Question

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\tan t \mathbf{i}+\left(t+\frac{1}{t}\right) \mathbf{j}-\ln (t+1) \mathbf{k}$$

Short answer

The second derivative, \(\mathbf{r}''(t)\), is: $$\mathbf{r}''(t) = 2\sec^2 t \tan t \mathbf{i} + \frac{2}{t^3} \mathbf{j} + \frac{1}{(t+1)^2} \mathbf{k}$$ The third derivative, \(\mathbf{r}'''(t)\), is: $$\mathbf{r}'''(t) = \left(4\sec^4 t + 4\sec^2 t \tan^3 t\right) \mathbf{i} - \frac{6}{t^4} \mathbf{j} - \frac{2}{(t+1)^3} \mathbf{k}$$

Step-by-step solution

Find the first derivative of each component function

Differentiate \(\tan t\), \(t + \frac{1}{t}\), and \(-\ln(t+1)\) with respect to \(t\). First derivative of \(\tan t\): $$\frac{d}{dt} (\tan t) = \sec^2 t$$ First derivative of \(t + \frac{1}{t}\): $$\frac{d}{dt} \left(t + \frac{1}{t}\right) = 1 - \frac{1}{t^2}$$ First derivative of \(-\ln(t+1)\): $$\frac{d}{dt} (-\ln(t+1)) = -\frac{1}{t+1}$$

Write the first derivative vector function

Now that we have the first derivatives of each component function, we can combine them to form the first derivative of the vector function, \(\mathbf{r}'(t)\): $$\mathbf{r}'(t) = \sec^2 t \mathbf{i} + \left(1 - \frac{1}{t^2}\right) \mathbf{j} - \frac{1}{t+1} \mathbf{k}$$

Find the second derivative of each component function

Differentiate the first derivative of each component function with respect to \(t\). Second derivative of \(\tan t\): $$\frac{d^2}{dt^2} (\tan t) = \frac{d}{dt} (\sec^2 t) = 2\sec^2 t \tan t$$ Second derivative of \(t + \frac{1}{t}\): $$\frac{d^2}{dt^2} \left(t + \frac{1}{t}\right) = \frac{d}{dt} \left(1 - \frac{1}{t^2}\right) = \frac{2}{t^3}$$ Second derivative of \(-\ln(t+1)\): $$\frac{d^2}{dt^2} (-\ln(t+1)) = \frac{d}{dt} \left(-\frac{1}{t+1}\right) = \frac{1}{(t+1)^2}$$

Write the second derivative vector function

Now that we have the second derivatives of each component function, we can combine them to form the second derivative of the vector function, \(\mathbf{r}''(t)\): $$\mathbf{r}''(t) = 2\sec^2 t \tan t \mathbf{i} + \frac{2}{t^3} \mathbf{j} + \frac{1}{(t+1)^2} \mathbf{k}$$

Find the third derivative of each component function

Differentiate the second derivative of each component function with respect to \(t\). Third derivative of \(\tan t\): $$\frac{d^3}{dt^3} (\tan t) = \frac{d}{dt} \left(2\sec^2 t \tan t\right) = 4\sec^4 t + 4\sec^2 t \tan^3 t$$ Third derivative of \(t + \frac{1}{t}\): $$\frac{d^3}{dt^3} \left(t + \frac{1}{t}\right) = \frac{d}{dt} \left(\frac{2}{t^3}\right) = -\frac{6}{t^4}$$ Third derivative of \(-\ln(t+1)\): $$\frac{d^3}{dt^3} (-\ln(t+1)) = \frac{d}{dt} \left(\frac{1}{(t+1)^2}\right) = -\frac{2}{(t+1)^3}$$

Write the third derivative vector function

Now that we have the third derivatives of each component function, we can combine them to form the third derivative of the vector function, \(\mathbf{r}'''(t)\): $$\mathbf{r}'''(t) = \left(4\sec^4 t + 4\sec^2 t \tan^3 t\right) \mathbf{i} - \frac{6}{t^4} \mathbf{j} - \frac{2}{(t+1)^3} \mathbf{k}$$

Key concepts

Derivative of Vector Functions

A vector function is made up of multiple component functions, each associated with a basis vector, such as \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). This means if you want to find the derivative of a vector function, you must differentiate each of its component functions separately with respect to the variable, usually \( t \).
The derivative of a vector function \( \mathbf{r}(t) \) is written as \( \mathbf{r}'(t) \), and it represents the instantaneous rate of change of the vector function. You handle differentiation of vector functions by treating them like a set of coordinates:

• Differentiate \( \tan t \) with respect to \( t \), resulting in \( \sec^2 t \).
• Differentiate \( t + \frac{1}{t} \), giving \( 1 - \frac{1}{t^2} \).
• Differentiate \( -\ln(t+1) \), resulting in \( -\frac{1}{t+1} \).

After finding these individual derivatives, you construct the vector derivative \( \mathbf{r}'(t) \) by combining these results: \[ \mathbf{r}'(t) = \sec^2 t \mathbf{i} + \left(1 - \frac{1}{t^2}\right) \mathbf{j} - \frac{1}{t+1} \mathbf{k} \] This process might seem like standard calculus, but it's important to remember that the operations must be performed on each component function individually.

Second Derivative

The second derivative of a vector function is the derivative of its first derivative. Essentially, you're determining how the rate at which the vector function is changing is itself changing. To find the second derivative of a vector function \( \mathbf{r}(t) \), we differentiate \( \mathbf{r}'(t) \). For the exercise provided, we compute:

• The second derivative of \( \sec^2 t \) is \( 2\sec^2 t \tan t \).
• The second derivative of \( 1 - \frac{1}{t^2} \) gives \( \frac{2}{t^3} \).
• The second derivative of \( -\frac{1}{t+1} \) results in \( \frac{1}{(t+1)^2} \).

By combining these results, we get the second derivative vector function: \[ \mathbf{r}''(t) = 2\sec^2 t \tan t \mathbf{i} + \frac{2}{t^3} \mathbf{j} + \frac{1}{(t+1)^2} \mathbf{k} \] Understanding the second derivative is important because it offers insights into the concavity and acceleration of the vector path.

Third Derivative

The third derivative of a vector function extends the idea further by taking the derivative of the second derivative. This tells us how the rate of acceleration is changing over time, sometimes related to jerk in physics if the vector function represents a spatial path. For the function in this exercise, we calculate:

• The third derivative of \( 2\sec^2 t \tan t \) is found by differentiating again, resulting in \( 4\sec^4 t + 4\sec^2 t \tan^3 t \).
• The third derivative of \( \frac{2}{t^3} \) is \( -\frac{6}{t^4} \).
• The third derivative of \( \frac{1}{(t+1)^2} \) yields \( -\frac{2}{(t+1)^3} \).

These derivatives combine to give the third derivative vector function: \[ \mathbf{r}'''(t) = \left(4\sec^4 t + 4\sec^2 t \tan^3 t\right) \mathbf{i} - \frac{6}{t^4} \mathbf{j} - \frac{2}{(t+1)^3} \mathbf{k} \] Considering the third derivative is useful in engineering and physics contexts where it helps predict the smoothness of transitions along a path.