# Multivariable Arc Length Problem: Weird Form with Parameterization

In summary: However, in your calculation, you have squared the individual components of ##c'(t)##, which is incorrect. Instead, you should square the entire vector, which will give you the correct formula of ##ds = \sqrt{1 + 4e^{2t} + 4e^{4t}} \, dt##. This will then simplify to ##ds = \sqrt{(1 + 2e^{2t})^2} \, dt = 1 + 2e^{2t} \, dt##.
Homework Statement
See Attachment
Relevant Equations
ds=sqrt(dx^2+dy^2+dz^2)
Arc Length = integral on C of ds
Problem: See Attachment. Parts (a) & (b) are clear, but my confusion arises in (c)-- I feel like there is a much simpler form. While technically my answer is correct, there must be something I'm missing.

I parameterized the curve C=(t, e^2t, e^2t) and got c'(t)=(1,2e^2t,2e^2t), which should be correct since I got the answer for (b). I said ds=sqrt(||c'(t)||)

So ds=sqrt(1+4e^2t+4e^4t) =sqrt((1+2e^2t)^2)=1 + 2e^2t. The limits are from t=0 to t=x_0.

Integral(1+2e^2t) dt = t+e^2t, and I'll evaluate at the limits to get:

x_0 + e^(2x_0) - 1. So if x_0=ln(W+1)^(1/3), then e^(2x_0)= (W+1)^2/3. But still, why ask for it in this crazy form? Or is it just a silly excuse to

So ds=sqrt(1+4e^2t+4e^4t) =sqrt((1+2e^2t)^2)=1 + 2e^2t.
Stop right there, check your formula, my clac book says that

$$ds = \sqrt{\left( \tfrac{dx}{dt}\right) ^2 +\left( \tfrac{dy}{dt}\right) ^2 +\left( \tfrac{dz}{dt}\right) ^2 } \, dt$$

etotheipi
benorin said:
Stop right there, check your formula, my clac book says that

$$ds = \sqrt{\left( \tfrac{dx}{dt}\right) ^2 +\left( \tfrac{dy}{dt}\right) ^2 +\left( \tfrac{dz}{dt}\right) ^2 } \, dt$$

I agree on the formula, but not sure where I went wrong. Since it's arc length and the path is parameterized as c(t)= (t, 2e^t, e^(2t)) and c'(t)= (1, 2e^(t), 2e^(2t)) = (dx/dt, dy/dt, dz/dt). So squaring all terms and summing up gives: 1 + 4e^(2t) + 4e^(4t). So then I took the square root, simplified to 1+2e^2t, and then tacked on the dt. Where did I go wrong?

benorin said:
Stop right there, check your formula, my clac book says that

##ds = \sqrt{\left( \tfrac{dx}{dt}\right) ^2 +\left( \tfrac{dy}{dt}\right) ^2 +\left( \tfrac{dz}{dt}\right) ^2 } \, dt##
The above is equivalent to ##ds = \sqrt{dx^2 + dy^2 + dz^2}##

## 1. What is a multivariable arc length problem?

A multivariable arc length problem is a mathematical problem that involves finding the length of a curve in a two or more dimensional space.

## 2. What is the "weird form" in a multivariable arc length problem?

The "weird form" in a multivariable arc length problem refers to the unconventional or non-standard form of the curve's equation or parameterization.

## 3. What is parameterization in the context of a multivariable arc length problem?

Parameterization is the process of representing a curve or surface in terms of one or more parameters, usually in the form of equations. In a multivariable arc length problem, parameterization is used to find the length of a curve by expressing it in terms of a single parameter.

## 4. How is a multivariable arc length problem solved?

To solve a multivariable arc length problem, the curve's equation or parameterization is first determined. Then, the integral of the square root of the sum of the squares of the derivatives of the parameters is evaluated over the given interval. This integral gives the length of the curve.

## 5. Why are multivariable arc length problems important?

Multivariable arc length problems are important in various fields of science and engineering, such as physics, astronomy, and computer graphics. They allow us to calculate the length of complex curves and surfaces, which is essential in understanding and analyzing real-world phenomena and designing efficient and accurate models and simulations.

• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
2K
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
973
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Differential Equations
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
2K