Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) when \(\mathbf{r}(t)=\left\langle t^{10}, 8 t, \cos t\right\rangle\)

Short answer

Answer: The second derivative of the vector function \(\mathbf{r}(t)\) is \(\mathbf{r}''(t) = \left\langle 90t^8, 0, -\cos t \right\rangle\).

Step-by-step solution

Find the first derivative of r(t)

To obtain the first derivative, we will take the derivative of each component separately. Derivative of t^10 is 10t^9, derivative of 8t is 8, and derivative of cos(t) is -sin(t). So, the first derivative is: \(\mathbf{r}'(t) =\left\langle 10t^9, 8, -\sin t \right\rangle\)

Find the second derivative of r(t)

Now, we will take the derivative of each component in \(\mathbf{r}'(t)\) to compute the second derivative. The derivative of \(10t^9\) is \(90t^8\), the derivative of 8 is 0, and the derivative of \(-\sin(t)\) is \(-cos(t)\). Therefore, the second derivative is: \(\mathbf{r}''(t) = \left\langle 90t^8, 0, -\cos t \right\rangle\)

Key concepts

Vector Calculus

Vector calculus is an extension of ordinary calculus to spaces where the variables are vectors rather than scalars. It forms the basis of the mathematical field of vector analysis and has significant applications in physics and engineering. In this context, functions are defined by multiple variables and outputs, each with its own dimension, often representing distinct physical quantities. For example, as seen in the exercise, a vector-valued function like \( \mathbf{r}(t) \) with components that depend on the parameter \( t \) can describe a curve in three-dimensional space.

Key concepts in vector calculus include operations like differentiation and integration applied to vector fields, along with specialized tools such as gradient, divergence, and curl. Understanding how these operations work on vector-valued functions is crucial to analyzing many physical phenomena, such as fluid flow or electromagnetic fields.

Derivatives of Vector Functions

The derivative of a vector function is a fundamental concept in calculus. It measures how the function changes with respect to a change in an independent variable, such as time. For a vector-valued function like \( \mathbf{r}(t) = \left\langle t^{10}, 8t, \cos t\right\rangle \), the derivative is taken component-wise. This approach means applying traditional differentiation rules to each component of the vector as if it were an individual scalar function.

As exhibited in the exercise, the first derivative of the vector function \( \mathbf{r}(t) \) is given by \( \mathbf{r}'(t) = \left\langle 10t^9, 8, -\sin t \right\rangle \). This differential operation provides us with another vector-valued function that represents the velocity of an object moving along the path defined by \( \mathbf{r}(t) \) if we interpret \( t \) as time.

Second Derivative

The second derivative of a vector function gives deeper insight into the behavior of the function. For motion, it represents the acceleration of an object. To compute the second derivative, you simply differentiate the components of the first derivative once again. In the given exercise, after finding \( \mathbf{r}'(t) \), the second derivative \( \mathbf{r}''(t) = \left\langle 90t^8, 0, -\cos t \right\rangle \) is determined.

An important aspect to remember is that while a zero scalar second derivative suggests linear motion, a zero component in a vector second derivative implies that there is no acceleration in that particular dimension. The second derivative can help identify points of inflection, curvature, and even help in optimizing parameters for certain processes in multidimensional settings.

Calculus

Calculus is a branch of mathematics focused on rates of change (differential calculus) and accumulation of quantities (integral calculus). It's a powerful tool that allows for the modeling and analysis of dynamic systems. In essence, calculus extends the algebraic processes to operate on non-linear and dynamic systems which can change infinitesimally.

The exercise provided is an application of differential calculus on a vector-valued function, one of the many ways calculus can be applied. Mastering the calculations of derivatives and integrals is not only necessary for solving mathematical problems but essential for understanding and describing physical systems in natural sciences and engineering. With each derivative taken, we get closer to understanding the instantaneous rates of change and patterns of variation in complex systems.