Optimization
Optimization comes from the same root as "optimal". "Optimal" means the highest. When you do the optimization process, that is when you are "making it best" to maximize everything and to achieve optimal results, a set of parameters is the base for the selection of the best element for a given system.
Integration
Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integration and differentiation are inverse operators. Differential equations give a relation between a function and its derivative.
Application of Integration
In mathematics, the process of integration is used to compute complex area related problems. With the application of integration, solving area related problems, whether they are a curve, or a curve between lines, can be done easily.
Volume
In mathematics, we describe the term volume as a quantity that can express the total space that an object occupies at any point in time. Usually, volumes can only be calculated for 3-dimensional objects. By 3-dimensional or 3D objects, we mean objects that have length, breadth, and height (or depth).
Area
Area refers to the amount of space a figure encloses and the number of square units that cover a shape. It is two-dimensional and is measured in square units.
![### Arc Length Parametrization of a Vector Function
**Problem Statement:**
Find an arc length parametrization of \(\mathbf{r}(t) = \left( e^t \sin(t), e^t \cos(t), 2e^t \right)\).
**Solution:**
The arc length parametrization \(\mathbf{r}_1(s)\) is given by the following expression:
\[
\mathbf{r}_1(s) = \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \sin \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \right), \cos \left( \ln \left( 1 + \frac{s}{\sqrt{6}} \right) \right), 2 \right)
\]
**Explanation:**
Here, \(\mathbf{r}(t)\) describes a vector function with three components. The goal is to find its arc length parametrization, which essentially reparametrizes the curve in terms of the arc length \(s\), making \(s\) the new parameter that measures distance along the curve.
**Detailed Breakdown of Steps (Illustrative):**
1. **Parameterize the Curve:**
Given the original function \(\mathbf{r}(t)\), calculate the magnitude of the derivative, \(\left| \mathbf{r}'(t) \right|\), which represents the rate of change of the vector function's length.
2. **Integral Calculation:**
Integrate the magnitude of the derivative to find the arc length as a function of \(t\), denoted by \(s(t)\).
3. **Invert the Relation:**
Solve for \(t\) in terms of the arc length \(s\) to get the inverse function, \(t(s)\).
4. **Reparametrize the Function:**
Substitute \(t(s)\) back into the original vector function to obtain the arc length parametrization \(\mathbf{r}_1(s)\).
The resulting parametrization ensures that the parameter \(s\) directly corresponds to the arc length measured along the curve from a fixed starting point. This reparametrization simplifies computations involving the length of paths and is particularly useful in applications concerning physical distances and curvatures.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6647b19a-77ee-47ef-8a0e-540e0cdeed5a%2F8cf06502-3b75-4560-9812-f9ca1d271b19%2Fz87n9r4.png&w=3840&q=75)

Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 5 images









