Question

Find the derivative of the given function. $$ \vec{r}(t)=\left\langle\frac{1}{t}, \frac{2 t-1}{3 t+1}, \tan t\right\rangle $$

Short answer

\( \vec{r}'(t) = \left\langle -\frac{1}{t^2}, \frac{5}{(3t + 1)^2}, \sec^2 t \right\rangle \)

Step-by-step solution

Differentiate Each Component

To find the derivative of the vector function \( \vec{r}(t) \), differentiate each component function with respect to \( t \). The vector function given is \( \vec{r}(t)=\left\langle\frac{1}{t}, \frac{2t-1}{3t+1}, \tan t\right\rangle \).

Step 1.1: Differentiate First Component

The first component is \( \frac{1}{t} \). To differentiate \( \frac{1}{t} \), rewrite it as \( t^{-1} \). The derivative of \( t^{-1} \) with respect to \( t \) is \( -t^{-2} \), or \( -\frac{1}{t^2} \).

Step 1.2: Differentiate Second Component

The second component is \( \frac{2t - 1}{3t + 1} \). Use the quotient rule for derivatives: if \( f(t) = \frac{u(t)}{v(t)} \), then \( f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{(v(t))^2} \). Here, \( u(t) = 2t - 1 \) and \( v(t) = 3t + 1 \). Compute the derivatives: \( u'(t) = 2 \), \( v'(t) = 3 \), giving: \[ \frac{(2)(3t + 1) - (2t - 1)(3)}{(3t + 1)^2} = \frac{6t + 2 - (6t - 3)}{(3t + 1)^2} = \frac{5}{(3t + 1)^2} \].

Step 1.3: Differentiate Third Component

The third component is \( \tan t \). The derivative of \( \tan t \) is \( \sec^2 t \).

Write the Derivative Vector

Combine the derivatives of each individual component to form the derivative of the vector function. The derivative of the vector function \( \vec{r}(t) \) is: \[ \vec{r}'(t) = \left\langle -\frac{1}{t^2}, \frac{5}{(3t + 1)^2}, \sec^2 t \right\rangle \].

Key concepts

Derivative of Vector Functions

Vector functions are mathematical expressions that associate a vector to every point of a parameter, often time. Understanding how to derive these functions is crucial for understanding their behavior and properties. In vector calculus, much like differentiating scalar functions, we also differentiate vector functions component by component. For example, if we have a vector function \( \vec{r}(t) = \left\langle f(t), g(t), h(t) \right\rangle \), its derivative \( \vec{r}'(t) \) is found by differentiating each component function separately:

• The derivative of \( f(t) \) is \( f'(t) \)
• The derivative of \( g(t) \) is \( g'(t) \)
• The derivative of \( h(t) \) is \( h'(t) \)

Therefore, the derivative vector is \( \vec{r}'(t) = \left\langle f'(t), g'(t), h'(t) \right\rangle \). Differentiation of vector functions is widely used in physics and engineering, such as in describing the motion and forces on objects.

Differentiation Rules

To differentiate vector functions effectively, we often apply established differentiation rules: the product rule, chain rule, and quotient rule. These rules simplify the differentiation of more complex components:

• Product Rule: Used when differentiating products of functions. If \( u(t) \) and \( v(t) \) are functions, the derivative of their product is \( (uv)' = u'v + uv' \).


• Chain Rule: Used for composite functions. If \( h(t) = f(g(t)) \), then \( h'(t) = f'(g(t)) \, g'(t) \).


• Quotient Rule: Used for differentiating quotients of functions. Given that \( f(t) = \frac{u(t)}{v(t)} \), its derivative is \( f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{v(t)^2} \).

In the example of the exercise, the quotient rule is used to differentiate the second component of the vector function, while simple power rule derivatives are applied to the first and chain rule connects to the third component. Mastering these rules makes dealing with complex vector functions manageable and insightful.

Tangent Function Derivative

The tangent function, \( \tan t \), is a fundamental trigonometric function often encountered in calculus. Its behavior is important across mathematics and applied sciences. The derivative of the tangent function helps us understand how the slope of the tangent changes at each point along a curve, and is crucial for calculating rates of change in oscillatory and periodic functions. The derivative of \( \tan t \) is \( \sec^2 t \). This outcome derives from the trigonometric identity that links tangent and secant:\[ \frac{d}{dt} \tan t = \sec^2 t \]Understanding this derivative helps simplify more complex expressions involving trigonometric identities. It emerges in solving problems related to wave functions, oscillations, and the behavior of light and sound. Recognizing and applying the derivative of the tangent function perfectly complements the skills needed in calculus involving either trigonometric or vector functions.