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 3.8.80a
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.)
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)

1
First, understand that the problem involves finding the equations of the tangent and normal lines to the curve at a specific point. The curve given is x⁴ = 2x² + 2y², and the point is (x₀, y₀) = (2, 2).
To find the equation of the tangent line, we need to calculate the derivative of the curve with respect to x, which involves implicit differentiation since y is also a function of x. Start by differentiating both sides of the equation x⁴ = 2x² + 2y² with respect to x.
Apply implicit differentiation: For the left side, differentiate x⁴ to get 4x³. For the right side, differentiate 2x² to get 4x, and differentiate 2y² using the chain rule to get 4y(dy/dx). Set the derivatives equal: 4x³ = 4x + 4y(dy/dx).
Solve for dy/dx, which represents the slope of the tangent line at the point (x₀, y₀). Substitute x = 2 and y = 2 into the differentiated equation to find the specific slope at the given point.
Once you have the slope of the tangent line, use the point-slope form of a line equation, y - y₀ = m(x - x₀), where m is the slope you found, to write the equation of the tangent line. For the normal line, use the negative reciprocal of the tangent slope as the slope of the normal line and apply the point-slope form again.

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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 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, one can use the point-slope form as well, substituting the point coordinates and the normal slope, which is derived from the tangent slope.
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Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations that define y implicitly in terms of x, rather than explicitly as y = f(x). This method involves differentiating both sides of the equation with respect to x and applying the chain rule when differentiating terms involving y. It is particularly useful for finding derivatives of curves defined by equations like x⁴ = 2x² + 2y², where y cannot be easily isolated.
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