Chapter 12: Problem 11
In Problems 9-12, use the total differential dz to approximate the change in z as \((x, y)\) moves from \(P\) to \(Q .\) Then use a calculator to find the corresponding exact change \(\Delta z\) (to the accuracy of your calculator). See Example 3. $$ z=\ln \left(x^{2} y\right) ; P(-2,4), Q(-1.98,3.96) $$
Short Answer
Step by step solution
Recognize the Function
Compute the Partial Derivatives
Evaluate Partial Derivatives at Point P
Calculate Total Differential dz
Compute Exact Change \( \, \Delta z \, \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
In our given function, \( z = \ln(x^2 y) \), partial derivatives allow us to see how \( z \) is affected by changes in either \( x \) or \( y \) independently.
To find the partial derivative with respect to \( x \), \( x \) is considered a variable, and \( y \) is treated as a constant. Similarly, when finding the partial derivative with respect to \( y \), \( y \) is variable and \( x \) is constant.
This process essentially "slices" the function along one of its input axes, giving us a derivative that reflects the rate of change in one direction.
- The partial derivative with respect to \( x \) is \( \frac{2}{x} \).
- The partial derivative with respect to \( y \) is \( \frac{1}{y} \).
Total Differential
For a function \( z = f(x, y) \), the total differential \( dz \) can be calculated using:
\[ dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy \]
This equation tells us that the change in \( z \) depends on both \( dx \) and \( dy \), which are the changes in \( x \) and \( y \) respectively.
In our example, we calculated \( dz \) using values of partial derivatives at point \( P \) and small changes \( dx = 0.02 \) and \( dy = -0.04 \). This gave an approximate change \( dz = -0.03 \).
- This calculation uses linear approximation to estimate change over a small region close to point \( P \).
- The result helps quickly predict the behavior of \( z \) without needing complex computations.
Exact Change Calculation
To calculate the exact change \( \Delta z \), you find \( z \) at both the initial point \( P \) and the new point \( Q \). The difference \( \Delta z = z_Q - z_P \) gives the exact amount \( z \) has changed.
For \( z = \ln(x^2 y) \), we calculated:
- At \( P(-2, 4) \), \( z_P = \ln(16) \approx 2.7726 \).
- At \( Q(-1.98, 3.96) \), \( z_Q = \ln(15.6352) \approx 2.7495 \).
This exact value shows us precisely how much \( z \) changes as we move from \( P \) to \( Q \). It’s essential for validating the accuracy of our total differential, confirming if our approximation was close to the real outcome.