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Magazine Chapter 12: Problem 7
Determine the domain and the component functions of the given function. $$ \begin{aligned} &2 \mathbf{F}-3 \mathbf{G}, \text { where } \mathbf{F}(t)=t \mathbf{i}+t^{2} \mathbf{j}+t^{3} \mathbf{k} \text { and } \mathbf{G}(t)=\cos t \mathbf{i}+ \\ &\sin t \mathbf{j}+\mathbf{k} \end{aligned} $$
Short Answer
Expert verified
Domain: \( \mathbb{R} \). Components: \( 2t - 3\cos t \), \( 2t^2 - 3\sin t \), \( 2t^3 - 3 \).
Step by step solution
01
Understand the Given Functions
The given functions are vector functions, specifically \( \mathbf{F}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \) and \( \mathbf{G}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + \mathbf{k} \). The goal is to find \( 2\mathbf{F} - 3\mathbf{G} \).
02
Identify the Domain of the Functions
The domain of a function consists of all values of \( t \) for which the function is defined. For \( \mathbf{F}(t) \), there are no restrictions on \( t \), so its domain is all real numbers, \( \mathbb{R} \). For \( \mathbf{G}(t) \), since sine and cosine are defined for all \( t \), its domain is also \( \mathbb{R} \). Thus, the domain of \( 2\mathbf{F} - 3\mathbf{G} \) is the intersection, which is also \( \mathbb{R} \).
03
Compute \( 2\mathbf{F} \)
Multiply each component of \( \mathbf{F}(t) \) by 2: \( 2\mathbf{F}(t) = 2t \mathbf{i} + 2t^2 \mathbf{j} + 2t^3 \mathbf{k} \).
04
Compute \( 3\mathbf{G} \)
Multiply each component of \( \mathbf{G}(t) \) by 3: \( 3\mathbf{G}(t) = 3\cos t \mathbf{i} + 3\sin t \mathbf{j} + 3\mathbf{k} \).
05
Subtract \( 3\mathbf{G} \) from \( 2\mathbf{F} \)
Combine the results from Step 3 and Step 4: \[(2t \mathbf{i} + 2t^2 \mathbf{j} + 2t^3 \mathbf{k}) - (3\cos t \mathbf{i} + 3\sin t \mathbf{j} + 3\mathbf{k}) = (2t - 3\cos t) \mathbf{i} + (2t^2 - 3\sin t) \mathbf{j} + (2t^3 - 3) \mathbf{k}.\]
06
Identify the Component Functions
The component functions of the result are: \( (2t - 3\cos t) \) for \( \mathbf{i} \), \( (2t^2 - 3\sin t) \) for \( \mathbf{j} \), and \( (2t^3 - 3) \) for \( \mathbf{k} \). These functions determine each component of the vector function \( 2\mathbf{F} - 3\mathbf{G} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Functions
Vector functions are a fascinating part of vector calculus. Unlike scalar functions, which output a single value, vector functions provide a vector as an output. This means that for every input, typically represented by a parameter like \( t \), the result is a vector with several components.
For instance, consider \( \mathbf{F}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \). Here, \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors in the x, y, and z directions, respectively.
For instance, consider \( \mathbf{F}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \). Here, \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors in the x, y, and z directions, respectively.
- \( t \mathbf{i} \) - This is the x-component, varying linearly with \( t \).
- \( t^2 \mathbf{j} \) - The y-component, having a quadratic relationship with \( t \).
- \( t^3 \mathbf{k} \) - The z-component, which is cubic in \( t \).
Function Domain
The domain of a vector function is crucial since it tells us where the function is defined. For a function to be valid, all its component functions must be defined.
In our example, both \( \mathbf{F}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \) and \( \mathbf{G}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + \mathbf{k} \) have component functions that do not impose any restrictions.
In our example, both \( \mathbf{F}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \) and \( \mathbf{G}(t) = \cos t \mathbf{i} + \sin t \mathbf{j} + \mathbf{k} \) have component functions that do not impose any restrictions.
- The polynomial terms \( t, t^2, \) and \( t^3 \) are always defined for real numbers.
- The sine and cosine functions have complete periodicity and are defined for all real \( t \).
Component Functions
Understanding component functions is essential for dissecting a vector function. Each component of a vector function corresponds to a scalar function.
In the example \( 2\mathbf{F} - 3\mathbf{G} \), each vector has three components, typically aligned with the standard axes: x, y, and z. For \( 2\mathbf{F} - 3\mathbf{G} \), these components become:
In the example \( 2\mathbf{F} - 3\mathbf{G} \), each vector has three components, typically aligned with the standard axes: x, y, and z. For \( 2\mathbf{F} - 3\mathbf{G} \), these components become:
- \( (2t - 3\cos t) \) - The x-component.
- \( (2t^2 - 3\sin t) \) - The y-component.
- \( (2t^3 - 3) \) - The z-component.
Vector Subtraction
Vector subtraction is a fundamental operation in vector calculus, involving the subtraction of corresponding components of two vectors.
For the operation \( 2\mathbf{F} - 3\mathbf{G} \), this means:
This process is analogous to regular arithmetic subtraction, but it extends into multiple dimensions, showing how complex transformations can be easily managed visually and algebraically.
For the operation \( 2\mathbf{F} - 3\mathbf{G} \), this means:
- Take the x-component of \( 2\mathbf{F} \) and subtract the x-component of \( 3\mathbf{G} \): \( 2t - 3\cos t \).
- Subtract the y-component of \( 3\mathbf{G} \) from the y-component of \( 2\mathbf{F} \): \( 2t^2 - 3\sin t \).
- Subtract the z-component of \( 3\mathbf{G} \) from the z-component of \( 2\mathbf{F} \): \( 2t^3 - 3 \).
This process is analogous to regular arithmetic subtraction, but it extends into multiple dimensions, showing how complex transformations can be easily managed visually and algebraically.
