Arc length of vector-valued function; am I starting right?

In summary: Also, your second to last line should be:∫[0,θ] 3sec^2(u)tan(u)duAnd then use a trig identity to simplify and integrate.
  • #1
RogerDodgr
20
0

Homework Statement


Given: R(t)= <(1/2)t^2, (4/3)t^(3/2), 2*sqrt(3)t>
Find: Arc length function s(t) where t_0 =0

Homework Equations


Is this the correct formula?
∫[0,t] sqrt( derivative^2 + derivative^2 +derivative^2) dt
∫[0,t] sqrt(t^2 + 4t + 12) dt

The Attempt at a Solution



∫[0,t] sqrt(t^2 + 4t + 12) dt

I am not looking for anyone to do my homework, but before I start a messy integral I just want to know if:
A) I am starting correctly, and
B) if their is an easier way to do this (or strategy)

Thanks.
 
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  • #2
[tex]s=\int_a^b|r'(t)|dt[/tex]

Your work is correct so far! Keep going ... what's your next step?
 
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  • #3
I apreciate your encouraging tone rocophysics; I was about to quit for the night.

∫[0,t] sqrt(t^2 + 4t + 12) dt
∫[0,t] sqrt((t+2)^2 + 8) dt
u=t+2 du=1
∫[0,t] sqrt((u)^2 + 8) du

then maybe trigonometric substitution.
 
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  • #4
[tex]\int_0^t\sqrt{t^2+4t+12}dt[/tex]

[tex]\int_0^t\sqrt{(t+2)^2+8}dt[/tex]

Go ahead and do trig sub after this step ...

[tex]t+2=\sqrt 8\tan\theta[/tex]
[tex]dt=\sqrt 9\sec^{2}\theta d\theta[/tex]

But don't forget to change your limits ...
 
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  • #5
I tried but I know there is at least one mistake somewhere; here's my attempt (each line is numbered 1-14):
http://www.sudokupuzzles.net/IMG_0030.jpg [Broken]
 
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  • #6
You forgot about your constant of 8 (line 10), and it should be to both your Integrals, not just one.
 
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1. What is the definition of arc length for a vector-valued function?

The arc length of a vector-valued function is the distance traveled along the curve by the particle represented by the function. It can be thought of as the length of a string that is wrapped around the curve.

2. How do you calculate the arc length of a vector-valued function?

To calculate the arc length of a vector-valued function, you need to integrate the magnitude of the derivative of the function with respect to the parameter. This integral is typically expressed as a definite integral with the lower and upper limits being the initial and final values of the parameter.

3. Can I use any parameter to calculate the arc length of a vector-valued function?

Yes, you can use any parameter that is used to define the function. However, it is important to note that different parameters may result in different values for the arc length, but they will represent the same distance traveled along the curve.

4. Is there a specific formula for finding the arc length of a vector-valued function?

Yes, there is a specific formula for finding the arc length of a vector-valued function. It is known as the arc length formula and is given by: L = ∫ab |𝛄(t)| dt, where a and b are the lower and upper limits of the parameter, and |𝛄(t)| is the magnitude of the derivative of the function with respect to the parameter.

5. Are there any practical applications of calculating the arc length of a vector-valued function?

Yes, there are many practical applications of calculating the arc length of a vector-valued function. Some examples include calculating the distance traveled by a particle in motion, determining the length of a wire or rope that follows a certain path, and finding the perimeter of a curve in a 3D space.

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