Problem 44 - University Calculus: Early Transcendentals

Question

Find all points on the curve \(y=\cot x, 0

Step-by-Step Solution

Answer

The tangent line is parallel to the line \( y = -x \) at the point \( (\frac{\pi}{2}, 0) \).

Step 1: Differentiate the Function

To find where the tangent is parallel to a given line, we first need the derivative of the function. The derivative of the function will give us the slope of the tangent line. The function given is \( y = \cot x \). The derivative of \( \cot x \) is \( y' = -\csc^2 x \).

Step 2: Determine the Slope of the Given Line

The line given is \( y = -x \), which is in the slope-intercept form \( y = mx + b \). Comparing it to \( y = mx + b \), we see the slope \( m \) is \(-1\).

Step 3: Set Derivative Equal to Slope of the Line

Since we want the slope of the tangent line to be equal to \(-1\), set the derivative equal to \(-1\): \[-\csc^2 x = -1\] This simplifies to \[\csc^2 x = 1\].

Step 4: Solve for x

The equation \( \csc^2 x = 1 \) implies \( \csc x = \pm 1 \). Therefore, \( \sin x = \pm 1 \). Within the interval \( 0 < x < \pi \), \( \sin x = 1 \) at \( x = \frac{\pi}{2} \). However, \( \sin x = -1 \) does not occur within \( 0 < x < \pi \). Therefore, the only solution is \( x = \frac{\pi}{2} \).

Step 5: Calculate the Point on the Curve

Substitute \( x = \frac{\pi}{2} \) back into the original function to find the point on the curve: \[ y = \cot \left(\frac{\pi}{2}\right) = 0 \] Hence, the point is \( \left( \frac{\pi}{2}, 0 \right) \).

Step 6: Sketch the Curve and Tangents

Draw the curve of \( y = \cot x \) between \( 0 < x < \pi \). At \( x = \frac{\pi}{2} \), sketch the tangent line, which has a slope of \(-1\). The equation of this tangent line will be \( y = -x + \frac{\pi}{2} \). Clearly label both the curve and the tangent.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used in calculus when we need to find the derivative of a function that is not explicitly solved for one of the variables. In many scenarios, the function may be given in a form where variables are intermingled, such as an equation involving both \( x \) and \( y \).

To differentiate an equation implicitly, follow these steps:

• Differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \) (i.e., applying the chain rule when differentiating terms with \( y \)).
• When differentiating a term involving \( y \), use the chain rule: if you differentiate \( y \), replace it with \( y' \) (or \( \frac{dy}{dx} \)).
• Solve for \( y' \), which represents the slope of the tangent line at any point \( (x, y) \) on the curve.

By using implicit differentiation, you can handle more complex equations accurately, which is particularly useful in problems involving curves defined by implicit equations.

Derivative of Trigonometric Functions

Trigonometric functions have their own specific rules for differentiation that are vital in calculus, especially when dealing with periodic functions. In the given problem, the function \( y = \cot x \) requires the derivative of the cotangent function.

The derivative of \( \cot x \) is given by:

• \( \frac{d}{dx}[\cot x] = -\csc^2 x \)

This means that the rate of change or slope of the cotangent function at any point is \( -\csc^2 x \). Knowing the derivative formulae of trigonometric functions like sine, cosine, and tangent is essential, as they frequently appear in calculus problems.

Derivatives of trigonometric functions help determine the behavior of curves, determine maxima or minima, and find tangent lines, as in the example problem.

Tangent Line Equation

The equation of a tangent line to a curve gives us a linear approximation of the curve at a specific point. Here's how you find this equation when given a function:

• First, compute the derivative of the function to find the slope of the tangent at the desired point.
• Substitute the point's \( x \)-value into the derivative to get the slope \( m \) of the tangent line at that point.
• Use the point-slope form of a linear equation: \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point of tangency.

In the example above, we set the derivative of the function \( y = \cot x \) equal to \(-1\) because we want the tangent's slope to be parallel to \( y = -x \). This led to the point \( (\frac{\pi}{2}, 0) \). Thus, the tangent line equation becomes \( y = -x + \frac{\pi}{2} \).

Understanding how to find and write the tangent line equations is crucial for analyzing the local behavior of functions, estimating values, and solving applied mathematics problems.