Question

B Berechnen Sie das totale Differential für die Funktionen a) \(z=\sqrt{1-x^{2}-y^{2}}\) b) \(z=x^{2}+y^{2}\) c) \(f(x, y, z)=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\)

Short answer

The total differentials for the functions are: \na) \(dz = - \frac{{x}}{{\sqrt{1-x^{2}-y^{2}}}}dx - \frac{{y}}{{\sqrt{1-x^{2}-y^{2}}}}dy\), \nb) \(dz = 2xdx + 2ydy\), \nc) \(df = - \frac{{x}}{{(x^{2}+y^{2}+z^{2})^{3/2}}}dx - \frac{{y}}{{(x^{2}+y^{2}+z^{2})^{3/2}}}dy - \frac{{z}}{{(x^{2}+y^{2}+z^{2})^{3/2}}}dz.\

Step-by-step solution

Calculate Total Differential for Function a

The total differential dz of z given by \(z=\sqrt{1-x^{2}-y^{2}}\) is calculated by differentiating z with respect to each independent variable, x and y, and summing the result retaining the infinitesimal changes dx and dy: \n\(dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy\).\nFirst, calculate the partial derivatives: \n\(\frac{{\partial z}}{{\partial x}} = - \frac{{x}}{{\sqrt{1-x^{2}-y^{2}}}}\) and\(\frac{{\partial z}}{{\partial y}} = - \frac{{y}}{{\sqrt{1-x^{2}-y^{2}}}}\).\nThen, substitute these into the dz equation for the total differential.

Calculate Total Differential for Function b

The total differential dz of z given by \(z=x^{2}+y^{2}\) is calculated similarly to the first part of Function a. However, notice that the function is simpler this time. First, calculate the partial derivatives: \n\(\frac{{\partial z}}{{\partial x}} = 2x\) and\(\frac{{\partial z}}{{\partial y}} = 2y\).\nSubstitute these into the dz equation for the total differential.

Calculate Total Differential for Function c

For the function \(f(x, y, z)=\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}\), the total differential df is calculated by differentiating f with respect to each independent variable, x, y, and z, and summing the results: \n\(df = \frac{{\partial f}}{{\partial x}}dx + \frac{{\partial f}}{{\partial y}}dy + \frac{{\partial f}}{{\partial z}}dz\).\nAgain compute the partial derivatives, substitute these into the df equation for the total differential.

Key concepts

Partial Derivatives

Partial derivatives are essential in understanding how multivariable functions change with respect to one variable while keeping the other variables constant. Imagine you have a surface representing a function of two variables. If we "slice" the surface along one axis, we get a curve that can be analyzed similarly to a function of a single variable. The slope of this curve at any point is called the partial derivative. For example, for the function \(z = \sqrt{1-x^2-y^2}\), the partial derivative with respect to \(x\) describes how \(z\) changes as \(x\) changes, while \(y\) is held fixed. Mathematically, it's calculated as \(\frac{\partial z}{\partial x} = - \frac{x}{\sqrt{1-x^2-y^2}}\). Remember:

• Partial derivatives give us a way to analyze changes in one variable at a time in a function of multiple variables.
• They are foundational in calculating the total differential, which estimates how function values change with small changes in each variable.

Partial derivatives are a stepping stone to more complex calculations in multivariable calculus, making them particularly useful in fields like physics, engineering, and economics.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus, such as differentiation and integration, to functions of several variables. Unlike single-variable calculus that deals with functions along a line, multivariable calculus tackles functions over planes and spaces. This allows us to handle more complex systems and phenomena.A key aspect of multivariable calculus is its ability to model physical and real-world problems where multiple factors influence an outcome. For instance, consider the function \(f(x, y, z) = \frac{1}{\sqrt{x^2 + y^2 + z^2}}\), which might represent a field around a point. This is a clear example where understanding the effects of changes in all three variables \((x, y, z)\) is necessary.In multivariable calculus, we often:

• Calculate partial derivatives to analyze the change concerning each variable.
• Use techniques like the total differential to approximate how functions change with small increments in variables.
• Solve real-life problems involving multiple varying factors, such as in thermodynamics or fluid dynamics.

Overall, multivariable calculus opens new avenues for analyzing and solving problems that are beyond the scope of single-variable calculus.

Differentiation Techniques

Differentiation techniques in multivariable calculus allow us to explore how small changes in variables affect multivariable functions. In these scenarios, differentiation isn't just about slopes, but about sensitivity and influences of multiple factors.**Steps to differentials in multivariable functions:** Unlike single-variable functions where we find a derivative, in multivariable settings, total differentials become important:

• First, we find the partial derivatives with respect to all independent variables.
• Then, combine these derivatives with infinitesimal changes (e.g., \(dx, dy, dz\)) to find the total differential.
• For example, consider \(z = x^2 + y^2\); the total differential is \(dz = 2x\,dx + 2y\,dy\).

This technique is central in predicting how functions behave under slight perturbations in any of their variables.Differentiation in the context of multivariable calculus is crucial in fields like optimization, where understanding variable influences can improve decisions or designs. Whether it's optimizing a cost function in economics or analyzing stress in mechanical components, these techniques provide the insight needed to inform critical decisions.