Question

Find the domain of the vector-valued function. \(\mathbf{r}(t)=\mathbf{F}(t)-\mathbf{G}(t)\) where \(\mathbf{F}(t)=\ln t \mathbf{i}+5 t \mathbf{j}-3 t^{2} \mathbf{k}, \quad \mathbf{G}(t)=\mathbf{i}+4 t \mathbf{j}-3 t^{2} \mathbf{k}\)

Short answer

The domain of the function \(\mathbf{r}(t)\) is \(t > 0\).

Step-by-step solution

Understanding the given vector-valued functions

First, we need to understand the vector-valued functions given. We have \(\mathbf{F}(t)=\ln t \mathbf{i}+5 t \mathbf{j}-3 t^{2} \mathbf{k}\) and \(\mathbf{G}(t)=\mathbf{i}+4 t \mathbf{j}-3 t^{2} \mathbf{k}\). The function \(\mathbf{F}(t)\) is defined for \(t > 0\) because of the log function. The function \(\mathbf{G}(t)\) is defined for all \(t\) since there is no value of \(t\) for which it would be undefined.

Determine the resulting function \(\mathbf{r}(t) = \mathbf{F}(t) - \mathbf{G}(t)\)

We are asked to find the domain of the result of the subtraction of vector \(\mathbf{G}(t)\) from vector \(\mathbf{F}(t)\). Subtracting these two vectors will yield the new function: \[ \mathbf{r}(t) = (\ln(t) \mathbf{i} +5t \mathbf{j} -3t^2 \mathbf{k}) - (\mathbf{i} +4t \mathbf{j} -3t^2 \mathbf{k})\]To simplify the result, combine like terms: \[ \mathbf{r}(t) = (\ln(t)-1) \mathbf{i} +(5t-4t) \mathbf{j} \]

Determine the domain of the resulting function \(\mathbf{r}(t)\)

Because the new function \(\mathbf{r}(t)\) is a product of subtraction of the original functions, the domain of \(\mathbf{r}(t)\) is a set of all real values that are in the domains of both \(\mathbf{F}(t)\) and \(\mathbf{G}(t)\). Hence, the domain of \(\mathbf{r}(t)\) is the set of all \(t\) such that \(t > 0\), because \(\mathbf{F}(t)\) restricts \(\mathbf{r}(t)\)'s domain to greater than 0. We also have no upper limit on \(t\) in any of the components of \(\mathbf{r}(t)\).

Key concepts

Domain of a Function

When studying the domain of a function, we're analyzing the complete set of input values (often represented by 'x' or 't' in equations) that will yield valid output values without causing any contradictions such as division by zero or taking the logarithm of a negative number.

In our case, the vector-value function involves a logarithmic component in \( \mathbf{F}(t) = \ln t \mathbf{i} + 5t \mathbf{j} -3t^2 \mathbf{k} \), which highlights the importance of understanding restrictions imposed by logarithmic functions. Since logarithms are only defined for positive real numbers, \( \mathbf{F}(t) \) can only accept positive values for 't'. This narrows down the range of 't' values that we can use as input for our function, which ultimately defines the domain.

Vector Calculus

Vector calculus extends the traditional calculus to consider functions that yield vector results rather than scalar results. Vector functions, like \( \mathbf{F}(t) \) and \( \mathbf{G}(t) \) from our exercise, are characterized by having vector outputs, consisting of multiple scalar components, each potentially being a function of the variable 't'.

When subtracting vector functions as \( \mathbf{r}(t) = \mathbf{F}(t) - \mathbf{G}(t) \), the process involves subtracting corresponding components from each other. This operation is akin to dealing with multiple simultaneous scalar functions and keeping track of which inputs lead to well-defined outputs for all components. In essence, vector calculus combines the complexities of multivariable calculus with the intricacies of vector operations.

Logarithmic Functions

Logarithmic functions, such as \( \ln(t) \) that appears in \( \mathbf{F}(t) \) of the given exercise, have very particular properties. They are the inverse operations of exponentiation, meaning that they effectively 'undo' what exponents do. As such, their domains are strictly limited to positive real numbers.

The function \( \ln(t) \) is defined only for \( t > 0 \), since the natural logarithm of a non-positive number is not a real number. This has a direct impact on the domain of any vector-valued function, or indeed any function, that includes a logarithmic component. Understanding logarithmic functions and their domains is essential when trying to determine the domain of a more complex function that incorporates logarithmic elements.