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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 3.85
Chapter 3, Problem 3.85

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.


xy + 2x - 5y = 2, (3, 2)

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1
First, we need to find the derivative of the given equation with respect to x. The equation is xy + 2x - 5y = 2. Use implicit differentiation to differentiate both sides of the equation with respect to x.
Differentiate the left side: For the term xy, use the product rule: d(xy)/dx = x(dy/dx) + y. For the term 2x, the derivative is 2. For the term -5y, the derivative is -5(dy/dx).
Combine the derivatives: x(dy/dx) + y + 2 - 5(dy/dx) = 0. Rearrange this to solve for dy/dx, which represents the slope of the tangent line.
Substitute the given point (3, 2) into the derivative to find the slope of the tangent line at that point. This will give you the specific slope value needed for the tangent line equation.
Use the point-slope form of a line, y - y1 = m(x - x1), where m is the slope found in the previous step and (x1, y1) is the point (3, 2), to write the equation of the tangent line. For the normal line, use the negative reciprocal of the tangent slope and the same point to find its equation.

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Key Concepts

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

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function defined implicitly by an equation involving both x and y. In this case, we differentiate the equation xy + 2x - 5y = 2 with respect to x, treating y as a function of x. This allows us to find dy/dx, which is essential for determining the slope of the tangent line at a specific point.
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Finding The Implicit Derivative

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. To find the equation of the tangent line, we use the point-slope form of a line, which requires the slope (found from the derivative) and the coordinates of the point of tangency.
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Slopes of Tangent Lines

Normal Line

The normal line to a curve at a given point is a line that is perpendicular to the tangent line at that point. Its slope is the negative reciprocal of the slope of the tangent line. To find the equation of the normal line, we again use the point-slope form, substituting the coordinates of the point and the slope of the normal line, which is derived from the tangent line's slope.
Recommended video:
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Slopes of Tangent Lines
Related Practice
Textbook Question

Moving along a parabola A particle moves along the parabola y = x² in the first quadrant in such a way that its x-coordinate (measured in meters) increases at a steady 10 m/sec. How fast is the angle of inclination θ of the line joining the particle to the origin changing when x = 3 m?

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Textbook Question

Slopes on the graph of the tangent function Graph y = tan x and its derivative together on (−π/2, π/2). Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers.

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Textbook Question

Differentiating Implicitly


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


x²(x – y)² = x² – y²

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Textbook Question

The devil’s curve (Gabriel Cramer, 1750) Find the slopes of the devil’s curve y⁴ – 4y² = x⁴ – 9x² at the four indicated points.


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Textbook Question

Normal lines to a parabola Show that if it is possible to draw three normal lines from the point (a, 0) to the parabola x = y² shown in the accompanying diagram, then a must be greater than 1/2. One of the normal lines is the x-axis. For what value of a are the other two normal lines perpendicular?


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Textbook Question

In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.


x³/² + 2y³/² = 17, (1, 4)

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