Question

Consider the vector-valued function \(\mathbf{r}(t)=\left(e^{t} \sin t\right) \mathbf{i}+\left(e^{t} \cos t\right) \mathbf{j}\). Show that \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime \prime}(t)\) are always perpendicular to each other.

Short answer

The dot product of \( \mathbf{r}(t) \) and \( \mathbf{r}^{\prime\prime}(t) \) is zero which implies that these two vectors are always perpendicular

Step-by-step solution

Calculate the first derivative

Firstly, compute the first derivative of \( \mathbf{r}(t) \): \(\mathbf{r}^{\prime}(t) = \frac{d}{dt}\mathbf{r}(t)\) which further simplifies to \(\mathbf{r}^{\prime}(t) = \left(e^{t} \cos t + e^{t} \sin t \right) \mathbf{i} + \left( e^{t} \cos t - e^{t} \sin t \right) \mathbf{j}\)

Compute the second derivative

Then, follow the same process to obtain the second derivative of \(\mathbf{r}(t)\): \(\mathbf{r}^{\prime\prime}(t) = \frac{d^2}{dt^2}\mathbf{r}(t)\) which results in \(\mathbf{r}^{\prime\prime}(t) = \left(2e^{t} \cos t \right) \mathbf{i} - \left(2e^{t} \sin t \right) \mathbf{j}\)

compute the dot product

Compute the dot product of the vectors \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime\prime}(t)\). The dot product is given by \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime\prime}(t) = \left(e^{t} \sin t\right)*\left(2e^{t} \cos t\right) + \left(e^{t} \cos t\right)* \left(-2e^{t} \sin t\right)\)

Simplify the dot product

Simplify the expression: \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime\prime}(t) = 2e^{2t} \sin t \cos t - 2e^{2t} \sin t \cos t = 0 \)

Key concepts

Vector Calculus

The subject of vector calculus is integral to understanding fields within physics and engineering, such as electromagnetism and fluid dynamics. It's a branch of mathematics that deals with vector fields and operations that can be performed on these fields, including differentiation and integration. The vectors we discuss in vector calculus are not just numbers, but quantities that have both magnitude and direction. For instance, the vector-valued function \( \mathbf{r}(t)=\left(e^{t} \sin t\right) \mathbf{i}+\left(e^{t} \cos t\right)\mathbf{j} \) represents a vector whose tail is at the origin of a coordinate system and whose head is located at the point given by the output of the function at time \(t\) .

This function gives a vector for each value of \(t\) , producing a curve in the 2D space spanned by \(\mathbf{i}\) and \(\mathbf{j}\) , which represent the unit vectors in the horizontal and vertical directions, respectively. Understanding such vector-valued functions and their derivatives, which represent how the function changes at any point in time, is a key aspect of vector calculus.

Dot Product

The dot product, or scalar product, is a fundamental operation in vector calculus, as it measures the extent to which two vectors in the same space point in the same direction. It combines two vectors to produce a scalar. The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) , denoted as \(\mathbf{a} \cdot \mathbf{b}\) , is calculated as \(a_x b_x + a_y b_y + a_z b_z\) , where \(a_x, a_y, a_z\) and \(b_x, b_y, b_z\) are the components of \(\mathbf{a}\) and \(\mathbf{b}\) respectively. Importantly, if the dot product of two nonzero vectors is zero, this indicates that they are perpendicular (orthogonal) to each other.

In the exercise provided, to prove that the vectors \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime \prime}(t)\) are always perpendicular, we compute their dot product and find it to be zero. This is a powerful application of the dot product in vector calculus; it allows us to understand geometric relationships between different functions and their derivatives.

Derivatives

Derivatives play a critical role in calculus, representing rates of change. In a physical context, if a vector-valued function like \(\mathbf{r}(t)\) represents the position of a particle over time, then its derivative \(\mathbf{r}^{\prime}(t)\) , gives us the particle’s velocity — how fast and in what direction it's moving at any point in time. Similarly, the second derivative, \(\mathbf{r}^{\prime \prime}(t)\) , as seen in the exercise, corresponds to the particle’s acceleration.

Taking the derivative of a vector-valued function involves differentiating each component of the vector with respect to \(t\) . For example, the first derivative in the exercise is obtained by differentiating both components with the product and chain rules of differentiation. This process can be repeated to find higher-order derivatives, such as the second derivative \(\mathbf{r}^{\prime \prime}(t)\) , which reveals additional layers of how the function's behavior evolves over time.