Skip to main content
Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.5.47b
Chapter 3, Problem 3.5.47b

In Exercises 47 and 48, find an equation for


(b) the horizontal tangent line to the curve at Q.


<IMAGE>

Verified step by step guidance
1
Step 1: Identify the given curve equation, which is y = 4 + cot(x) - 2csc(x). The goal is to find the horizontal tangent line at point Q. A horizontal tangent occurs when the derivative of the curve, dy/dx, equals zero.
Step 2: Compute the derivative of the curve y = 4 + cot(x) - 2csc(x) with respect to x. Use the derivatives of trigonometric functions: d(cot(x))/dx = -csc^2(x) and d(csc(x))/dx = -csc(x)cot(x).
Step 3: Substitute the derivatives into dy/dx. The derivative becomes dy/dx = -csc^2(x) + 2csc(x)cot(x). Simplify this expression to make it easier to analyze.
Step 4: Set dy/dx = 0 to find the x-coordinate where the tangent is horizontal. Solve the equation -csc^2(x) + 2csc(x)cot(x) = 0 for x. This involves factoring or using trigonometric identities.
Step 5: Once the x-coordinate is found, substitute it back into the original curve equation y = 4 + cot(x) - 2csc(x) to find the corresponding y-coordinate. The equation of the horizontal tangent line will be y = [value of y at Q].

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Horizontal Tangent Line

A horizontal tangent line to a curve at a point indicates that the slope of the tangent at that point is zero. This occurs when the derivative of the function at that point equals zero. Finding a horizontal tangent involves solving for when the derivative of the function is zero, which helps identify points where the curve has a local maximum or minimum.
Recommended video:
05:13
Slopes of Tangent Lines

Derivative of Trigonometric Functions

The derivative of a function provides the slope of the tangent line at any point on the curve. For trigonometric functions like cotangent and cosecant, the derivatives are -csc^2(x) and -csc(x)cot(x), respectively. Understanding these derivatives is crucial for finding where the slope of the tangent line is zero, which is necessary for identifying horizontal tangents.
Recommended video:
6:04
Introduction to Trigonometric Functions

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are potential locations for local maxima, minima, or points of inflection. In the context of finding horizontal tangents, critical points are where the derivative equals zero, indicating a potential horizontal tangent line at those points.
Recommended video:
04:50
Critical Points
Related Practice
Textbook Question

By computing the first few derivatives and looking for a pattern, find the following derivatives.


a. d⁹⁹⁹/dx⁹⁹⁹ (cos x)

Textbook Question

Faster than a calculator Use the approximation (1 + x)ᵏ ≈ 1 + kx to estimate the following.


a. (1.0002)⁵⁰

Textbook Question

Quadratic approximations


b. Find the quadratic approximation to f(x) = 1/(1 − x) at x = 0.

Textbook Question

Consider the function f graphed here. The domain of f is the interval [−4, 6] and its graph is made of line segments joined end to end.


" style="" width="350">


b. Graph the derivative of f. The graph should show a step function.

Textbook Question

Suppose that the functions f and g and their derivatives with respect to x have the following values at x = 0 and x = 1.


" style="" width="250">


Find the derivatives with respect to x of the following combinations at the given value of x.


b. f(x)g³(x), x = 0

Textbook Question

The folium of Descartes (See Figure 3.27)


b. At what point other than the origin does the folium have a horizontal tangent line?