91Ó°ÊÓ

In mathematics, a multivariable function is essentially a function that depends on more than one variable. These types of functions are often represented as \( f(x, y, z, \ldots) \), where each variable can vary independently. In our exercise, the function given is \( m(t, s) = s \ln t + 3.75s + 14.96 \), which depends on two variables: \( t \) and \( s \). This means that the function's value changes as either \( t \) or \( s \) changes. Such functions are crucial in modeling real-world scenarios where multiple factors affect the outcome.

Multivariable functions are a natural extension of single-variable functions and allow us to analyze systems with more complexity. They are used across a broad range of fields including economics, engineering, and physics. Understanding how to manipulate these functions, like how to compute partial derivatives, is essential for diving deeper into the analysis of multivariable systems.

In the context of multivariable functions, a partial derivative represents the rate at which the function changes with respect to one of the variables, holding all other variables constant. In our example, finding the partial derivative of \( m(t, s) \) with respect to \( t \) means we calculate how \( m \) changes as \( t \) slightly changes, while \( s \) is held constant.

To compute \( m_t \), the partial derivative of \( m(t, s) \) with respect to \( t \), we focus only on the terms involving \( t \). The function \( s \ln t \) contributes \( \frac{s}{t} \) when we differentiate with respect to \( t \). Since \( 3.75s \) and \( 14.96 \) do not depend on \( t \), they are considered constants and differentiate to zero. Thus, \( m_t = \frac{s}{t} \). This expression shows how the function changes with \( t \) for a fixed \( s \).

Partial derivatives are particularly useful when analyzing functions involving multiple variables, as they provide insight into how each variable independently influences the function's value.

When we take the partial derivative of a multivariable function with respect to \( s \), we are concerned with how the function changes when \( s \) varies and \( t \) is held constant. This is known as \( m_s \). In the function \( m(t, s) = s \ln t + 3.75s + 14.96 \), we differentiate this with respect to \( s \).

For the term \( s \ln t \), the partial derivative with respect to \( s \) is \( \ln t \), since \( t \) acts as a constant multiplier. The term \( 3.75s \) contributes \( 3.75 \) when differentiated with respect to \( s \), and the constant \( 14.96 \) remains unchanged. Therefore, \( m_s = \ln t + 3.75 \).

This provides us with an expression indicating the function's sensitivity to changes in \( s \), which is crucial in scenarios where changes in specific variables need to be tracked separately.

Evaluating partial derivatives at specific values provides a clear, numerical insight into the behavior of the function at those points. In our exercise, we evaluated \( m_s \) at \( t = 3 \), which means we wanted to know how \( m \) changes with respect to \( s \) specifically when \( t \) is 3.

To find this, we plug \( t = 3 \) into the expression \( m_s = \ln t + 3.75 \). Thus, we get \( m_s \) at \( t = 3 \) is \( \ln(3) + 3.75 \). Using a calculator, we find that \( \ln(3) \approx 1.0986 \), so \( m_s \) at \( t = 3 \approx 4.8486 \).

This calculated result tells us the instantaneous rate of change of the function with respect to \( s \) at that specific point, which is incredibly useful for practical applications in fields such as physics and economics, where understanding how systems react to changes in certain conditions can influence decisions and predictions.