Curve Parametrization Analysis Finding Unit Tangent Vector And Length

Introduction

In the realm of calculus and vector analysis, understanding curves parametrized by vector functions is crucial for various applications, ranging from physics simulations to computer graphics. This article delves into a comprehensive analysis of the curve parametrized by the vector function r(t) = <3t^2, 2t^3>, where t belongs to the interval [0, √3]. We will explore key properties of this curve, including the derivation of the unit tangent vector T(t) and the calculation of the curve's length. This exploration will provide a deeper understanding of how to analyze and characterize curves defined in vector form. Understanding these concepts is fundamental for advanced studies in mathematics, engineering, and related fields.

(a) Finding the Unit Tangent Vector T(t)

The unit tangent vector, denoted as T(t), provides essential information about the direction of a curve at a given point. It is a vector of unit length that is tangent to the curve at the point corresponding to the parameter value t. To find T(t), we first need to compute the derivative of the position vector r(t) with respect to t, which gives us the tangent vector r'(t). Then, we normalize r'(t) to obtain the unit tangent vector.

The given position vector is r(t) = <3t^2, 2t^3>. To find the tangent vector r'(t), we differentiate each component of r(t) with respect to t:

r'(t) = d/dt <3t^2, 2t^3> = <d/dt (3t^2), d/dt (2t^3)> = <6t, 6t^2>

Now that we have the tangent vector r'(t), we need to find its magnitude, denoted as ||r'(t)||. The magnitude of a vector <x, y> is given by √(x^2 + y^2). Therefore,

||r'(t)|| = √((6t)^2 + (6t²⁾2) = √(36t^2 + 36t^4) = √(36t^2(1 + t^2)) = 6|t|√(1 + t^2)

Since t ∈ [0, √3], t is non-negative, so |t| = t. Thus,

||r'(t)|| = 6t√(1 + t^2)

Finally, the unit tangent vector T(t) is obtained by dividing the tangent vector r'(t) by its magnitude ||r'(t)||:

T(t) = r'(t) / ||r'(t)|| = <6t, 6t^2> / (6t√(1 + t^2)) = <1, t> / √(1 + t^2)

Thus, the unit tangent vector T(t) for the curve parametrized by r(t) = <3t^2, 2t^3> is <1/√(1 + t^2), t/√(1 + t^2)>. This vector indicates the direction of the curve at any point t within the interval [0, √3]. The unit tangent vector is a critical concept in differential geometry and is used to define other important properties of curves, such as curvature and torsion. Understanding the unit tangent vector is essential for analyzing the behavior of curves in space.

(b) Finding the Length of the Curve

The arc length of a curve provides a measure of the total distance traversed along the curve between two points. For a curve parametrized by a vector function r(t) over an interval [a, b], the arc length L is given by the integral of the magnitude of the derivative of r(t) with respect to t, from a to b. In other words,

L = ∫[a, b] ||r'(t)|| dt

In our case, the curve is parametrized by r(t) = <3t^2, 2t^3>, and the interval is [0, √3]. We have already found the derivative r'(t) = <6t, 6t^2> and its magnitude ||r'(t)|| = 6t√(1 + t^2). Now, we need to compute the integral of ||r'(t)|| over the interval [0, √3]:

L = ∫[0, √3] 6t√(1 + t^2) dt

To evaluate this integral, we can use a substitution method. Let u = 1 + t^2. Then, du = 2t dt, and dt = du / (2t). When t = 0, u = 1 + 0^2 = 1. When t = √3, u = 1 + (√3)^2 = 1 + 3 = 4. Thus, the integral becomes:

L = ∫[1, 4] 6t√u (du / (2t)) = ∫[1, 4] 3√u du = 3 ∫[1, 4] u^(1/2) du

Now, we can integrate u^(1/2) with respect to u:

∫ u^(1/2) du = (2/3)u^(3/2) + C

So, the arc length L is:

L = 3 [(2/3)u^(3/2)] [from 1 to 4] = 2[u^(3/2)] [from 1 to 4] = 2(4^(3/2) - 1^(3/2))

4^(3/2) = (√4)^3 = 2^3 = 8, and 1^(3/2) = 1. Therefore,

L = 2(8 - 1) = 2(7) = 14

Thus, the length of the curve parametrized by r(t) = <3t^2, 2t^3> over the interval [0, √3] is 14 units. This result demonstrates how calculus can be used to find the exact length of a curved path, which is a fundamental concept in various fields such as physics, engineering, and computer graphics. Calculating arc length is a crucial skill in understanding the geometry of curves and their properties.

Conclusion

In this article, we have conducted a thorough analysis of the curve parametrized by the vector function r(t) = <3t^2, 2t^3> over the interval [0, √3]. We successfully derived the unit tangent vector T(t) = <1/√(1 + t^2), t/√(1 + t^2)>, which provides the direction of the curve at any point t within the interval. Additionally, we calculated the length of the curve to be 14 units, demonstrating the application of integration to find the arc length of a parametrized curve. These calculations are essential for understanding the geometric properties of curves and have wide-ranging applications in various scientific and engineering disciplines.

The concepts and techniques discussed in this article are fundamental to understanding vector calculus and its applications. The ability to find unit tangent vectors and arc lengths is crucial for further studies in differential geometry, physics, and engineering. By mastering these concepts, students and professionals can better analyze and solve problems involving curves and motion in space. Continued practice and application of these techniques will further solidify understanding and proficiency in this area of mathematics.