Question

Evaluate the given limit. $$ \lim _{h \rightarrow 0} \frac{\vec{r}(t+h)-\vec{r}(t)}{h}, \text { where } \vec{r}(t)=\left\langle t^{2}, t, 1\right\rangle $$

Short answer

The limit evaluates to \( \langle 2t, 1, 0 \rangle \).

Step-by-step solution

Substitute the Expression for \( \vec{r}(t) \)

The vector function is given by \( \vec{r}(t) = \langle t^2, t, 1 \rangle \). Plug this into the limit expression: \[ \lim _{h \rightarrow 0} \frac{\vec{r}(t+h)-\vec{r}(t)}{h} = \lim _{h \rightarrow 0} \frac{\langle (t+h)^2, (t+h), 1 \rangle - \langle t^2, t, 1 \rangle}{h}. \]

Calculate the Difference \( \vec{r}(t+h) - \vec{r}(t) \)

Find \( \vec{r}(t+h) = \langle (t+h)^2, t+h, 1 \rangle \) and compute the difference: \[ (t+h)^2 - t^2 = t^2 + 2th + h^2 - t^2 = 2th + h^2, \]\[ t+h-t = h, \]Subtract the vectors: \[ \vec{r}(t+h) - \vec{r}(t) = \langle 2th + h^2, h, 0 \rangle. \]

Form the Limit Expression

Substitute the difference from Step 2 back into the limit:\[ \lim _{h \rightarrow 0} \frac{\langle 2th + h^2, h, 0 \rangle}{h} = \lim _{h \rightarrow 0} \langle \frac{2th + h^2}{h}, \frac{h}{h}, \frac{0}{h} \rangle. \]

Simplify Each Component

Divide each component of the vector by \( h \): 1. \[ \frac{2th + h^2}{h} = 2t + h, \]2. \[ \frac{h}{h} = 1, \]3. \[ \frac{0}{h} = 0. \]Now, the expression simplifies to: \[ \lim _{h \rightarrow 0} \langle 2t + h, 1, 0 \rangle. \]

Evaluate the Limit

As \( h \rightarrow 0 \), substitute \( h = 0 \) into the simplified expression: \[ \langle 2t + 0, 1, 0 \rangle = \langle 2t, 1, 0 \rangle. \] Thus, the evaluated limit is \( \langle 2t, 1, 0 \rangle \).

Key concepts

Limit Evaluation

Evaluating limits is a fundamental concept, particularly when dealing with vector functions. It involves finding the value that a function approaches as the input reaches a certain point. In the context of vector functions, we examine the behavior of each component in the vector as we approach the limit. Think of it like zooming in on a path, trying to determine the exact direction it's heading as you get closer to a specific point.

For the expression \( \lim _{h \rightarrow 0} \frac{\vec{r}(t+h)-\vec{r}(t)}{h} \), this means analyzing the behavior as \( h \) gets closer to 0. It’s akin to finding the derivative of the function, which describes the instantaneous rate of change. The goal is to simplify the expression by dividing each term by \( h \), and then observing what happens as \( h \, \rightarrow \, 0 \).

• First, plug the function into the expression.
• Next, expand and simplify each component separately.
• Finally, simplify further to reveal the resulting vector as \( h \) approaches 0.

This process helps in determining the derivative, indicating both magnitude and direction of change.

Derivative of a Vector Function

The derivative of a vector function gives insights into how the function changes at each point in the space it inhabits. It is essentially a vector itself, specifying the rate and direction of change from point to point along the path of the original vector function. This is crucial in fields like physics and engineering, where motion or forces can be described in terms of vector calculus.

The vector function \( \vec{r}(t) = \langle t^2, t, 1 \rangle \) is a simple example. The derivative represents how each component of the vector changes with respect to \( t \). To find this derivative, we follow a structured approach:

• First, evaluate the change in each component when a small increment \( h \) is added to \( t \), and divide by \( h \).
• Next, simplify each term separately.
• As \( h \rightarrow 0 \), observe how these terms tend towards a finite value.

This results in the derivative vector \( \langle 2t, 1, 0 \rangle \), which describes how the function \( \vec{r}(t) \) evolves as \( t \) changes, providing a precise and instant snapshot of movement or force.

Vector Function

Vector functions are a way to describe mathematical objects that have both direction and magnitude. They typically represent a position along a path described in three-dimensional space. For instance, the vector function \( \vec{r}(t) = \langle t^2, t, 1 \rangle \) maps the input \( t \) to a point in 3D space. This vector expression represents a path, tracing the movement or change in a system over a domain.

• Each component in the vector function, in this case \( t^2 \), \( t \), and \( 1 \), represents a dimension in space.
• The input \( t \) can be understood as time or any parameter tracing the position along this path.
• Understanding vector functions is essential because they offer a more complete picture of movement and change compared to scalar functions.

When working with vector calculus, it’s important to understand both the individual components and the overall vector picture. These functions can describe everything from an orbiting satellite's path to the force acting on a bridge.