Parameterized curve vs. Integral(Please look at my result)

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In summary, by setting \vec{u} = \frac{l-m}{|l-m|}, we can show that the arc length from l to m is always less than or equal to the length of the straight line joining l and m.
  • #1
Superboy1234
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Homework Statement



Hello just found this forum and have question if I have done my homework correctly

Anyway let [tex]\theta: I \to \mathbb{R}^3[/tex] be a parameterized curve.
Next assume that [tex][a,b] \subset I [/tex] and put [tex]\theta(a) = l[/tex](1) and [tex]\theta(b) = m[/tex]

Then show for any constant vector [tex]\vec{u}[/tex],[tex]|\vec{u}| = 1[/tex].

that the inequality

[tex](m-l) \cdot \vec{u} = \int_{a}^{b} \vec{\theta(t)} \cdot \vec{u(t)} \ dt \leq \int_{a}^{b} |\vec{\theta'(t)}| dt[/tex](1)

The Attempt at a Solution



Solution:
If I look at the dot-product of [tex]\frac{d}{dt}(\vec{\theta} \cdot \vec{u}}) = \vec{\theta(t)^{\prime}} \cdot \vec{u(t)}[/tex]

Next by Fundamental theorem of Calculus leftside of (1) is true.

The inequality however is a bit tricky for me. I seem to remember an inequality
which states that the dot product of two vectors a and b can be written as
[tex]\vec{a} \cdot \vec{b} \leq |\vec{a}| \cdot |\vec{b}|[/tex] which transferred to Our situation gives
[tex]\vec{\theta'(t)} \cdot \vec{u(t)} \leq |\vec{\theta'(t)}| \cdot |\vec{u(t)}|[/tex] and since [tex]|\vec{u}| = 1[/tex], then this is equivalent to [tex]\theta'(t) \cdot \vec{u(t)} \leq |\theta'(t)| [/tex] and finally by the rules regarding our Riemann Integrals [tex]\int_{a}^{b} f \leq \int_{a}^{b} g[/tex], then the integrals [tex]\int_{a}^{b} \ldots dt \leq \int_{a}^{b} \ldots \ dt[/tex] is also true.

There is a second part as well

If I set [tex]\vec{u} = \frac{l-m}{|l-m|}[/tex]

Then show that [tex]|\theta(b) - \theta(a)| \leq \int_{a}^{b} |\vec{\theta(t)^{\prime}| dt[/tex]

If I insert [tex]\vec{u}[/tex] into lefthand side of the inequality I get [tex](l-m) \cdot \frac{l-m}{|l-m|} = |l-m|[/tex] since [tex](l-m) \cdot (l-m) = |l-m|^2[/tex]

From this it should follow that the length of the curve from l to m is a straight line joining these points. But how exactly does it do that?
Is it something to do with if I on my curve draw a straight line from l to m and use this line as a side in a triangle and by using pythagoras then claim that line previously mentioned always will be shorter than the arc length from l to m?

Cheers Superboy
 
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  • #2
The Attempt at a Solution Yes, you are correct. By setting \vec{u} = \frac{l-m}{|l-m|}, the left hand side of the inequality becomes |l - m|, which is the length of the straight line joining l and m. Therefore, the inequality states that the arc length from l to m is always less than or equal to the length of the straight line joining l and m. You can prove this by considering a triangle with vertices at l, m, and some point \theta(t) on the curve. The two shorter sides of the triangle will then be the straight line joining l and m, and the arc length from l to m, respectively. The third side will be the line joining \theta(t) and either l or m (whichever is closer). By the triangle inequality, we know that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. Hence, this implies that the length of the straight line joining l and m is greater than or equal to the arc length from l to m.
 

Related to Parameterized curve vs. Integral(Please look at my result)

1. What is a parameterized curve?

A parameterized curve is a type of mathematical function that describes the path of a point. It is defined by a set of equations, with one or more parameters, that determine the coordinates of the point at any given time.

2. What is an integral?

An integral is a mathematical concept that represents the area under a curve, or the accumulation of a quantity over an interval. It is often used to solve problems in physics, engineering, and other fields.

3. How are parameterized curves and integrals related?

Parameterized curves and integrals are related because the integral of a parameterized curve represents the area under the curve. The integral can also be used to find the length of a parameterized curve.

4. What are the benefits of using a parameterized curve over an integral?

One benefit of using a parameterized curve is that it allows for a more precise representation of a curved path. It also allows for more complex curves to be described and analyzed. However, integrals are often used to solve real-world problems and find practical solutions.

5. Can a parameterized curve and an integral have different results?

Yes, a parameterized curve and an integral can have different results. This is because a parameterized curve describes the path of a point, while an integral represents the accumulation of a quantity. Depending on the specific equations and parameters used, the results may differ.

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