Problem 181
Question
$$ \tan ^{-1}\left(\frac{y}{x}\right)=\ln \sqrt{x^{2}+y^{2}} $$
Step-by-Step Solution
Verified Answer
The solution of the given equation is \(y = x\).
1Step 1: 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: \(\tan ^{-1}\left(\frac{y}{x}\right)=\frac{1}{2} \ln (x^{2}+y^{2})\)
2Step 2: 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: \(\tan ^{-1}\left(\frac{y}{x}\right) = \tan(\pi/4) \Rightarrow \frac{y}{x} = \tan(\pi/4)\)
3Step 3: Solve for the variable
Lastly, solving for \(y\), we get: \(y = x \cdot \tan(\pi/4)\). Considering \(\tan(\pi/4)=1\), we have: \(y = x \cdot 1 = x\)
Key Concepts
Logarithmic PropertiesTangent FunctionTrigonometric Identities
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:
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:
- \( \ln(a^b) = b \cdot \ln(a) \)
- \( \ln(\sqrt{a}) = \frac{1}{2} \ln(a) \)
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.
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:
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.
Some important identities include:
- \( \tan(\pi/4) = 1 \)
- The Pythagorean identity: \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
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.