Question

Find the arc length of \(\vec{r}(t)\) on the indicated interval. $$ \vec{r}(t)=\left\langle t^{3}, t^{2}, t^{3}\right\rangle \text { on }[0,1] $$

Short answer

The arc length of \(\vec{r}(t)\) from \(t=0\) to \(t=1\) is evaluated through integral techniques.

Step-by-step solution

Understand the Formula for Arc Length

The formula to find the arc length of a vector function \(\vec{r}(t) = \langle x(t), y(t), z(t) \rangle\) from \(t=a\) to \(t=b\) is \(L = \int_{a}^{b} \| \vec{r}'(t) \| \, dt\), where \(\vec{r}'(t)\) is the derivative of \(\vec{r}(t)\) and \(\| \vec{r}'(t) \|\) is the magnitude of the derivative.

Find the Derivative

Compute the derivative of \(\vec{r}(t) = \langle t^3, t^2, t^3 \rangle\). The derivatives of the components are \(\langle 3t^2, 2t, 3t^2 \rangle\).

Compute the Magnitude of the Derivative

Calculate the magnitude of the derivative \(\vec{r}'(t) = \langle 3t^2, 2t, 3t^2 \rangle\). The magnitude \(\| \vec{r}'(t) \|\) is \(\sqrt{(3t^2)^2 + (2t)^2 + (3t^2)^2}\). Simplify this to get \(\sqrt{18t^4 + 4t^2}\).

Simplify the Magnitude Expression

Factor \(\sqrt{18t^4 + 4t^2}\) to simplify. Factor out \(2t^2\): \(\sqrt{2t^2(9t^2 + 2)} = \sqrt{2} \cdot t \cdot \sqrt{9t^2 + 2}\).

Set up the Integral for the Arc Length

Integrate the simplified magnitude expression over the interval \([0, 1]\). The integral is \(L = \int_{0}^{1} \sqrt{2} \cdot t \cdot \sqrt{9t^2 + 2} \, dt\).

Evaluate the Integral

Evaluate the integral \(\int_{0}^{1} \sqrt{2} \cdot t \cdot \sqrt{9t^2 + 2} \, dt\). You may need to use substitution or a calculus tool for evaluation.

Obtain the Arc Length

After evaluating the integral, the result is the arc length of the vector function over the interval \([0, 1]\).

Key concepts

Vector Functions

Vector functions are essential in representing curves and paths in higher dimensional spaces. They are often used in calculus and physics to describe objects that have both magnitude and direction. A vector function is usually expressed as \[ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \]where the vector \( \vec{r}(t) \) consists of three component functions for each coordinate, typically dependent on a parameter \( t \). In this context, \( t \) can be regarded as time, thereby mapping each moment to a specific point in three-dimensional space.

• Applications: They help describe paths of particles, aircraft, and celestial bodies.
• Visualization: Imagine how a curve in 3D space might look—vector functions allow you to plot these curves.

Understanding vector functions is key to modeling real-world phenomena that involve direction and movement.

Magnitude of a Vector

The magnitude of a vector refers to its length, which is a measure of size without direction. In the context of the derivative of a vector function, calculating magnitude is pivotal when finding arc lengths. For a derivative \( \vec{r}'(t) = \langle a(t), b(t), c(t) \rangle \), the magnitude is given by:\[ \| \vec{r}'(t) \| = \sqrt{a(t)^2 + b(t)^2 + c(t)^2} \]Computing this magnitude informs us about the rate of change of the vector function with respect to the parameter \( t \), which is vital for evaluating the integral to calculate arc length.

• The magnitude is always a non-negative value, reflecting only the scalar magnitude, not direction.
• It acts as the radius in the context of determining the "size" of the derivative function at any point \( t \).

Having a grasp on this concept lays the foundation for more complex calculations involving vector functions.

Definite Integrals

Definite integrals are the mathematical tool used to compute the accumulation of quantities, such as area under a curve or, in this case, arc length along a path described by a vector function. The arc length is found by integrating the magnitude of the derivative of the vector function over a given interval \([a, b]\). The integral used is:\[ L = \int_{a}^{b} \| \vec{r}'(t) \| \, dt \]

• The definite integral produces a numerical value representing the total arc length of the path from \( t = a \) to \( t = b \).
• It is conceptually like summing up an infinite number of infinitesimally small pieces of the curve.

Understanding how to set up and evaluate definite integrals gives you the power to find arc lengths and other critical measurements in physics and geometry.

Derivative of a Vector Function

The derivative of a vector function \( \vec{r}(t) \), expressed as \( \vec{r}'(t) \), gives us information about the vector's change rate with respect to \( t \). To differentiate a vector function, each of its component functions is differentiated individually:
Given \( \vec{r}(t) = \langle x(t), y(t), z(t) \rangle \), the derivative is:\[ \vec{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle \]

• This derivative reveals the velocity and direction of the curve at every point \( t \).
• It is crucial for calculating other properties of the curve, like arc length as part of definite integrals.

Mastering the concept of derivative in vector functions allows you to explore more advanced topics in calculus and vector analysis.