Problem 111 - Algebra and Trigonometry

Question

Find the average rate of change of \(y=\tan \left(\sec ^{-1} x\right)\) over the interval \(\left[\frac{\sqrt{10}}{3}, \sqrt{10}\right] .\)

Step-by-Step Solution

Answer

\( \frac{2\sqrt{10}}{5} \)

Step 1 - Identify the formula for average rate of change

The average rate of change of a function over an interval \[ [a, b] \] is given by \[ \frac{f(b) - f(a)}{b - a} \]. Here, the function is \[ y = \tan(\text{sec}^{-1} x) \], and the interval is \[ \bigg[ \frac{\sqrt{10}}{3}, \sqrt{10} \bigg] \].

Step 2 - Calculate \[ f(a) \]

To find \[ f(a) \], calculate \[ y = \tan(\text{sec}^{-1} \left( \frac{\sqrt{10}}{3} \right)) \]: \[ \text{sec}^{-1} \left( \frac{\sqrt{10}}{3} \right) = \theta \] such that \[ \sec(\theta) = \frac{\sqrt{10}}{3} \]. Using the trigonometric identity \[ \tan(\theta) = \sqrt{\sec^2(\theta) - 1} \], we get \[ \tan(\theta) = \sqrt{ \left( \frac{\sqrt{10}}{3} \right)^2 - 1 } = \sqrt{ \frac{10}{9} - 1 } = \sqrt{ \frac{10 - 9}{9} } = \sqrt{ \frac{1}{9} } = \frac{1}{3} \]. Therefore, \[ y = \frac{1}{3} \].

Step 3 - Calculate \[ f(b) \]

Now find \[ f(b) \]: \[ y = \tan(\text{sec}^{-1} (\sqrt{10})) \]: \[ \text{sec}^{-1}(\sqrt{10}) = \theta \] such that \[ \sec(\theta) = \sqrt{10} \]. Using \[ \tan(\theta) = \sqrt{\sec^2(\theta) - 1} \], we get \[ \tan(\theta) = \sqrt{10 - 1} = \sqrt{9} = 3 \]. Therefore, \[ y = 3 \].

Step 4 - Apply the average rate of change formula

Using \[ f(a) = \frac{1}{3} \] and \[ f(b) = 3 \], and the interval endpoints \[ a = \frac{\sqrt{10}}{3} \] and \[ b = \sqrt{10} \], apply the formula: \[ \frac{f(b) - f(a)}{b - a} = \frac{3 - \frac{1}{3}}{\sqrt{10} - \frac{\sqrt{10}}{3}} = \frac{\frac{9}{3} - \frac{1}{3}}{\frac{3\sqrt{10} - \sqrt{10}}{3}} = \frac{\frac{8}{3}}{\frac{2\sqrt{10}}{3}} = \frac{8}{2\sqrt{10}} = \frac{4}{\sqrt{10}} = \frac{4\sqrt{10}}{10} = \frac{2\sqrt{10}}{5} \].

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are used to determine the angles when the values of trigonometric functions are known.
These are particularly useful in calculus and geometry.
Common inverse trigonometric functions include \(\text{sec}^{-1}(x)\), \(\text{tan}^{-1}(x)\), and \(\text{cos}^{-1}(x)\).
For example, \(\text{sec}^{-1}(x)\) gives an angle whose secant is x.
If \(\text{sec}(\theta) = x\), then \(\text{sec}^{-1}(x) = \theta\).
Understanding how to use these functions helps solve many trigonometric and calculus problems.

Tangent Function

The tangent function, denoted as \(\tan(x)\), gives the ratio of the sine and cosine of an angle.
This means \(\tan(x) = \frac{\text{sin}(x)}{\text{cos}(x)}\).
In other words, in a right triangle, the tangent of an angle is the ratio of the opposite side to the adjacent side.
This function is periodic with a period of \(\frac{\text{\text{π}}}{2}\), meaning it repeats its values every 90 degrees or \(\frac{\text{π}}{2}\) radians.
Understanding this function is crucial for solving different types of problems in trigonometry and calculus.

Secant Function

The secant function, denoted as \(\text{sec}(x)\), is the reciprocal of the cosine function, so \( \text{sec}(x) = \frac{1}{\text{cos}(x)} \).
It is undefined when \(\text{cos}(x) = 0\), because you cannot divide by zero.
The secant function helps in many applications involving triangles and circles.
It is periodic with a period of \(\frac{2\text{π}}{2}\), the same as the cosine function.
When solving problems involving angles and sides of triangles, the secant function often appears alongside other functions like sine, cosine, and tangent.