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Chapter 3: Problem 74

In Exercises 73 and \(74,\) find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x )\) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y )\) . How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. $$ x^{3}+y^{2}=\sin ^{2} y $$

Short Answer

Expert verified
\(\frac{dy}{dx} = \frac{-3x^2}{2y - \sin(2y)}\) and \(\frac{dx}{dy} = \frac{\sin(2y) - 2y}{3x^2}\) are negative reciprocals, showing perpendicularity.

Step by step solution

01

Differentiate Implicitly with respect to x

Given the equation \(x^3 + y^2 = \sin^2 y\), differentiate both sides with respect to \(x\). This gives: 1. Differentiate \(x^3\) to get \(3x^2\).2. Use the chain rule on \(y^2\) to get \(2y \frac{dy}{dx}\).3. Differentiate \(\sin^2 y\) using the chain rule to get \(2\sin y \cos y \frac{dy}{dx}\) or \(\sin(2y) \frac{dy}{dx}\) using the double angle formula.This results in the equation: \[ 3x^2 + 2y \frac{dy}{dx} = \sin(2y) \frac{dy}{dx}. \]
02

Solve for dy/dx

Rearrange the terms in the equation to isolate \(\frac{dy}{dx}\):\[ 2y \frac{dy}{dx} - \sin(2y) \frac{dy}{dx} = -3x^2. \]Factor out \(\frac{dy}{dx}\):\[ \frac{dy}{dx} (2y - \sin(2y)) = -3x^2. \]Now, solve for \(\frac{dy}{dx}\):\[ \frac{dy}{dx} = \frac{-3x^2}{2y - \sin(2y)}. \]
03

Differentiate Implicitly with respect to y

Differentiate the equation \(x^3 + y^2 = \sin^2 y\) with respect to \(y\):1. Differentiate \(x^3\) using the chain rule to obtain \(3x^2 \frac{dx}{dy}\).2. Differentiate \(y^2\) to get \(2y\).3. Differentiate \(\sin^2 y\) to get \(2\sin y \cos y\) or \(\sin(2y)\).This gives the equation:\[ 3x^2 \frac{dx}{dy} + 2y = \sin(2y). \]
04

Solve for dx/dy

Rearrange the terms to isolate \(\frac{dx}{dy}\):\[ 3x^2 \frac{dx}{dy} = \sin(2y) - 2y. \]Then solve for \(\frac{dx}{dy}\):\[ \frac{dx}{dy} = \frac{\sin(2y) - 2y}{3x^2}. \]
05

Analyze the Relationship

The expressions for \(\frac{dy}{dx}\) and \(\frac{dx}{dy}\) are reciprocals except for a negative sign. Specifically,\[ \frac{dy}{dx} \cdot \frac{dx}{dy} = -1, \] indicating that these derivatives are negative reciprocals of each other. Geometrically, this means that the tangent lines in each respective perspective are perpendicular to each other.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Chain Rule in Implicit Differentiation
When working with implicit differentiation, the chain rule is an essential tool. It helps us differentiate composite functions, which is crucial when variables are not explicitly separated. In this context, we have both functions of \(x\) and \(y\) within the equation \(x^3 + y^2 = \sin^2 y\).
  • The chain rule allows us to tackle each part of an equation that involves a function in terms of another function. For instance, when differentiating \(y^2\) in terms of \(x\), we treat \(y\) as a function of \(x\), resulting in the term \(2y \frac{dy}{dx}\).
  • Similarly, to differentiate \(\sin^2 y\) with respect to \(x\), we apply the chain rule: \(2 \sin y \cos y \frac{dy}{dx}\) or \(\sin(2y) \frac{dy}{dx}\), thanks to the double angle formula.
  • This application is pivotal because directly differentiating each part accurately represents how small changes in \(x\) affect \(y\), and vice versa.
By using the chain rule effectively, we gather the necessary components to solve for \(\frac{dy}{dx}\) and \(\frac{dx}{dy}\) iteratively, which is crucial for understanding implicit relationships between the variables.
Geometric Interpretation of Reciprocal Derivatives
The relationship between \(\frac{dy}{dx}\) and \(\frac{dx}{dy}\) can be understood geometrically through how lines behave in the Cartesian plane. When these derivatives are negative reciprocals, it signifies a very special geometric relationship.
  • Concept of Negative Reciprocals: If two lines have slopes \(m_1\) and \(m_2\), and their product is \(-1\), \(m_1 \cdot m_2 = -1\), these lines are perpendicular to each other.
  • Interpretation on the Graph: For instance, if \(\frac{dy}{dx}\) is a slope of a line tangent to a curve in the \(x\)-perspective, \(\frac{dx}{dy}\) reveals how the same curve behaves in the \(y\)-perspective.
  • This means that the tangent line at any point on the curve has a direct reciprocal relationship with the normal line, indicating perpendicularity, reflecting a particular symmetry or orthogonality based on how we manipulate the relationships of each variable.
This geometric understanding leads us to the conclusion that implicit differentiation not only calculates slopes but also uncovers deeper geometric properties of curves.
The Significance of Reciprocal Derivatives
Reciprocal derivatives, particularly when they involve a negative sign, unveil intriguing mathematical relationships that are significant beyond mere calculations. They relate two functions' interaction and reveal underlying mathematical truths.
  • Reciprocal Relationship: In our example, the product \(\frac{dy}{dx} \cdot \frac{dx}{dy} = -1\) showcases that each derivative is the negative reciprocal of the other.
  • Contextual Analysis: This means that even as \(x\) changes with respect to \(y\) or \(y\) with respect to \(x\), the rate at these changes maintain a constant opposition, interpreted as perpendicularity in geometric terms.
  • Wider Implications: Having knowledge of this reciprocal relationship is crucial in fields such as physics, engineering, and more, where understanding the interplay between rates of change could impact real-world applications and analyses.
This exploration of reciprocal derivatives solidifies the foundational understanding of relationships within implicit differentiation, providing critical insight into the geometric and functional qualities of complex equations.

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Most popular questions from this chapter

In Exercises \(87-94\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=2 t^{2}+3, \quad y=t^{4}, \quad t=-1 $$

Use a CAS to perform the following steps on the parametrized curves in Exercises \(113-116 .\) a. Plot the curve for the given interval of \(t\) values. b. Find \(d y / d x\) and \(d^{2} y / d x^{2}\) at the point \(t_{0}\) . c. Find an equation for the tangent line to the curve at the point defined by the given value \(t_{0} .\) Plot the curve together with the tangent line on a single graph. $$ \begin{array}{l}{x=2 t^{3}-16 t^{2}+25 t+5, \quad y=t^{2}+t-3, \quad 0 \leq t \leq 6} \\ {t_{0}=3 / 2}\end{array} $$

In Exercises \(81-86,\) find a parametrization for the curve. the ray (half line) with initial point \((-1,2)\) that passes through the point \((0,0)\)

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g .\) The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g .\) By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\) a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts \((b)\) and \((c)\) . b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a 100 -cm pendulum is moved from a location were \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001\) sec. Find \(d g\) and estimate the value of \(g\) at the new location.

In Exercises \(47-56,\) 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. $$ x^{2} y^{2}=9, \quad(-1,3) $$
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