Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

4. Applications of Derivatives

Implicit Differentiation

Calculus

4. Applications of Derivatives

Implicit Differentiation

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4:16 minutes

Problem 3.7.33

Textbook Question

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

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.

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.

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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Related Practice

Textbook Question

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

x²/³ + y²/³ = 1

Textbook Question

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

y² – 2x = 1 – 2y

Textbook Question

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

2√y = x – y

Textbook Question

In Exercises 29 and 30, find the slope of the curve at the given points.

y² + x² = y⁴ – 2x at (–2,1) and (–2,–1)

Textbook Question

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.

y = 2 sin(πx – y), (1,0)

Textbook Question

Parallel tangent lines Find the two points where the curve x² + xy + y² = 7 crosses the x-axis, and show that the tangent lines to the curve at these points are parallel. What is the common slope of these tangent lines?

Textbook Question

The eight curve Find the slopes of the curve y⁴ = y² – x² at the two points shown here.

Textbook Question

The folium of Descartes (See Figure 3.27)

b. At what point other than the origin does the folium have a horizontal tangent line?

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Slopes, Tangent Lines, and Normal LinesIn 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)

Key Concepts

Implicit Differentiation

Tangent Line

Normal Line

Watch next

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)