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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 3.7.33
Chapter 3, Problem 3.7.33

Slopes, Tangent Lines, and Normal Lines


In Exercises 31–40, verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.


x²y² = 9, (–1,3)

Verified step by step guidance
1
First, verify that the point (-1, 3) is on the curve by substituting x = -1 and y = 3 into the equation x²y² = 9. Calculate (-1)²(3)² and check if it equals 9.
To find the tangent line, start by differentiating the equation x²y² = 9 implicitly with respect to x. Use the product rule and chain rule to differentiate both sides.
After differentiating, solve for dy/dx to find the slope of the tangent line at the point (-1, 3). Substitute x = -1 and y = 3 into the derivative to find the specific slope at that point.
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 (-1, 3), 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. Substitute this new slope and the point (-1, 3) into the point-slope form to find the equation of the normal line.

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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 the equation x²y² = 9, both x and y are intertwined, requiring implicit differentiation to find dy/dx, which is essential for determining the slope of the tangent line at a given 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 curve at that point. For the curve x²y² = 9 at the point (-1, 3), the tangent line can be found using the derivative obtained through implicit differentiation.
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Slopes of Tangent Lines

Normal Line

A normal line to a curve at a given point is perpendicular to the tangent line at that point. The slope of the normal line 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 41–44, determine whether the piecewise-defined function is differentiable at x = 0.


g(x) = { x²/³, x ≥ 0

x¹/³, x < 0

Textbook Question

Find the derivatives of the functions in Exercises 1–42.


𝔂 = x³ - 3 (x² + π²)

1
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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 lateral surface area S = πr√(r² + h²) of a right circular cone when the radius changes from r₀ to r₀ + dr and the height does not change

Textbook Question

a. Graph the function


ƒ(x) = { x², -1 ≤ x < 0

{ -x², 0 ≤ x ≤ 1.


b. Is ƒ continuous at x = 0?

c. Is ƒ differentiable at x = 0?


Give reasons for your answers.

Textbook Question

Find the derivatives of the functions in Exercises 1–42.


𝔂 = 1 x² csc 2

2 x

Textbook Question

Is there a value of b that will make


g(x) = { x + b, x < 0

cos x, x ≥ 0


continuous at x = 0? Differentiable at x = 0? Give reasons for your answers.