91Ó°ÊÓ

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as one of its input variables changes, while keeping other variables constant. If we have a function \( z = f(x, y) \), then:

• \( \frac{\partial z}{\partial x} \) describes how \( z \) changes with respect to \( x \).
• \( \frac{\partial z}{\partial y} \) describes how \( z \) changes with respect to \( y \).

In our problem, we know the values of the partial derivatives at a specific point: \( f_x(4,8) = 3 \) indicates that a small increase in \( x \) while holding \( y \) constant will increase \( z \) by about 3 times the increase in \( x \). Similarly, \( f_y(4,8) = -1 \) implies that an increase in \( y \) will decrease \( z \) by about the same amount.

Multivariable calculus extends the concepts of calculus to functions of more than one variable. In our exercise, we deal with a function \( z = f(x, y) \), which depends on two variables, \( x \) and \( y \). This introduces new challenges and tools for exploring how these variables interact with the function's rate of change.
To study how \( z \) changes with respect to these variables, we must use the concept of partial derivatives, as each input affects the output differently. This problem also employs the chain rule, a critical concept in multivariable calculus, to deduce the rate of change of \( z \) in terms of another variable, \( t \). This is crucial when the variables \( x \) and \( y \) are themselves functions of another variable, as with \( x = t^2 \) and \( y = t^3 \) in this scenario.
The ability to compute such derivatives allows us to understand and model real-world situations, such as predicting changes in a system with multiple influencing factors.

Rate of change is a significant concept in calculus describing how a quantity varies with respect to another. In single-variable functions, it is depicted by the derivative. For multivariable functions like \( z = f(x, y) \), the chain rule helps determine the rate of change with respect to a parameter that influences both variables, such as \( t \) in our exercise.
Here’s how it works:

• Consider \( x = t^2 \) and \( y = t^3 \), meaning both \( x \) and \( y \) change as \( t \) changes.
• We apply the chain rule: \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \).
• This lets us express how \( z \) changes as \( t \) changes, even though \( z \) is not a direct function of \( t \).
• In the solved problem, substituting the derivatives and the values of \( t \), \( x \), and \( y \) revealed that \( \frac{dz}{dt} = 0 \). Therefore, at \( t = 2 \), changes in \( x \) and \( y \) due to changes in \( t \) exactly balance each other out, resulting in no net change in \( z \).

Understanding rates of change is crucial for analyzing dynamic systems where multiple variables are interdependent.