Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

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 \(3 .\) \(z=x^{2}-5 x y+y ; P(2,3), Q(2.03,2.98)\)

Step-by-Step Solution

Answer

\( dz \approx -0.15 \); exact \( \Delta z \approx -0.0408 \).

Step 1: Find the partial derivatives

To find the total differential, first, compute the partial derivatives of the function. For the function \( z = x^2 - 5xy + y \), we need the partial derivatives with respect to \( x \) and \( y \).\( \frac{\partial z}{\partial x} = 2x - 5y \)\( \frac{\partial z}{\partial y} = -5x + 1 \).

Step 2: Calculate partial derivatives at point P

Evaluate the partial derivatives at point \( P(2, 3) \):\( \frac{\partial z}{\partial x} \bigg|_P = 2(2) - 5(3) = 4 - 15 = -11 \)\( \frac{\partial z}{\partial y} \bigg|_P = -5(2) + 1 = -10 + 1 = -9 \).

Step 3: Use the total differential formula

Use the formula for the total differential \( dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \) to approximate the change.Here, \( dx = 2.03 - 2 = 0.03 \) and \( dy = 2.98 - 3 = -0.02 \).Thus, compute:\( dz = (-11)(0.03) + (-9)(-0.02) = -0.33 + 0.18 = -0.15 \).

Step 4: Calculate exact value of \( \Delta z \)

Calculate \( z \) at points \( P \) and \( Q \):- For \( P(2,3) \):\( z_P = 2^2 - 5 \times 2 \times 3 + 3 = 4 - 30 + 3 = -23 \).- For \( Q(2.03, 2.98) \):\( z_Q = (2.03)^2 - 5 \times 2.03 \times 2.98 + 2.98 \approx 4.1209 - 30.1417 + 2.98 = -23.0408 \).\( \Delta z = z_Q - z_P = -23.0408 + 23 = -0.0408 \).

Key Concepts

Partial Derivatives

Partial derivatives play a critical role in understanding how a function changes as its variables change. When we have a function with more than one variable, like in our case with function \(z = x^2 - 5xy + y\), we determine how the function changes when altering one variable while keeping the others constant.
Let's break it down:

• Partial derivative with respect to \(x\): This tells us how \(z\) changes as \(x\) changes, with \(y\) held constant. Calculated as \(\frac{\partial z}{\partial x} = 2x - 5y\).
• Partial derivative with respect to \(y\): This indicates how \(z\) changes as \(y\) changes, keeping \(x\) constant. This is found by \(\frac{\partial z}{\partial y} = -5x + 1\).

These expressions help us understand the sensitivity of \(z\) to changes in \(x\) and \(y\), guiding us in the approximation process.

Approximation

Approximation using the total differential is a method to estimate how a function changes near a known point. Imagine you're looking at changes at point \(P(2,3)\) to another point \(Q(2.03, 2.98)\).
From above, we know:

• Change in \(x (dx)\): \(2.03 - 2 = 0.03\)
• Change in \(y (dy)\): \(2.98 - 3 = -0.02\)

Using these, we use the total differential formula \(dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy\) to approximate changes.

• The partial derivatives at \(P\) give us slopes (\(-11\) for \(x\) and \(-9\) for \(y\)).
• These slopes are then multiplied by their respective small changes to estimate the total change in \(z\).

This expertise in approximation is fundamental in calculus, providing us with tools to predict changes with reasonable accuracy, especially in complex systems.

Calculus Problem Solving

Solving calculus problems involves a clear understanding of the relationships among the variables involved and the function itself. In our exercise, the primary goal was to use calculus tools to understand how a small move in variables \(x\) and \(y\) affects \(z\).
To solve such problems, follow these steps:

• Calculate partial derivatives to understand each variable's individual effect.
• Substitute specific values into these derivatives to tailor your estimation to a precise point.
• Apply the total differential to find the approximate change in the function.
• Validate by computing the exact change using straightforward calculations when possible.

Each step in a calculus problem builds upon the last. So, attention to detail at each stage is crucial for overall accuracy. Effective problem solving in calculus not only requires mathematical skills but also strategic planning in the approach.

Exact Change Calculation

Exact change calculation gives us the precise difference between two function values at specific points. After estimating using differentials, it's essential to find this exact value for confirmation and as a comparison.
Here's how you calculate it:

• Evaluate function \(z\) at point \(P(2,3)\), yielding \(z_P = -23\).
• Similarly, compute at \(Q(2.03, 2.98)\) which gives \(z_Q \approx -23.0408\).
• The exact change \(\Delta z = z_Q - z_P\), which in this case is \(-0.0408\).

These calculations reconcile our approximations with precise data. Exact calculations are important benchmarks, enabling verification of other analytical or computational models in calculus.