Textbook Question
In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
f(x) = √(x + 1), (8, 3)
229
views
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.77
Find the points on the curve y = tan x, -π/2 < x < π/2, where the normal line is parallel to the line y = -x/2. Sketch the curve and normal lines together, labeling each with its equation.
Verified step by step guidance1
Identify the slope of the given line y = -x/2. The slope is -1/2.
Find the derivative of the curve y = tan(x) to determine the slope of the tangent line at any point x. The derivative is dy/dx = sec^2(x).
Since the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the tangent line's slope. Set the slope of the normal line equal to -1/2 to find the x-values where this condition holds.
Solve the equation -1/(sec^2(x)) = -1/2 to find the x-values. This simplifies to sec^2(x) = 2, which further simplifies to cos^2(x) = 1/2.
Solve for x in the interval -π/2 < x < π/2 where cos^2(x) = 1/2. These x-values are the points where the normal line is parallel to y = -x/2. Use these x-values to find the corresponding y-values on the curve y = tan(x).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative of a Function
The derivative of a function at a point gives the slope of the tangent line to the curve at that point. For the function y = tan(x), the derivative is y' = sec^2(x). Understanding this concept is crucial for finding where the tangent line has a specific slope, which is necessary for determining where the normal line is parallel to a given line.
Recommended video:
06:30
Derivatives of Other Trig Functions
Normal Line to a Curve
A normal line to a curve at a given point is a line perpendicular to the tangent line at that point. If the slope of the tangent line is m, the slope of the normal line is -1/m. In this problem, we need to find where the normal line has a slope of -1/2, which means the tangent line must have a slope of 2.
Recommended video:
05:13Slopes of Tangent Lines
Parallel Lines
Parallel lines have the same slope. In this problem, the normal line must be parallel to the line y = -x/2, which has a slope of -1/2. By setting the slope of the normal line equal to -1/2, we can find the points on the curve where this condition is satisfied, helping us solve the problem.
Recommended video:
05:13Slopes of Tangent Lines
Related Practice
Textbook Question
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
x³ + y³ = 18xy
316
views
Textbook Question
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
x²y + xy² = 6
252
views
Textbook Question
Find the derivatives of the functions in Exercises 19–40.
q = sin(t / (√t + 1))
260
views
Textbook Question
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
x = sec y
285
views
Textbook Question
Derivative Calculations
In Exercises 1–8, given y = f(u) and u = g(x), find dy/dx = f'(g(x)) g'(x).
y = sin u, u = 3x + 1
298
views