Problem 36 - Calculus Single Variable

Question

Find the tangent line to the graph of \(y=f(x)\) at \(P\). \(f(x)=\csc (x) / x, P=(\pi / 6,12 / \pi)\)

Step-by-Step Solution

Answer

The equation of the tangent line is \( y - \frac{12}{\pi} = m(x - \frac{\pi}{6}) \) where \( m \) is found from Step 5.

Step 1: Identify the Function and Derivative

The function given is \( f(x) = \frac{\csc(x)}{x} \). To find the tangent line, we first need the derivative of \( f(x) \), denoted as \( f'(x) \).

Step 2: Apply the Quotient Rule

The derivative of \( \frac{u}{v} \) is given by \( \frac{v'u - uv'}{v^2} \). Here, \( u = \csc(x) \) and \( v = x \). We calculate \( u' = -\csc(x)\cot(x) \) and \( v' = 1 \).

Step 3: Compute the Derivative

Plugging into the quotient rule formula, we have: \[ f'(x) = \frac{x(-\csc(x)\cot(x)) - \csc(x) \cdot 1}{x^2} = \frac{-x\csc(x)\cot(x) - \csc(x)}{x^2}. \] Simplify: \[ f'(x) = \frac{-\csc(x)(x\cot(x) + 1)}{x^2}. \]

Step 4: Evaluate the Derivative at the Point

Now, evaluate \( f'(x) \) at \( x = \frac{\pi}{6} \): \[ f'\left(\frac{\pi}{6}\right) = \frac{-\csc\left(\frac{\pi}{6}\right)\left(\frac{\pi}{6}\cot\left(\frac{\pi}{6}\right) + 1\right)}{\left(\frac{\pi}{6}\right)^2}. \] We know \( \csc\left(\frac{\pi}{6}\right) = 2 \) and \( \cot\left(\frac{\pi}{6}\right) = \sqrt{3} \). Plug these values in to compute.

Step 5: Simplify the Derivative Result

Substitute the trigonometric values: \[ f'\left(\frac{\pi}{6}\right) = \frac{-2\left(\frac{\pi}{6} \cdot \sqrt{3} + 1\right)}{\left(\frac{\pi}{6}\right)^2}. \] Simplify \[ = \frac{-2\left(\frac{\pi \sqrt{3}}{6} + 1\right)}{\frac{\pi^2}{36}} = \frac{-72\left(\frac{\pi \sqrt{3}}{6} + 1\right)}{\pi^2}. \]

Step 6: Determine the Equation of the Tangent Line

The equation of a line is \( y - y_1 = m(x - x_1) \). With \( m = f'\left(\frac{\pi}{6}\right) \), \( x_1 = \frac{\pi}{6} \), and \( y_1 = \frac{12}{\pi} \). Substitute these into the line equation to get: \[ y - \frac{12}{\pi} = f'\left(\frac{\pi}{6}\right)\left(x - \frac{\pi}{6}\right). \] Compute the slope from Step 5 to complete this equation.

Key Concepts

quotient rule

The quotient rule is used to differentiate functions that are expressed as a quotient, or ratio, of two other functions. This is particularly useful if you have a function in the form of \(\frac{u}{v}\), where both \(u\) and \(v\) are themselves functions of \(x\). The rule for finding the derivative \( \left( \frac{d}{dx} \right) \) of a quotient \(\frac{u}{v}\) is:

• \( \frac{v'u - uv'}{v^2} \)

In simpler terms, you differentiate the numerator (\(u\)) and the denominator (\(v\)) separately to find \(u'\) and \(v'\). Then, plug those into the quotient rule formula. Simply put, it's a way to break down more complex functions into manageable chunks. For the function \(f(x) = \frac{\csc(x)}{x}\), we identify \(u = \csc(x)\) and \(v = x\), then apply the quotient rule formula to find \(f'(x)\). This method is crucial when dealing with complex ratios, especially those involving trigonometric functions.

trigonometric functions

Trigonometric functions arise often in different fields of mathematics and engineering. In calculus, they frequently appear in both the functions that need differentiation and their derivatives. Common trigonometric functions include sine (\(\sin\)), cosine (\(\cos\)), tangent (\(\tan\)), and their reciprocals like cosecant (\(\csc\)), secant (\(\sec\)), and cotangent (\(\cot\)).

• The cosecant function, \(\csc(x)\), is the reciprocal of the sine function (\(\csc(x) = \frac{1}{\sin(x)}\)).
• The cotangent function, \(\cot(x)\), is the reciprocal of the tangent function (\(\cot(x) = \frac{1}{\tan(x)}\)).

Understanding these functions helps us in computing derivatives that contain them. For example, knowing the derivative of \(\csc(x)\) is \(-\csc(x)\cot(x)\) is vital when applying the quotient rule in the exercise. Evaluating these trigonometric functions at specific angle values, like \(\frac{\pi}{6}\), equips us with the necessary numeric results to substitute into formulas.

derivative evaluation

Derivative evaluation is the process of finding the derivative of a function at a specific point. This is crucial when you need to determine the slope of the tangent line to the graph at a given point \((x, y)\). The tangent line's slope is equal to the value of the derivative at that point.

• First, calculate the general derivative of the function using appropriate differentiation rules such as the quotient rule.
• Second, substitute the specific \(x\) value of interest, in this case, \(x = \frac{\pi}{6}\), into the derivative to find the slope \(m\) of the tangent line at that point.

For example, with \(f(x) = \frac{\csc(x)}{x}\), we found that the derivative \(f'(x)\) is \(\frac{-\csc(x)(x\cot(x) + 1)}{x^2}\). Evaluating this at \(x = \frac{\pi}{6}\) allows us to find the slope at the point \(P\). This slope \(f'\left(\frac{\pi}{6}\right)\) helps complete the formula for the tangent line's equation: \(y - y_1 = m(x - x_1)\). Knowing how to evaluate derivatives is essential for translating the abstract notion of a derivative into practical geometric interpretations.