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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 3.86
Chapter 3, Problem 3.86

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.


(y - x)² = 2x + 4, (6, 2)

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First, recognize that the given equation \((y - x)^2 = 2x + 4\) is implicitly defined. To find the tangent line, we need to differentiate this equation with respect to \(x\).
Apply implicit differentiation to \((y - x)^2 = 2x + 4\). Differentiate both sides with respect to \(x\), remembering to use the chain rule for the left side: \(2(y - x)(\frac{dy}{dx} - 1) = 2\).
Solve the differentiated equation for \(\frac{dy}{dx}\) to find the slope of the tangent line at any point \((x, y)\). Substitute \(x = 6\) and \(y = 2\) into the differentiated equation to find the specific slope at the point \((6, 2)\).
Use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the slope found in the previous step and \((x_1, y_1)\) is the point \((6, 2)\), to write the equation of the tangent line.
To find the normal line, use the fact that the slope of the normal line is the negative reciprocal of the slope of the tangent line. Use the point-slope form again with this new slope to write the equation of the normal line at the point \((6, 2)\).

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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 functions that are not explicitly solved for one variable in terms of another. In this problem, the equation (y - x)² = 2x + 4 involves both x and y, requiring implicit differentiation to find dy/dx, which is essential for determining the slope of the tangent line at the given point.
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Finding The Implicit Derivative

Tangent Line

The tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. Its slope is equal to the derivative of the curve at that point. For the equation (y - x)² = 2x + 4 at the point (6, 2), the tangent line can be found using the derivative obtained through implicit differentiation.
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Slopes of Tangent Lines

Normal Line

The normal line to a curve at a given point is perpendicular to the tangent line at that point. Its slope is the negative reciprocal of the slope of the tangent line. Once the slope of the tangent line is determined, the normal line can be calculated, providing insight into the geometric properties of the curve at the specified point.
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Slopes of Tangent Lines
Related Practice
Textbook Question

In Exercises 41–58, find dy/dt.


y = (t⁻³/⁴ sin(t))⁴/³

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Textbook Question

Show that the line y = mx + b is its own tangent line at any point (x₀, mx₀ + b).

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Textbook Question

In Exercises 41–58, find dy/dt.


y = tan²(sin³(t))

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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 surface area S = 6x² of a cube when the edge lengths change from x₀ to x₀ + dx

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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

Implicit Differentiation


In Exercises 43–50, find by implicit differentiation.


xy + 2x + 3y = 1

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