Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\left\langle t^{2}+1, t+1,1\right\rangle$$

Short answer

Answer: The second derivative of the vector function is \(\mathbf{r}^{\prime\prime}(t) = \left\langle 2, 0, 0 \right\rangle\) and the third derivative is \(\mathbf{r}^{\prime\prime\prime}(t) = \left\langle 0, 0, 0 \right\rangle\).

Step-by-step solution

Compute the first derivative of \(\mathbf{r}(t)\)#

Let's compute the first derivative of the function by finding the derivative of each component with respect to \(t\): $$\mathbf{r}^{\prime}(t) = \left\langle\frac{d}{dt}(t^{2}+1), \frac{d}{dt}(t+1), \frac{d}{dt}(1)\right\rangle$$ Using the basic differentiation rules, we get: $$\mathbf{r}^{\prime}(t) = \left\langle 2t, 1, 0 \right\rangle$$

Compute the second derivative of \(\mathbf{r}(t)\)#

Now let's find the second derivative by differentiating each component of the first derivative with respect to \(t\): $$\mathbf{r}^{\prime\prime}(t) = \left\langle\frac{d}{dt}(2t), \frac{d}{dt}(1), \frac{d}{dt}(0)\right\rangle$$ Using the basic rules of differentiation, we obtain: $$\mathbf{r}^{\prime\prime}(t) = \left\langle 2, 0, 0 \right\rangle$$

Compute the third derivative of \(\mathbf{r}(t)\)#

Finally, let's compute the third derivative by differentiating each component of the second derivative with respect to \(t\): $$\mathbf{r}^{\prime\prime\prime}(t) = \left\langle\frac{d}{dt}(2), \frac{d}{dt}(0), \frac{d}{dt}(0)\right\rangle$$ Again, applying the basic rules of differentiation, we get: $$\mathbf{r}^{\prime\prime\prime}(t) = \left\langle 0, 0, 0 \right\rangle$$ So, the second derivative of the vector function \(\mathbf{r}(t)\) is \(\mathbf{r}^{\prime\prime}(t) = \left\langle 2, 0, 0 \right\rangle\), and the third derivative is \(\mathbf{r}^{\prime\prime\prime}(t) = \left\langle 0, 0, 0 \right\rangle\).

Key concepts

Derivatives of Vector Functions

Derivatives of vector functions help us understand how vector quantities change as their input variables change. A vector function, such as \( \mathbf{r}(t) = \langle t^2 + 1, t + 1, 1 \rangle \), has multiple components that are each functions of a variable—in this case, \( t \). Differentiating vector functions involves taking the derivative of each component separately. This means if you have a vector \( \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle \), you find its derivative by calculating \( \mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle \). For the given exercise, we find the first derivative by differentiating each component of \( \mathbf{r}(t) \) with respect to \( t \), which gives us \( \mathbf{r}'(t) = \langle 2t, 1, 0 \rangle \).
This process of component-wise differentiation allows us to explore how each dimension of the vector changes independently, providing insight into the behavior of the vector over the parameter \( t \).
Understanding the derivatives of vector functions is essential in physics and engineering, where these functions often represent trajectories or velocities.

Differentiation Rules

Differentiation rules are the fundamental principles guiding how derivatives are calculated. The basic rules used when differentiating vector functions are the same as those used for scalar functions. Here are a few key rules:

• **Power Rule**: \( \dfrac{d}{dt}[t^n] = nt^{n-1} \)
• **Constant Rule**: \( \dfrac{d}{dt}[c] = 0 \), where \( c \) is a constant.
• **Sum Rule**: \( \dfrac{d}{dt}[f(t) + g(t)] = f'(t) + g'(t) \)

The power rule and the constant rule were used in this exercise. For instance, when differentiating \( t^2 + 1 \), you apply the power rule to \( t^2 \) to get \( 2t \) and the constant rule to \( 1 \) to get \( 0 \). The final derivative is \( 2t \), illustrating how these rules simplify complex calculations. By applying these simple differentiation rules, we can compute the derivatives needed to analyze the behavior of vector functions. This approach is efficient and practical for both straightforward and intricate functions.

Higher-Order Derivatives

Higher-order derivatives are derivatives of a derivative. In vector calculus, we often look for the first, second, and sometimes higher derivatives to understand the behavior of motion or change. The first derivative \( \mathbf{r}'(t) \) gives the velocity of the vector. The second derivative \( \mathbf{r}''(t) \) represents acceleration, which shows how velocity is changing over time. Finally, if we go a step further, the third derivative \( \mathbf{r}'''(t) \) can represent the rate of change of acceleration, often used in advanced dynamic analysis.
For the function \( \mathbf{r}(t) = \langle t^2 + 1, t + 1, 1 \rangle \), the second derivative was calculated as \( \mathbf{r}''(t) = \langle 2, 0, 0 \rangle \). This means that the acceleration in vector's direction is constant for the first component and zero for the others. Then, the third derivative, \( \mathbf{r}'''(t) = \langle 0, 0, 0 \rangle \), suggests no change in acceleration.
These higher-order derivatives are crucial for understanding complex motion and are indispensable tools in fields like mechanics, robotics, and any area involving dynamic systems.