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A vector function is a function that takes one or more variables and returns a vector. In the exercise, the vector function is denoted as \( \mathbf{r}(t) \), where \( t \) is the parameter. This function consists of component functions for each basis vector \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \):

• \( x(t) = \cos t^2 \), which corresponds to the \( \mathbf{i} \) component.
• \( y(t) = \sin t^2 \), which corresponds to the \( \mathbf{j} \) component.
• \( z(t) = e^{-t} \), which corresponds to the \( \mathbf{k} \) component.

Each component is a scalar function in terms of \( t \), but combined they represent a vector at each point in time \( t \). By understanding vector functions, we can analyze paths and motions in space. Vectors provide essential ways to represent directions and magnitudes in multi-dimensional spaces. It's important for calculus students to recognize how vector functions translate to real-world motions and changes.

Smoothness in calculus refers to how nice and 'well-behaved' a function is within the scope of calculus operations. For a vector function like \( \mathbf{r}(t) \), smoothness means that the function is continuously differentiable. This implies that:

• All the component functions have derivatives.
• These derivatives themselves are continuous functions.

To check for smoothness, we differentiate each component and confirm the continuity of these derivatives. In our exercise:

• \( x'(t) = -2t \sin t^2 \)
• \( y'(t) = 2t \cos t^2 \)
• \( z'(t) = -e^{-t} \)

All derivatives are continuous for all \( t \). Thus, we conclude that \( \mathbf{r}(t) \) is a smooth vector function. Smooth vector functions are significant in mathematical modeling and engineering since they ensure the outputs are feasible and physically possible paths or motions.

Differentiability of a function at a point implies that its derivative exists at that point. For vector functions like \( \mathbf{r}(t) \), it means each component can be differentiated:

• \( x(t) = \cos t^2 \) gives \( x'(t) = -2t \sin t^2 \)
• \( y(t) = \sin t^2 \) gives \( y'(t) = 2t \cos t^2 \)
• \( z(t) = e^{-t} \) gives \( z'(t) = -e^{-t} \)

Every component has a derivative which implies \( \mathbf{r}(t) \) is differentiable everywhere.Differentiability is crucial for calculus because it allows us to determine rates of change and analyze how dynamic systems behave over time. It's a cornerstone for concepts like velocity and acceleration in physics, where examining changes in vector quantities is necessary.

Continuity of a function means there are no abrupt changes or jumps in the values of the function. A function is continuous if for a small change in input, the output changes only slightly. For vector functions, each component must be continuous to ensure overall function continuity.In our vector function \( \mathbf{r}(t) \), each component is in terms of well-known continuous functions (cosine, sine, and exponential function). Specifically:

• \( \cos t^2 \) and \( \sin t^2 \) are continuous for all \( t \).
• \( e^{-t} \) is continuous for all \( t \).

Thus, the function \( \mathbf{r}(t) \) is continuous.Continuity is fundamental to calculus because it ensures that a function behaves predictably, without sudden jumps. Continuous functions are integral in mathematical modeling, ensuring that solutions to equations behave similarly to real-world phenomena.