Problem 22 - APEX Calculus

Question

In Exercises \(19-22,\) functions \(z=f(x, y), x=g(s, t)\) and \(y=h(s, t)\) are given. (a) Use the Multivariable Chain Rule to compute \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) (b) Evaluate \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\) at the indicated \(s\) and \(t\) values.$ 12. $$ z=e^{-\left(x^{2}+y^{2}\right)}, \quad x=t, \quad y=s t^{2} ; \quad s=1, t=1 $$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial s} = -2e^{-2}, \quad \frac{\partial z}{\partial t} = -6e^{-2} \) at \(s=1, t=1\).

Step 1: Identify Functions

We are given the functions \(z = f(x, y) = e^{-(x^2 + y^2)}\), \(x = g(s, t) = t\), and \(y = h(s, t) = st^2\). We need these to compute the derivatives.

Step 2: Apply Chain Rule for \(\frac{\partial z}{\partial s}\)

Using the Multivariable Chain Rule, we have: \[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \].Calculate each partial derivative:- \( \frac{\partial z}{\partial x} = -2xe^{-(x^2 + y^2)} \)- \( \frac{\partial z}{\partial y} = -2ye^{-(x^2 + y^2)} \)- \( \frac{\partial x}{\partial s} = 0 \) (since \(x = t\))- \( \frac{\partial y}{\partial s} = t^2 \) (since \(y = st^2\))

Step 3: Calculate \(\frac{\partial z}{\partial s}\)

Substitute the derivatives found into the chain rule expression:\[ \frac{\partial z}{\partial s} = -2xe^{-(x^2 + y^2)} \cdot 0 + -2ye^{-(x^2 + y^2)} \cdot t^2 = -2 y t^2 e^{-(x^2 + y^2)} \].

Step 4: Apply Chain Rule for \(\frac{\partial z}{\partial t}\)

Using the Chain Rule, we have: \[ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \].Calculate each partial derivative:- \( \frac{\partial x}{\partial t} = 1 \) (since \(x = t\))- \( \frac{\partial y}{\partial t} = 2st \) (since \(y = st^2\))

Step 5: Calculate \(\frac{\partial z}{\partial t}\)

Substitute the derivatives found into the chain rule expression:\[ \frac{\partial z}{\partial t} = -2xe^{-(x^2 + y^2)} \cdot 1 + -2ye^{-(x^2 + y^2)} \cdot 2st \]\[ = -2xe^{-(x^2 + y^2)} - 4yst e^{-(x^2 + y^2)} \].

Step 6: Evaluate at \(s = 1, t = 1\)

Set \(x = t = 1\) and \(y = st^2 = 1^3 = 1\). Evaluate the expressions:\[ \frac{\partial z}{\partial s} = -2 \cdot 1 \cdot 1^2 e^{-(1^2 + 1^2)} = -2 \cdot e^{-2} \]\[ \frac{\partial z}{\partial t} = -2 \cdot 1 \cdot e^{-2} - 4 \cdot 1 \cdot 1 \cdot e^{-2} = -6e^{-2} \].

Key Concepts

Partial Derivatives

When working with functions of several variables, partial derivatives represent the rate of change of the function concerning one variable while keeping others constant. This is like zooming in on just one dimension without letting the others interfere. In our exercise, we deal with a function \( z = f(x, y) \) and we want to find out how \( z \) changes as the variables \( s \) and \( t \) vary. Partial derivatives help calculate how each separate variable affects the overall change.In the step-by-step solution provided, the derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) were computed first. These represent how the function \( z \) changes with \( x \) and \( y \) individually:

• \( \frac{\partial z}{\partial x} = -2xe^{-(x^2 + y^2)} \)
• \( \frac{\partial z}{\partial y} = -2ye^{-(x^2 + y^2)} \)

The strategies used to simplify these expressions, like treating other variables as constants, are typical when finding partial derivatives.

Chain Rule

The Chain Rule is like a bridge that connects the change in one variable to another—in essence, it links together partial derivatives. When dealing with functions of multiple variables, such as \( z = f(x, y) \), where \( x \) and \( y \) depend on other variables \( s \) and \( t \), the Multivariable Chain Rule helps compute derivatives like \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \).To apply the Chain Rule for \( \frac{\partial z}{\partial s} \), the expression becomes:\[ \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \]Similarly, for \( \frac{\partial z}{\partial t} \):\[ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \]The Chain Rule requires taking the derivative with respect to each intermediate step and leads to understanding how the mixing of variables affects the direction and magnitude of change. It elegantly combines various rates of change into one comprehensive formula.

Functions of Several Variables

In the context of multivariable calculus, functions with several variables mean that the output depends on two or more inputs. Think of a weather forecast model where the temperature prediction \( z \) is based on several factors such as time \( t \) and location \( s \).In our exercise, the function \( z = e^{-(x^2 + y^2)} \) is influenced by \( x \) and \( y \), while \( x \) and \( y \) themselves are defined by \( t \) and \( s \). This layered dependency creates a cascade where each variable adjustment triggers changes in the subsequent layers.Understanding functions involving multiple variables is essential as it helps model complex systems and phenomena where multiple factors interplay. The solution demonstrates this by showing how subtle changes in \( s \) and \( t \) translate through the layers of functions to affect \( z \). By grasping these relations, we can make more insightful predictions about the behavior of such systems.