Question

Find the radius of curvature at the indicated value. \(y=\tan x,\) at \(x=\pi / 4\)

Short answer

The radius of curvature at \(x=\pi/4\) is \(\frac{5\sqrt{5}}{4}\).

Step-by-step solution

Differentiate the Function

To find the radius of curvature, we first need to find the first derivative of the given function. The function is \(y = \tan x\). The derivative of \(\tan x\) is \(y' = \sec^2 x\).

Find the Second Derivative

Now, let's calculate the second derivative. The derivative of \(\sec^2 x\) is obtained using the chain rule, resulting in \(y'' = 2 \sec^2 x \cdot \sec x \tan x = 2 \sec^3 x \tan x\).

Evaluate Derivatives at Given Point

Substitute \(x = \pi/4\) in the first and second derivatives. We have \(y' = \sec^2(\pi/4) = 2\) and \(y'' = 2 \sec^3(\pi/4) \tan(\pi/4) = 4\).

Use the Radius of Curvature Formula

The formula for the radius of curvature \(R\) is given by \(R = \frac{(1 + (y')^2)^{3/2}}{|y''|}\). Substitute \(y' = 2\) and \(y'' = 4\) into the formula and calculate: \[R = \frac{(1 + 2^2)^{3/2}}{|4|} = \frac{(1 + 4)^{3/2}}{4} = \frac{5^{3/2}}{4}.\]

Simplify and Calculate the Numerical Value

Calculate \(5^{3/2}\) which is \(\sqrt{5^3} = \sqrt{125} = 5\sqrt{5}\). Thus, the radius of curvature at \(x=\pi/4\) is \(\frac{5\sqrt{5}}{4}\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. This process is crucial for analyzing and understanding the behavior of curves. In the given exercise, where we need to find the radius of curvature, the first step is differentiation. The function provided is \(y = \tan x\).
To find the first derivative \(y'\), we use the rule for differentiating the tangent function:

• The derivative of \(\tan x\) is \(\sec^2 x\).

The first derivative \(y' = \sec^2 x\) tells us how steep the curve is at any point \(x\).
Next, for further calculations such as curvature, we find the second derivative, which involves applying the chain rule to \(\sec^2 x\). This step gives us \(y'' = 2 \sec^2 x \cdot \sec x \tan x = 2 \sec^3 x \tan x\).
Getting the second derivative correctly is important as it provides information about the concavity and the curvature of the tangent curve.

Trigonometric Functions

Trigonometric functions like tangent, sine, and cosine appear frequently in calculus and are essential for solving various mathematical problems, like those involving curves. In this exercise, we work with the tangent function, \(y = \tan x\). This function describes a repeating pattern of ups and downs known as periodicity.
Key points to remember about trigonometric functions include:

• They have infinite repetitions, so they don't have a point where they flatline; instead, they continually oscillate between values.
• Tangent, specifically, has vertical asymptotes, where the function isn't defined, and it experiences rapid changes around those points.

When dealing with trigonometric derivatives, knowing their derivatives by heart helps. For tangent, the derivative is the square of the secant function, \(\sec^2 x\), another trigonometric function. At \(x = \pi/4\), the secant function, \(\sec \theta\), simplifies calculations due to its specific values at common angles.
Using these properties makes solving curves like \(y = \tan x\) more approachable.

Calculus Problem Solving

Problem-solving in calculus often involves a step-by-step approach where initially complex expressions are simplified using derivatives and known formulas. In this exercise, the goal is to find the radius of curvature of a curve at a specific point. Curvature tells us how sharply a curve bends at a particular point.
The process involves:

• Finding both the first and second derivatives of the function, which describe how the function changes.
• Substituting the derivatives at the given point \(x = \pi / 4\) to evaluate them in real number terms.
• Using the formula for the radius of curvature \(R = \frac{(1 + (y')^2)^{3/2}}{|y''|}\) which combines the derivatives into a single, meaningful measure.

In our example, substituting \(y' = 2\) and \(y'' = 4\) into the formula, we calculate \[R = \frac{(1 + 2^2)^{3/2}}{|4|} = \frac{5^{3/2}}{4}\]. This result indicates the radius of curvature, essentially describing the size of the curve's 'circle' at that point. Solving such problems develops a deeper understanding of how functions behave and change, which is a cornerstone of calculus.