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Chapter 4: Problem 181

$$ \tan ^{-1}\left(\frac{y}{x}\right)=\ln \sqrt{x^{2}+y^{2}} $$

Short Answer

Expert verified
The solution of the given equation is \(y = x\).

Step by step solution

01

Rewrite the equation using logarithmic properties

First, note that \(\ln \sqrt{x^{2}+y^{2}} = \frac{1}{2} \ln (x^{2}+y^{2})\). Now, the equation reads as: \(\tan ^{-1}\left(\frac{y}{x}\right)=\frac{1}{2} \ln (x^{2}+y^{2})\)
02

Apply the tangent properties

One of the properties of tangent function says that \(\tan(\pi/4)=1\). So, let's divide \(y/x\) by \(1\), we get: \(\tan ^{-1}\left(\frac{y/x}{1}\right)=\frac{1}{2} \ln (x^{2}+y^{2})\). This can be rewritten as: \(\tan ^{-1}\left(\frac{y}{x}\right) = \tan(\pi/4) \Rightarrow \frac{y}{x} = \tan(\pi/4)\)
03

Solve for the variable

Lastly, solving for \(y\), we get: \(y = x \cdot \tan(\pi/4)\). Considering \(\tan(\pi/4)=1\), we have: \(y = x \cdot 1 = x\)

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Key Concepts

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

Logarithmic Properties
Logarithmic properties are key when dealing with expressions involving logarithms. One useful property is the way we can simplify expressions under a square root by converting them into a natural logarithm.

For example, consider the expression \( \ln \sqrt{x^{2}+y^{2}} \). This is equivalent to \( \frac{1}{2} \ln (x^{2}+y^{2}) \) because the square root is the same as raising to the power of \( \frac{1}{2} \).

Remember these essential rules for working with logarithms:
  • \( \ln(a^b) = b \cdot \ln(a) \)
  • \( \ln(\sqrt{a}) = \frac{1}{2} \ln(a) \)
Understanding these properties allows us to simplify complex expressions and solve equations more easily.
Tangent Function
The tangent function is one of the fundamental trigonometric functions, often expressed as \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \) in a right-angled triangle.

The inverse tangent, or \( \tan^{-1}(z) \), returns the angle \( \theta \) such that \( \tan(\theta) = z \). This is important for converting calculations back to angle measures.

A key property of \( \tan \) is \( \tan(\pi/4) = 1 \). This means if \( \theta = \pi/4 \), the opposite and adjacent sides are equal.
  • If \( \tan^{-1}(\frac{y}{x}) = \frac{1}{2} \ln(x^2 + y^2) \), understanding \( \tan(\pi/4) = 1 \) helps solve for variables.
  • Knowing these trigonometric properties simplifies equations involving inverse tangents.
Trigonometric Identities
Trigonometric identities are formulas involving trigonometric functions that are true for each angle. They are crucial in simplifying expressions and solving complex equations.

Some important identities include:
  • \( \tan(\pi/4) = 1 \)
  • The Pythagorean identity: \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
These identities allow transformation and simplification, providing different ways to view the same trigonometric expression.

By applying identities like \( \tan(\pi/4) = 1 \), we see that \( \frac{y}{x} = \tan(\pi/4) \) leads to \( y = x \). Understanding these relationships eases the process of solving problems connected to trigonometric functions.

Knowing identities by heart is vital for mathematical problem-solving, especially in calculus and geometry.

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Most popular questions from this chapter

$$ \text { Given } f(x)=x^{2} e^{x} \text { , find } f^{\prime}(0) \& f^{\prime}(1) \text { by first principles. } $$

$$ \text { Given } f(x)=\sin ^{-1} x \text { , find } f^{\prime}(0), f^{\prime}(-1) \& f^{\prime}(1) \text { by first principles. } $$

$$ y=x^{\ln x} $$

$$ y=x^{\frac{1}{x}} $$

$$ y=\sqrt[3]{\frac{x\left(x^{2}+1\right)}{\left(x^{2}-1\right)^{2}}} $$
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