Textbook Question
For Exercises 55 and 56, evaluate each limit by first converting each to a derivative at a particular x-value.
lim (x → 1) (x⁵⁰ − 1) / (x − 1)
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Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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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 3.7.42
Normal lines parallel to a line Find the normal lines to the curve xy + 2x – y = 0 that are parallel to the line 2x + y = 0.
Verified step by step guidance1
First, find the slope of the given line 2x + y = 0. Rewrite it in slope-intercept form (y = mx + b) to identify the slope. Solving for y gives y = -2x, so the slope (m) is -2.
Next, find the slope of the tangent line to the curve xy + 2x - y = 0. To do this, implicitly differentiate the equation with respect to x. Differentiate each term: d(xy)/dx + d(2x)/dx - d(y)/dx = 0.
Apply the product rule to differentiate xy: d(xy)/dx = x(dy/dx) + y. Differentiate the other terms: d(2x)/dx = 2 and d(y)/dx = dy/dx. Substitute these into the differentiated equation: x(dy/dx) + y + 2 - dy/dx = 0.
Solve for dy/dx to find the slope of the tangent line: dy/dx = (y + 2)/(1 - x). The slope of the normal line is the negative reciprocal of the tangent slope, so it is -(1 - x)/(y + 2).
Set the slope of the normal line equal to the slope of the given line (which is -2) and solve for x and y: -(1 - x)/(y + 2) = -2. This will give you the points on the curve where the normal lines are parallel to the given line.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Normal Line
A normal line to a curve at a given point is a line perpendicular to the tangent line at that point. To find the normal line, first determine the slope of the tangent line, then use the negative reciprocal of this slope for the normal line. This concept is crucial for understanding how to find lines that are perpendicular to a curve at a specific point.
Recommended video:
05:13
Slopes of Tangent Lines
Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. For the equation xy + 2x – y = 0, implicit differentiation helps find the slope of the tangent line by differentiating both sides with respect to x, treating y as a function of x. This is essential for determining the slope needed to find the normal line.
Recommended video:
05:14Finding The Implicit Derivative
Parallel Lines
Parallel lines have the same slope. To find normal lines to the curve that are parallel to a given line, such as 2x + y = 0, identify the slope of the given line. Since the slope of 2x + y = 0 is -2, the normal line to the curve must also have a slope of -2. This concept is key to ensuring the normal lines are parallel to the specified line.
Recommended video:
05:13Slopes of Tangent Lines
Related Practice
Textbook Question
Differential Estimates of Change
In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.
The change in the volume V = x³ of a cube when the edge lengths change from x₀ to x₀ + dx
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Textbook Question
In Exercises 19–22, find the slope of the curve at the point indicated.
y = (x − 1) / (x + 1), x = 0
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Textbook Question
Estimating height of a building A surveyor, standing 30 ft from the base of a building, measures the angle of elevation to the top of the building to be 75°. How accurately must the angle be measured for the percentage error in estimating the height of the building to be less than 4%?
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Textbook Question
One-Sided Derivatives
Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.
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Textbook Question
Graphs
Match the functions graphed in Exercises 27–30 with the derivatives graphed in the accompanying figures (a)–(d).
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