Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\left\langle 3 t^{12}-t^{2}, t^{8}+t^{3}, t^{-4}-2\right\rangle$$

Short answer

Question: Determine the second and third derivatives of the vector-valued function \(\mathbf{r}(t)=\left\langle 3t^{12}-t^2, t^8+t^3, t^{-4}-2 \right\rangle\). Answer: The second derivative is \(\mathbf{r}''(t)=\left\langle 396t^{10}-2, 56t^6+6t, 20t^{-6}\right\rangle\), and the third derivative is \(\mathbf{r}'''(t)=\left\langle 3960t^9, 336t^5+6, -120t^{-7}\right\rangle\).

Step-by-step solution

Compute the first derivative \(\mathbf{r}'(t)\)

Computing the first derivative \(\mathbf{r}'(t)\) by taking the derivatives of each component with respect to t: $$ \mathbf{r}'(t)=\left\langle \frac{d(3t^{12}-t^2)}{dt}, \frac{d(t^8+t^3)}{dt}, \frac{d(t^{-4}-2)}{dt} \right\rangle $$ Applying standard derivative rules, we get: $$ \mathbf{r}'(t)=\left\langle 36t^{11}-2t, 8t^7+3t^2, -4t^{-5}\right\rangle $$

Compute the second derivative \(\mathbf{r}''(t)\)

Computing the second derivative \(\mathbf{r}''(t)\) by taking the derivatives of each component of \(\mathbf{r}'(t)\) with respect to t: $$ \mathbf{r}''(t)=\left\langle \frac{d(36t^{11}-2t)}{dt}, \frac{d(8t^7+3t^2)}{dt}, \frac{d(-4t^{-5})}{dt} \right\rangle $$ Applying standard derivative rules, we get: $$ \mathbf{r}''(t)=\left\langle 396t^{10}-2, 56t^6+6t, 20t^{-6}\right\rangle $$

Compute the third derivative \(\mathbf{r}'''(t)\)

Computing the third derivative \(\mathbf{r}'''(t)\) by taking the derivatives of each component of \(\mathbf{r}''(t)\) with respect to t: $$ \mathbf{r}'''(t)=\left\langle \frac{d(396t^{10}-2)}{dt}, \frac{d(56t^6+6t)}{dt}, \frac{d(20t^{-6})}{dt} \right\rangle $$ Applying standard derivative rules, we get: $$ \mathbf{r}'''(t)=\left\langle 3960t^9, 336t^5+6, -120t^{-7}\right\rangle $$

Key concepts

Vector Functions

Vector functions are mathematical expressions that associate a vector with each input value. They are often written in terms of a parameter, like \( t \), and consist of a set of components that are functions of this parameter. For example, our given function is \( \mathbf{r}(t)=\langle 3 t^{12}-t^{2}, t^{8}+t^{3}, t^{-4}-2 \rangle \).

Each of these individual expressions within the brackets is a component of the vector function, and they describe how the vector changes with varying \( t \). This is useful in fields like physics, where vector functions can represent things like position, velocity, or acceleration over time.

To analyze a vector function, we often need to determine its derivatives. This helps us understand the behavior of the vector in terms of rates of change. As a result, vector functions are central in dynamics and fields that involve motion.

Calculus

Calculus is a branch of mathematics that deals with rates of change and the accumulation of quantities. It underpins much of modern science and engineering.

In our exercise, calculus allows us to find higher-order derivatives of a vector function. Higher-order derivatives refer to the derivatives of a derivative. For instance, starting from a position vector, the first derivative gives us velocity, the second provides acceleration, and further derivatives can be examined depending on the context.

• First Derivative: This gives the rate of change of the function, providing insight into its slope or speed.
• Second Derivative: This is the rate of change of the rate of change, often giving acceleration or curvature.
• Third Derivative: This further shows how the second derivative evolves and can be linked to concepts like jerk in physics.

By understanding calculus, students can apply mathematical rigor to analyze and predict the behavior of vector functions over time.

Derivative Rules

The process of finding derivatives, especially for vector functions, relies heavily on derivative rules from calculus. These rules guide us in finding how each component of a vector function changes with respect to a parameter.

Some key derivative rules include the power rule, product rule, and chain rule:

• Power Rule: Allows differentiation of functions of the form \( ax^n \), leading to \( nax^{n-1} \).
• Product Rule: Useful for functions that are products of two expressions; it states \( (uv)' = u'v + uv' \).
• Chain Rule: Essential for composite functions; it simplifies the differentiation process by breaking it into parts.

In the exercise, these rules are applied to each component of the vector function to find the first, second, and third derivatives. For instance, using the power rule on \( 3t^{12} \) in the vector's first component results in \( 36t^{11} \), simplifying the process of derivative calculation.

By mastering these rules, students can efficiently differentiate complex vector functions, a fundamental skill in calculus and its applications.