Problem 54 - Calculus Volume 3

Question

Find a tangent vector at the indicated value of \(t\). \(\mathbf{r}(t)=\cos (2 t) \mathbf{i}+2 \sin t \mathbf{j}+t^{2} \mathbf{k} ; t=\frac{\pi}{2}\)

Step-by-Step Solution

Answer

The tangent vector at \( t = \frac{\pi}{2} \) is \( \pi \mathbf{k} \).

Step 1: Find the Derivative of the Vector Function

To find the tangent vector, we first need the derivative of the vector function \( \mathbf{r}(t) \). Given \( \mathbf{r}(t) = \cos(2t) \mathbf{i} + 2 \sin(t) \mathbf{j} + t^2 \mathbf{k} \), calculate the derivative \( \mathbf{r}'(t) \) by differentiating each component with respect to \( t \).- The derivative of \( \cos(2t) \) is \(-2\sin(2t)\cdot 2 = -4\sin(2t)\).- The derivative of \( 2\sin(t) \) is \( 2\cos(t) \).- The derivative of \( t^2 \) is \( 2t \).Therefore, the derivative \( \mathbf{r}'(t) = -4\sin(2t) \mathbf{i} + 2\cos(t) \mathbf{j} + 2t \mathbf{k} \).

Step 2: Evaluate the Derivative at the Given \(t\)

With the derivative \( \mathbf{r}'(t) = -4\sin(2t) \mathbf{i} + 2\cos(t) \mathbf{j} + 2t \mathbf{k} \), substitute \( t = \frac{\pi}{2} \) into the expression:- \( -4\sin(2 \cdot \frac{\pi}{2}) \mathbf{i} = -4\sin(\pi) \mathbf{i} = 0 \mathbf{i} \) since \( \sin(\pi) = 0 \).- \( 2\cos(\frac{\pi}{2}) \mathbf{j} = 2\cdot0 \mathbf{j} = 0 \mathbf{j} \) since \( \cos(\frac{\pi}{2}) = 0 \).- \( 2(\frac{\pi}{2}) \mathbf{k} = \pi \mathbf{k} \).Thus, \( \mathbf{r}'\left(\frac{\pi}{2}\right) = 0 \mathbf{i} + 0 \mathbf{j} + \pi \mathbf{k} = \pi \mathbf{k} \).

Step 3: Write the Tangent Vector

The tangent vector at \( t = \frac{\pi}{2} \) is the value of \( \mathbf{r}'(t) \) at that point, which we just calculated as \( \pi \mathbf{k} \). This is the vector pointing in the direction of the curve at \( t = \frac{\pi}{2} \). It is purely in the \( k \) direction, with no components in the \( i \) or \( j \) directions.

Key Concepts

Tangent Vector

A tangent vector is a fundamental concept in vector calculus that helps us understand the direction a parametric curve is heading at a particular point. When we talk about a tangent vector, we are looking at the vector that touches the curve at one exact point and points in the same direction as the path of the curve at that point. This is important because it gives insight into how the curve moves through space at a given moment.
In our exercise, we found the tangent vector at a specific value of \( t \). Essentially, this involved finding the derivative of the vector function and evaluating it at the desired point. With \( t = \frac{\pi}{2} \), the tangent vector at this point was entirely in the \( k \) direction, indicating the curve's motion was straight upwards at this particular moment. Understanding tangent vectors is essential in various applications such as physics for understanding velocity and direction.

Derivative of Vector Function

The derivative of a vector function is crucial in determining how each component of a vector changes as its parameter changes. It's like peeling back layers to see how the vector as a whole behaves over time or space. By differentiating each component of a vector function with respect to its parameter, we can uncover this velocity-like behavior of the vector.
For the vector function \( \mathbf{r}(t) = \cos(2t) \mathbf{i} + 2 \sin(t) \mathbf{j} + t^2 \mathbf{k} \), differentiating each part with respect to \( t \) gives:

• - The \( \cos(2t) \) becomes \( -4\sin(2t) \).
• - The \( 2\sin(t) \) becomes \( 2\cos(t) \).
• - The \( t^2 \) becomes \( 2t \).

Thus, the derivative \( \mathbf{r}'(t) \) tells us precisely how the original vector function \( \mathbf{r}(t) \) changes with respect to \( t \). It is a new vector function pointing in the direction of the original curve's movement. This allows for predictions on speed and direction, serving as a foundational tool in calculus and mathematical physics.

Parametric Curves

Parametric curves offer a unique understanding of mathematical shapes by using parameters. Unlike conventional equations, which relate \( x \) and \( y \) directly, parametric equations define curves through a third variable, \( t \), known as the parameter. This allows each coordinate (such as \( x \), \( y \), and \( z \)) to be expressed as a function of \( t \), providing a comprehensive way to describe complex curves and surfaces.
For instance, the curve demonstrated in the vector function \( \mathbf{r}(t) = \cos(2t) \mathbf{i} + 2 \sin(t) \mathbf{j} + t^2 \mathbf{k} \) is defined such that as \( t \) changes, the curve traces out in three-dimensional space. This approach is particularly beneficial in modeling dynamics where time or another parameter plays a crucial role. By using parametric curves, you can easily visualize movements and paths in physics, computer graphics, and engineering designs as dynamic systems.

Vector Differentiation

Vector differentiation is a technique to analyze the changes in vector-valued functions. By taking the derivative of a vector function, one can deduce how the vector evolves as its input parameter varies. It's a process of finding the rate at which each vector component changes, which is similar to finding the slope but in multiple dimensions.
The process can be straightforward once each part of the vector is tackled individually. For instance, given \( \mathbf{r}(t) = \cos(2t) \mathbf{i} + 2 \sin(t) \mathbf{j} + t^2 \mathbf{k} \), each component is differentiate separately to form \( \mathbf{r}'(t) = -4\sin(2t) \mathbf{i} + 2\cos(t) \mathbf{j} + 2t \mathbf{k} \).

• Recognizing the similarity between this and derivative procedures in single-variable calculus simplifies the process.
• Vector differentiation generalizes these ideas into multiple dimensions.

Knowing how to differentiate vector functions becomes vital in fields like physics or engineering, providing comprehensive insights into forces, velocities, and other vector-based interactions across complex systems.