Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 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
- 10. Physics Applications of Integrals 2h 22m
4. Applications of Derivatives
Implicit Differentiation
Problem 3.8.82a
Textbook Question
79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>
a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)
(x²+y²)² = 25/3 (x²-y²); (x0,y0) = (2,-1) (lemniscate of Bernoulli)

1
First, understand the problem: We need to find the equations of the tangent and normal lines to the curve at a specific point. The curve is given by the equation \((x^2 + y^2)^2 = \frac{25}{3}(x^2 - y^2)\), and the point is \((x_0, y_0) = (2, -1)\).
To find the tangent line, we need the derivative of the curve at the point \((2, -1)\). Start by differentiating the given equation implicitly with respect to \(x\). This involves using the chain rule and implicit differentiation techniques.
After differentiating, substitute \(x = 2\) and \(y = -1\) into the derivative to find the slope \(m\) of the tangent line at the point \((2, -1)\).
Use the point-slope form of the equation of a line, \(y - y_0 = m(x - x_0)\), to write the equation of the tangent line. Here, \(m\) is the slope found in the previous step, and \((x_0, y_0)\) is \((2, -1)\).
The normal line is perpendicular to the tangent line. Therefore, its slope is the negative reciprocal of the tangent line's slope. Use the point-slope form again with this new slope to write the equation of the normal line at the point \((2, -1)\).

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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, which represents the instantaneous rate of change of the function. To find the equation of the tangent line, one typically uses the point-slope form of a line, incorporating the slope derived from the derivative.
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Normal Line
The normal line at a point on a curve is perpendicular to the tangent line at that same point. Its slope is the negative reciprocal of the slope of the tangent line. To find the equation of the normal line, one can use the point-slope form as well, substituting the point coordinates and the normal slope. This line represents the direction in which the curve is not changing at that point.
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Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. In cases like the given curve, where y is not isolated, one differentiates both sides of the equation with respect to x, applying the chain rule as necessary. This method allows for finding the derivative of y with respect to x, which is essential for determining the slopes of the tangent and normal lines.
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