# Arc length of vector function with trigonometric components

• Subdot
In summary, a particle moving on a curve according to the given equation traces out a path of length 3c2/a cos2t sin2t.
Subdot

## Homework Statement

Find the length of the path traced out by a particle moving on a curve according to the given equation during the time interval specified in each case.
r(t) = (c2/a)cos3t i + (c2/b)sin3t j

where i and j are the usual unit vectors, 0 $\leq$ t $\leq$ 2$\pi$, c2 = a2 - b2, and 0 < b < a

Also, r(t) is the position vector-valued function of t. Scalar functions will not be bolded.

## Homework Equations

I know that the arc length in this case is the integral of ||r'(t)|| = v(t) from 0 to 2$\pi$.

## The Attempt at a Solution

Differentiating gives: r'(t) = (-3c2/a)cos2t sin t i + (3c2/b)sin2t cos t j.

Therefore, v(t) = √{(9c4/a2)cos4t sin2t + (9c4/b2)sin4t cos2t)} = √{(9c4/a2)cos2t(cos2t sin2t) + (9c4/b2)sin2t(sin2t cos2t)} =

√{(9c4/a2)cos2t(0.25sin22t) + (9c4/b2)sin2t(0.25sin22t)} = √{(9c4/a2)cos2t(0.25sin22t) + (9c4/b2)sin2t(0.25sin22t)} =

(3c2/2)(sin2t)*√{(1/a2)cos2t + (1/b2)sin2t} = (3c2/(2ab))(sin2t)*√{(b2)cos2t + (a2)sin2t}

And that is where I get stuck. I'm not sure what the integral of that expression is, nor do I know what I can do to simplify it further. I would say that it has no simple solution, since the argument under the square root looks an awfully lot like an integral for an ellipse, and I have heard one needs something called an "elliptic integral" to solve. However, the answer for this problem is quite simple: (4a3-4b3)/(ab). Will someone please help me with this?

Last edited:
Hi Subdot!

(have a square-root: √ )
Subdot said:
(3c2/2)(sin2t)*Sqrt {(1/a2)cos2t + (1/b2)sin2t} = (3c2/(2ab))(sin2t)*Sqrt {(b2)cos2t + (a2)sin2t}

I'm not sure what the integral of that expression is …

Use standard trigonometric identities to write it as sin2t f(cos2t)

Hello! Oh! I get it now. I just use the identities: sin2 = .5 - .5cos2t and cos2 = .5 + .5cos2t after which I do a simple substitution. Thanks for that! On another related note, now it's obvious that I'm going to need to integrate it a step over the interval at a time.

[(1/(ab)) (.5b2 + (.5b2 - .5a2)cos(2t) + .5a2)1.5] from 0 to 2$\pi$. If I try to integrate it from 0 to .5$\pi$ then from .5$\pi$ to $\pi$ and so on I come up with (2a3 - 2b3)/(ab) + (2b3 - 2a3)/(ab). I'm guessing from the answer that I am justified in flipping that last fraction around to (2a^3 - 2b^3)/(ab). However, I'd like to know why, so I'll be able to do it in the future without knowing the answer. I think it's because the arc length must be positive and so I must flip (b3 - a3)/(ab) to (a3 - b3)/(ab) because b < a? So then do you take the absolute value of the integral as you move up the interval, not the integrand, when you find arc length?

The reason is that you factored cos(t)^2*sin(t)^2 outside of the sqrt as cos(t)*sin(t). It really should be |cos(t)*sin(t)|. So you have to break the integral where cos(t)*sin(t) changes sign.

Okay, I get it all now. I can't believe I missed that... Thanks both!

## 1. What is the formula for calculating the arc length of a vector function with trigonometric components?

The formula for calculating the arc length of a vector function with trigonometric components is given by: L = ∫√(x'(t)^2 + y'(t)^2) dt, where x'(t) and y'(t) are the derivatives of the x and y components of the vector function.

## 2. How is the arc length of a vector function with trigonometric components different from a regular arc length?

The arc length of a vector function with trigonometric components takes into account the changing direction of the vector as it moves along the curve, whereas a regular arc length only considers the magnitude of the vector.

## 3. Can the arc length of a vector function with trigonometric components be negative?

No, the arc length of a vector function with trigonometric components cannot be negative. It represents the distance traveled along the curve and distance is always a positive value.

## 4. Is there a simplified formula for calculating the arc length of a vector function with trigonometric components?

Yes, if the vector function has only one trigonometric component (e.g. only a sine or cosine term), then the arc length formula can be simplified to: L = ∫√(1 + y'(t)^2) dt.

## 5. How is the arc length of a vector function with trigonometric components used in real-life applications?

The arc length of a vector function with trigonometric components is used in various fields of science and engineering, such as physics, astronomy, and computer graphics. It can help determine the trajectory of a moving object, the shape of a curve, or the design of a 3D model.

Replies
2
Views
2K
Replies
1
Views
3K
Replies
2
Views
2K
Replies
4
Views
5K
Replies
3
Views
2K
Replies
2
Views
1K
Replies
12
Views
2K
Replies
4
Views
5K
Replies
5
Views
3K
Replies
4
Views
1K