α'(t) = (e^t, -e^-t, root2)

∫ [e^2u + e^-2u + 2]^(1/2) du, u = 0 to u = t.

Again, you might work some algebra on the square root. Arc length is usually denoted with ##s##.
 
  • #3
LCKurtz said:
With over 1500 posts to this group, I would say it is time you learned to type simple equations like yours in Latex to make them more readable for everyone.
Apparently you have forgotten algebra too. ##(a-b)^2\ne a^2+b^2##. Once you calculate the norm correctly, use algebra to make a perfect square under the square root.
Again, you might work some algebra on the square root. Arc length is usually denoted with ##s##.

Eh. You're right. 3.b is not a mystery anymore. I'll see what I can do with 5.
 
Last edited:
  • #4
For the arc length I got

s(t) = [itex] e^t - e^{-t} [/itex]
 
  • #5
Shackleford said:
For the arc length I got

s(t) = [itex]e^t - e^{-t} [/itex]

You can include the whole equation in itex brackets so it looks like this:

[itex]s(t) = e^t - e^{-t} [/itex]

Your problem asked for the arc length from between ##t=1## to ##t=3##. I think what you have calculated is the arc length measuring from ##0## to ##t##.
 
  • #6
LCKurtz said:
You can include the whole equation in itex brackets so it looks like this:

[itex]s(t) = e^t - e^{-t} [/itex]

Your problem asked for the arc length from between ##t=1## to ##t=3##. I think what you have calculated is the arc length measuring from ##0## to ##t##.

Thanks for the tip.

That's right. The second problem wants me to parametrize the curve in terms of arc length. In this case, I need to find t(s) to get [itex]α(t(s))[/itex].

I suppose I simply need to take the logarithm of both sides and go from there.
 
  • #7
Shackleford said:
Thanks for the tip.

That's right. The second problem wants me to parametrize the curve in terms of arc length. In this case, I need to find t(s) to get [itex]α(t(s))[/itex].

I suppose I simply need to take the logarithm of both sides and go from there.

Or, if you are comfortable with hyperbolic functions, you might notice$$
s(t)=2\sinh(t)$$and use the inverse hyperbolic sine function.
 
  • #8
LCKurtz said:
Or, if you are comfortable with hyperbolic functions, you might notice$$
s(t)=2\sinh(t)$$and use the inverse hyperbolic sine function.

I haven't dealt much with hyperbolic functions, but I did get

[itex] t(s) = sinh^{-1}\frac{s}{2}[/itex].

Why is the "s" so tiny?
 
  • #9
Shackleford said:
I haven't dealt much with hyperbolic functions, but I did get

[itex] t(s) = sinh^{-1}\frac{s}{2}[/itex].

Why is the "s" so tiny?

Maybe because the itex tag is for inline tex and it's trying to conserve space. If you put it on its own line with tex tags instead it looks like this$$
t(s) = sinh^{-1}\frac{s}{2}$$Then again, maybe not. It still looks small compared to the ##2##. Something to do with the default fonts I suppose.
 

What is the definition of arc length?

Arc length is defined as the distance along a curve or arc between two points, measured along the arc. It is typically denoted by the symbol "s".

What is the formula for calculating arc length?

The formula for calculating arc length is s = rθ, where "r" is the radius of the circle or arc, and "θ" is the central angle measured in radians.

How do you find the arc length of an angle?

To find the arc length of an angle, you can use the formula s = rθ, where "r" is the radius of the circle or arc, and "θ" is the measure of the angle in radians. Alternatively, you can use the formula s = 2πr(n/360), where "n" is the measure of the angle in degrees.

What is the relationship between arc length and central angle?

The arc length and central angle are directly proportional. This means that as the central angle increases, the arc length also increases, and vice versa.

What is the significance of "α from 0 to t" in arc length?

The notation "α from 0 to t" represents the starting and ending points of the curve or arc that is being measured for its length. The angle "α" represents the starting point, which is usually 0, and "t" represents the ending point, which can be any value within the range of the central angle.

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