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Approximate the value of \(f\) at the given point. $$ f(x, y)=\ln \left(x^{2}+y^{2}\right) ;(-0.03,0.98) $$

Short Answer

Expert verified
The approximate value of \( f(-0.03, 0.98) \) is \( -0.0396 \).

Step by step solution

01

Understand the Problem

We are given the function \( f(x, y) = \ln(x^2 + y^2) \) and asked to approximate its value at the point \((-0.03, 0.98)\). We'll do this using the values of \( x = -0.03 \) and \( y = 0.98 \).
02

Plug in the Values

Substitute \( x = -0.03 \) and \( y = 0.98 \) into the function. This gives us the expression \( f(-0.03, 0.98) = \ln((-0.03)^2 + (0.98)^2) \).
03

Calculate Inside the Logarithm

Compute the value inside the logarithm: \( (-0.03)^2 + (0.98)^2 = 0.0009 + 0.9604 = 0.9613 \).
04

Find the Natural Logarithm

Evaluate \( \ln(0.9613) \). Using a calculator, this approximates to \( \ln(0.9613) \approx -0.0396 \).
05

Write the Final Answer

Thus, the approximate value of \( f(-0.03, 0.98) \) is \( -0.0396 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Logarithms
Natural logarithms are fundamental in calculus and analysis. They are logarithms to the base of Euler's number, denoted as \( e \), which is approximately equal to 2.71828. Natural logarithms, expressed as \( \ln(x) \), are essential when dealing with growth processes, such as population growth or interest calculations. They arise naturally in these contexts due to their mathematical simplicity and properties.

When we compute a natural logarithm, we're essentially answering the question: "To what power must \( e \) be raised, to produce the given number?" For example, \( \ln(2.71828) = 1 \) because \( e^1 = e \). Similarly, finding the natural logarithm of a number less than 1, like in our exercise, will result in a negative value because \( e^x \) for negative \( x \) results in numbers less than 1.

In multivariable calculus, natural logarithms are often used to describe logarithmic surfaces and to compute integrals of exponential functions. They help to simplify expressions and reveal underlying patterns.
Function Approximation
Function approximation is a core concept in calculus, where an approximate value of a function at a specific point is determined. This can be beneficial when dealing with complex functions or when exact computation is difficult due to constraints like limited computational power.

There are several methods for function approximation such as Taylor series, polynomial approximations, and linear approximations. In the exercise, approximation allows us to evaluate \( f(x, y) \) by substituting nearby values and computing with those. By using a calculator or a series expansion method, we can find an estimated result instead of the exact one, which is often good enough for practical purposes.

Function approximation plays an essential role not just in theoretical tasks, but also in real-world applications such as computer graphics, engineering, and economic modeling. It simplifies the modeling of complex behaviors by using simpler, approximate expressions.
Coordinate Substitution
Coordinate substitution involves replacing the variables in a multivariable function with given specific coordinates. This is an essential technique when evaluating functions at specific points. By substitution, we're essentially transforming an abstract function description into a more concrete numerical task.

For our exercise, coordinate substitution was used to solve \( f(x, y) = \ln(x^2 + y^2) \) at \((-0.03, 0.98)\). We replaced \( x \) and \( y \) with their respective values to simplify the function, making it easier to compute its value. This process ensures more accurate evaluations and helps us understand how changes in one part of the coordinate system affect the overall solution.

Coordinate substitution is foundational in solving real-world problems, such as determining the outcome of physical processes at certain points, and is widely used in fields such as physics, engineering, and computer-aided design.

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Most popular questions from this chapter

It seems reasonable that an increase in taxation on a commodity would decrease the production of that commodity. The following argument supports that claim. Assume that all required derivatives exist. For any \(x \geq 0\), let \(P_{0}(x)\) be the profit before taxes on \(x\) units produced. Let \(P(x, t)\) denote the profit after taxes on \(x\) units produced with \(\operatorname{tax} t\) on each unit. Assume that at any tax rate \(t\) the company will maximize its profits by producing \(f(t)\) units so that $$ \begin{aligned} &\frac{\partial P}{\partial x}(f(t), t)=0 \\ &\frac{\partial^{2} P}{\partial x^{2}}(f(t), t)<0 \end{aligned} $$ (The conditions in (9) and (10) are just those required for the Second Derivative Test.) a. Show that \(P(x, t)=P_{0}(x)-t x\). b. Using (a), show that $$ \frac{\partial P}{\partial x}(x, t)=P_{0}^{\prime}(x)-t \quad \text { and } \quad \frac{\partial^{2} P}{\partial x^{2}}(x, t)=P_{0}^{\prime \prime}(x) $$ c. From \((9)\) and \((\mathrm{b})\), show that \(P_{0}^{\prime}(f(t))-t=0\). d. By differentiating both sides of the equation in (c) and by using (b) and (10), show that $$ f^{\prime}(t)=\frac{1}{P_{0}^{\prime \prime}(f(t))}=\frac{1}{\frac{\partial^{2} P}{\partial x^{2}}(f(t), t)}<0 $$ (Thus the production tends to decrease as the tax rate increases.)

We say that two surfaces are normal at a given point if their tangent planes at that point are perpendicular to one another.Show that the pair of surfaces are normal at the given point. \(x^{2}+y^{2}+z^{2}=16\) and \(z^{2}=x^{2}+y^{2} ;(2,2,2 \sqrt{2})\)

Let \(g\) be a differentiable function of one variable and let \(f(x, y)=x g(y / x)\). Show that every plane tangent to the graph of \(f\) passes through the origin.

$$ \text { Let } z=f(x-y) . \text { Show that } \frac{\partial z}{\partial x}=-\frac{\partial z}{\partial y} \text { . } $$

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(x, y)=x^{2}-e^{y^{2}} $$

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