Problem 20 - 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. $$ z=\cos \left(\pi x+\frac{\pi}{2} y\right), \quad x=s t^{2}, \quad y=s^{2} t ; \quad s=1, t=1 $$

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial s} = 2\pi\), \(\frac{\partial z}{\partial t} = 2\pi + \frac{\pi}{2}\).

Step 1: Identify the Functions and Variables

We have the function \( z = \cos\left( \pi x + \frac{\pi}{2} y \right) \), where \( x = s t^2 \) and \( y = s^2 t \). We need to find the partial derivatives \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \) using the Multivariable Chain Rule.

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

The chain rule gives us \(\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s}\).Calculate each required derivative:1. \( \frac{\partial z}{\partial x} = -\sin\left( \pi x + \frac{\pi}{2} y \right) \cdot \pi \)2. \( \frac{\partial z}{\partial y} = -\sin\left( \pi x + \frac{\pi}{2} y \right) \cdot \frac{\pi}{2} \)3. \( \frac{\partial x}{\partial s} = t^2 \)4. \( \frac{\partial y}{\partial s} = 2st \)Substitute these into the chain rule expression.

Step 3: Apply Multivariable Chain Rule for \(\frac{\partial z}{\partial t}\)

The chain rule gives us \(\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t}\).Calculate each required derivative:1. \( \frac{\partial z}{\partial x} = -\sin\left( \pi x + \frac{\pi}{2} y \right) \cdot \pi \)2. \( \frac{\partial z}{\partial y} = -\sin\left( \pi x + \frac{\pi}{2} y \right) \cdot \frac{\pi}{2} \)3. \( \frac{\partial x}{\partial t} = 2st \)4. \( \frac{\partial y}{\partial t} = s^2 \)Substitute these into the chain rule expression.

Step 4: Simplify and Substitute \(s = 1\) and \(t = 1\)

Calculate the expression for \(\frac{\partial z}{\partial s}\):\[\frac{\partial z}{\partial s} = -\sin\left( \pi x + \frac{\pi}{2} y \right) \cdot \pi \cdot t^2 + \left(-\sin\left( \pi x + \frac{\pi}{2} y \right) \cdot \frac{\pi}{2} \cdot 2st\right)\]Substitute \(x = 1 \cdot 1^2 = 1\), \(y = 1^2 \cdot 1 = 1\), so\[\pi x + \frac{\pi}{2} y = \pi + \frac{\pi}{2} = \frac{3\pi}{2}\]Hence:\[\frac{\partial z}{\partial s} = -\sin\left( \frac{3\pi}{2} \right) \cdot \pi \cdot 1 + (-\sin\left( \frac{3\pi}{2} \right) \cdot \pi)\]Simplify the expression.

Step 5: Evaluate \(\sin\left(\frac{3\pi}{2}\right)\)

The value of \(\sin\left(\frac{3\pi}{2}\right)\) is \(-1\). Using this, simplify:\[\frac{\partial z}{\partial s} = -(-1) \cdot \pi \cdot 1 + \left(-(-1) \cdot \ \frac{\pi}{2} \cdot 2 \right) = \pi + \pi\]Thus, \(\frac{\partial z}{\partial s} = 2\pi\).

Step 6: Calculate \(\frac{\partial z}{\partial t}\) with Substitution

Using the same \(\sin\) value:\[\frac{\partial z}{\partial t} = -\sin\left( \pi x + \frac{\pi}{2} y \right) \cdot \pi \cdot 2st + (-\sin\left( \pi x + \frac{\pi}{2} y \right) \cdot \frac{\pi}{2} \cdot s^2)\]Plug in \(s = 1\), \(t = 1\) giving us:\\[-(-1) \cdot \pi \cdot 2 + (-(-1) \cdot \frac{\pi}{2}) = 2\pi + \ \frac{\pi}{2}\]Thus, \(\frac{\partial z}{\partial t} = 2\pi + \ \frac{\pi}{2}\).

Key Concepts

Multivariable Chain Rule

The Multivariable Chain Rule is crucial to understanding how to differentiate a function that depends on multiple variables, each of which is itself a function of one or more independent variables. In our exercise, the function given is

• \( z = \cos(\pi x + \frac{\pi}{2} y) \)

and both \(x\) and \(y\) depend on \(s\) and \(t\) through:

• \( x = st^2 \)
• \( y = s^2 t \)

To find how \(z\) changes with respect to \(s\) and \(t\), we use the Multivariable Chain Rule. For the partial derivative \(\frac{\partial z}{\partial s}\), the rule provides us:

• \(\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s}\)

This means that to compute \(\frac{\partial z}{\partial s}\), you first find how \(z\) changes with \(x\) and \(y\) individually (i.e., these derivatives consider \(z\) as a single-variable function of \(x\) or \(y\)) and how \(x\) and \(y\) change with \(s\).

Similarly, for \(\frac{\partial z}{\partial t}\):

• \(\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t}\)

This procedure is fundamental when dealing with composite functions in multivariable calculus, aiding in evaluating partial derivatives efficiently.

Partial Derivatives

Partial Derivatives are key in multivariable calculus since they allow you to see how a function changes as one variable changes, while others are held constant. In essence, they give us a glimpse into the sensitivity of a function with respect to changes in its input variables. For our function \( z = \cos(\pi x + \frac{\pi}{2} y) \):

• \(\frac{\partial z}{\partial x} = -\sin(\pi x + \frac{\pi}{2} y) \cdot \pi\)
• \(\frac{\partial z}{\partial y} = -\sin(\pi x + \frac{\pi}{2} y) \cdot \frac{\pi}{2} \)

These derivatives illustrate how the output of our function \(z\) is impacted if we tweak \(x\) or \(y\) slightly. In this context, however, since \(x\) and \(y\) are themselves dependent on \(s\) and \(t\), it’s these relationships that help us understand the broader indirect effect on \(z\).

Partial derivatives provide vital insights when functions have multiple inputs, giving rise to more powerful analysis and applications, like optimization and the monitoring of rates of change.

Function Evaluation

Function Evaluation is a process where we calculate the value of the function for specific values of its variables. This step comes after finding the appropriate expressions for the derivatives in terms of \(x\) and \(y\). In our problem, once we have equations for \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t}\), substituting in specific values for \(s\) and \(t\) is essential:

• For \(s = 1\) and \(t = 1\), evaluate \(x\) and \(y\):
  • \(x = 1\cdot 1^2 = 1\)
  • \(y = 1^2 \cdot 1 = 1\)

Using these values, evaluate \(\pi x + \frac{\pi}{2} y\) which results in \(\frac{3\pi}{2}\). Then

• \(\frac{\partial z}{\partial s} = 2\pi\)
• \(\frac{\partial z}{\partial t} = 2\pi + \frac{\pi}{2}\)

This procedure highlights how to derive specific numbers from more complex expressions, allowing for practical applications and conclusions to be drawn based on theoretical frameworks.