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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.77
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.

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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).

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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.
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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.
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Slopes 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.
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Slopes of Tangent Lines
Related Practice
Textbook Question

Tolerance The height and radius of a right circular cylinder are equal, so the cylinder’s volume is V = πh³. The volume is to be calculated with an error of no more than 1% of the true value. Find approximately the greatest error that can be tolerated in the measurement of h, expressed as a percentage of h.

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Textbook Question

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Use implicit differentiation to find dy/dx in Exercises 1–14.


x³ + y³ = 18xy

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Textbook Question

Differentiating Implicitly


Use implicit differentiation to find dy/dx in Exercises 1–14.


x²y + xy² = 6

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Find the derivatives of the functions in Exercises 19–40.


q = sin(t / (√t + 1))

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Textbook Question

Find the value of a that makes the following function differentiable for all x-values.


g(x) = { ax, if x < 0

x² − 3x, if x ≥ 0

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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

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