Problem 198 - Calculus Volume 3

Question

Use the differential \(d z\) to approximate the change in \(z=\sqrt{4-x^{2}-y^{2}}\) as \((x, y)\) moves from point (1,1) to point (1.01,0.97) . Compare this approximation with the actual change in the function.

Step-by-Step Solution

Answer

The approximate change is 0.01414, and the actual change is approximately 0.014.

Step 1: Define the function and its change

The function given is \(z = \sqrt{4 - x^2 - y^2}\). We want to approximate the change in \(z\) as \((x, y)\) moves from point (1,1) to point (1.01,0.97).

Step 2: Compute partial derivatives

Compute the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). Using the chain rule for the derivative of a square root: \[ \frac{\partial z}{\partial x} = \frac{-x}{\sqrt{4 - x^2 - y^2}} \]\[ \frac{\partial z}{\partial y} = \frac{-y}{\sqrt{4 - x^2 - y^2}} \]

Step 3: Evaluate partial derivatives at initial point

Evaluate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) at the initial point \((x, y) = (1, 1)\).\[ \frac{\partial z}{\partial x} = \frac{-1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]\[ \frac{\partial z}{\partial y} = \frac{-1}{\sqrt{2}} = -\frac{1}{\sqrt{2}} \]

Step 4: Calculate the differential dz

The differential \(dz\) is given by:\[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \]Substitute \(dx = 1.01 - 1 = 0.01\) and \(dy = 0.97 - 1 = -0.03\):\[ dz = \left(-\frac{1}{\sqrt{2}}\right)\cdot 0.01 + \left(-\frac{1}{\sqrt{2}}\right)\cdot(-0.03) \]

Step 5: Simplify the differential expression

Simplify the expression for \(dz\):\[ dz = \frac{-0.01}{\sqrt{2}} + \frac{0.03}{\sqrt{2}} \]\[ dz = \frac{0.02}{\sqrt{2}} = 0.01414 \]

Step 6: Calculate the actual change in z

Find \(z\) at the initial point and at the new point to determine the actual change. At (1, 1):\[ z = \sqrt{4 - 1^2 - 1^2} = \sqrt{2} \]At (1.01, 0.97):\[ z = \sqrt{4 - 1.01^2 - 0.97^2} \approx \sqrt{4 - 1.0201 - 0.9409} \]\[ z = \sqrt{2.039} \approx 1.428 \]Thus, the actual change in \(z\) is \[ \Delta z = 1.428 - \sqrt{2} \approx 1.428 - 1.414 = 0.014 \]

Step 7: Compare approximation with actual change

The approximate change calculated with \(dz\) was 0.01414. The actual change \(\Delta z\) was approximately 0.014. Both values are very close, showing the differential provides an accurate approximation.

Key Concepts

Partial Derivatives

When understanding differential approximations, the concept of **partial derivatives** plays a crucial role. In calculus, a partial derivative represents how a function changes as only one of its many variables is allowed to change, keeping the others constant. This concept is particularly important in functions of multiple variables like our function \( z = \sqrt{4 - x^2 - y^2} \). For the given function, to find how \( z \) changes when \( x \) changes while \( y \) is held constant, we compute the partial derivative \( \frac{\partial z}{\partial x} \). Similarly, \( \frac{\partial z}{\partial y} \) tells us how \( z \) changes with respect to \( y \).

Partial derivatives help lay the groundwork for differential approximations by modeling the local linear behavior of functions. Just as with regular derivatives in single-variable calculus, in multivariable calculus, the partial derivative is essential in predicting how small changes in one variable can impact the function's overall value. In this exercise, using partial derivatives helps approximate the change in \( z \) as \( (x, y) \) moves slightly. Recognizing

• \( \frac{\partial z}{\partial x} = \frac{-x}{\sqrt{4 - x^2 - y^2}} \)
• \( \frac{\partial z}{\partial y} = \frac{-y}{\sqrt{4 - x^2 - y^2}} \)

apers the path to understanding the multifaceted curve the function takes and depicts how closely the differential approximates the actual change.

Chain Rule

Another cornerstone of differential approximations is the **Chain Rule**. In calculus, the chain rule provides a way to differentiate composite functions. It allows us to explore how changes in one set of variables affect another set indirectly through a function composition. This rule is very useful for functions like our \( z = \sqrt{4 - x^2 - y^2} \). Because of the function's complexity, particularly its nested structure of a square root containing a sum, applying the chain rule helps us break down the derivative into understandable parts.

The chain rule makes it possible to derive expressions for both \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), as shown in the solution:

• For \( x \), consider how the inside of the square root changes with respect to \( x \).
• For \( y \), consider how the inside of the square root changes with respect to \( y \).

These steps elaborately guide you in dissecting how small disturbances in \( x \) and \( y \) contribute to the alteration in \( z \).

The chain rule enables more clarity in understanding the interconnected effects of changes in input variables and how they propagate through complicated functions. It’s an indispensable tool for efficiently tackling problems involving multiple variables.

Actual Change Comparison

Differential approximations resemble a flashlight illuminating the intricate path of changing functions. But how accurate is this flashlight? This is where the **Actual Change Comparison** comes in. In this exercise, we compare the approximate change \(dz\) with the actual change in the function \( \Delta z \).

For small movements in \((x, y)\) from \((1, 1)\) to \((1.01, 0.97)\), the differential approximation gives us an estimated change:

• The differential, \(dz\), was found to be approximately 0.01414.

Meanwhile, the actual change \( \Delta z \) turned out to be about 0.014. These two values are incredibly close, showing the efficacy of differential approximations in this context.

Computing the actual change involves recalculating the function value at the new point and subtracting the old function value. This precision check validates the reliability of using differentials, encouraging us to trust this method in tackling similar problems with slight input variances. By seeing such minimal deviation between approximated and actual values, students reinforce their grasp of how and when to trust differential approximations for predicting function behavior. It's this comparison that builds confidence in the approximation tool, enhancing a student's analytical toolkit.