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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.8.50b
Chapter 3, Problem 3.8.50b

45–50. Tangent lines Carry out the following steps. <IMAGE>
b. Determine an equation of the line tangent to the curve at the given point.
(x²+y²)²=25/4 xy²; (1, 2)

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1
First, identify the given implicit equation of the curve: \((x^2 + y^2)^2 = \frac{25}{4} xy^2\). We need to find the derivative \(\frac{dy}{dx}\) using implicit differentiation.
Differentiate both sides of the equation with respect to \(x\). For the left side, use the chain rule: \(2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx})\). For the right side, apply the product rule: \(\frac{25}{4} (y^2 + 2xy \frac{dy}{dx})\).
Set the derivatives equal: \(2(x^2 + y^2)(2x + 2y \frac{dy}{dx}) = \frac{25}{4} (y^2 + 2xy \frac{dy}{dx})\).
Substitute the given point \((1, 2)\) into the differentiated equation to solve for \(\frac{dy}{dx}\). This will give the slope of the tangent line at the point.
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 \((x_1, y_1) = (1, 2)\), to write the equation of the tangent 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 differentiate equations where the dependent and independent variables are not explicitly separated. In this case, the equation (x² + y²)² = (25/4)xy² involves both x and y, requiring us to differentiate both sides with respect to x while treating y as a function of x. This method allows us to find dy/dx, which is essential for determining the slope of the tangent line.
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Finding The Implicit Derivative

Tangent Line Equation

The equation of a tangent line at a given point on a curve can be expressed using the point-slope form: y - y₀ = m(x - x₀), where (x₀, y₀) is the point of tangency and m is the slope at that point. Once the derivative (slope) is calculated using implicit differentiation, it can be substituted into this formula along with the coordinates of the point (1, 2) to find the specific equation of the tangent line.
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Equations of Tangent Lines

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. In the context of implicit differentiation, it allows us to differentiate terms involving y, which is a function of x. For example, when differentiating y², we apply the chain rule to obtain 2y(dy/dx), which is crucial for correctly finding the derivative of the given implicit equation.
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Intro to the Chain Rule
Related Practice
Textbook Question

Witch of Agnesi Let y(x²+4)=8 (see figure). <IMAGE>

b. Find equations of all lines tangent to the curve y(x²+4)=8 when y=1.

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Textbook Question

62–65. {Use of Tech} Graphing f and f'

b. Compute and graph f'.

f(x)=e^−x tan^−1 x on [0,∞)

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Textbook Question

Explain why or why not. Determine whether the following statements are true and give an explanation or counterexample.


b. ln(x + 1) + ln(x − 1) = ln(x² − 1), for all x.

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Textbook Question

{Use of Tech} Power and energy The total energy in megawatt-hr (MWh) used by a town is given by E(t) = 400t+2400/π sin πt/12, where t≥0 is measured in hours, with t=0 corresponding to noon.

b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day?

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Textbook Question

Use a graphing utility to graph the curve and the tangent line on the same set of axes.

y = (x + 5) / (x - 1); a = 3

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Textbook Question

21–30. Derivatives

b. Evaluate f'(a) for the given values of a.

f(t) = 3t⁴; a= -2, 2

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