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Chapter 3: Problem 27

In Exercises 27 and \(28,\) find the slope of the curve at the given points. \begin{equation} y^{2}+x^{2}=y^{4}-2 x \text { at }(-2,1) \text { and }(-2,-1) \end{equation}

Short Answer

Expert verified
At (-2, 1), slope = -1; at (-2, -1), slope = 1.

Step by step solution

01

Differentiate Implicitly

The given equation is \( y^2 + x^2 = y^4 - 2x \). To find the slope of the curve at a point, we need to differentiate both sides of the equation with respect to \( x \) using implicit differentiation. This gives:\[ \frac{d}{dx}(y^2) + \frac{d}{dx}(x^2) = \frac{d}{dx}(y^4) - \frac{d}{dx}(2x) \]Applying the chain rule:- \( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \)- \( \frac{d}{dx}(x^2) = 2x \)- \( \frac{d}{dx}(y^4) = 4y^3 \frac{dy}{dx} \)- \( \frac{d}{dx}(2x) = 2 \)So:\[ 2y \frac{dy}{dx} + 2x = 4y^3 \frac{dy}{dx} - 2 \]
02

Solve for \( \frac{dy}{dx} \)

Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \):\[ 2y \frac{dy}{dx} - 4y^3 \frac{dy}{dx} = -2 - 2x \]Factor out \( \frac{dy}{dx} \):\[ \frac{dy}{dx} (2y - 4y^3) = -2 - 2x \]Now, solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{-2 - 2x}{2y - 4y^3} \]
03

Evaluate the Derivative at Given Point (-2, 1)

Substitute \( x = -2 \) and \( y = 1 \) into the derivative:\[ \frac{dy}{dx} = \frac{-2 - 2(-2)}{2(1) - 4(1)^3} = \frac{-2 + 4}{2 - 4} = \frac{2}{-2} = -1 \]
04

Evaluate the Derivative at Given Point (-2, -1)

Substitute \( x = -2 \) and \( y = -1 \) into the derivative:\[ \frac{dy}{dx} = \frac{-2 - 2(-2)}{2(-1) - 4(-1)^3} = \frac{-2 + 4}{-2 + 4} = \frac{2}{2} = 1 \]

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Key Concepts

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

Slope of a Curve
When we talk about the "slope of a curve," we're essentially referring to how steep the curve is at a given point. It tells us how quickly the function is changing. To find the slope at a particular point, especially when dealing with implicit equations, we use derivatives. The derivative, denoted as \( \frac{dy}{dx} \), indicates the rate of change of the curve at that point. Here's how it connects:
  • The slope shows how much \( y \) changes for a small change in \( x \).
  • When the slope is positive, the curve rises to the right; when it's negative, it falls to the right.
  • With implicit differentiation, we find \( \frac{dy}{dx} \) without solving for \( y \) first, which is incredibly useful for curves defined implicitly, like \( y^2 + x^2 = y^4 - 2x \).
In short, finding the slope of a curve helps us understand the behavior of the function at specific points.
Chain Rule
The Chain Rule is a fundamental concept in calculus used to differentiate compositions of functions. In implicit differentiation, it allows us to differentiate expressions where one variable is a function of another, even if it's not explicitly solved for.Here's how the Chain Rule works:
  • Think of the variable \( y \) as a function of \( x \), so when you differentiate \( y^n \), you apply the chain rule: differentiate \( y^n \) with respect to \( y \), then multiply by \( \frac{dy}{dx} \).
  • In our exercise, when differentiating terms like \( y^2 \) and \( y^4 \), you use the chain rule to get \( 2y \frac{dy}{dx} \) and \( 4y^3 \frac{dy}{dx} \), respectively.
  • This technique helps us manage complex differentiations without having to simplify or rearrange the equations first.
When you think of the Chain Rule, remember it's your go-to tool for navigating through nested or composite functions smoothly.
Implicit Function Theorem
The Implicit Function Theorem provides a way to understand equations that aren't explicitly solved for one variable. It's particularly helpful when dealing with implicit differentiation, as it assures us that, locally, such equations can behave like explicit functions.Highlights of the Implicit Function Theorem:
  • It ensures that for equations of the form \( F(x,y) = 0 \), you can (under certain conditions) solve for one variable in terms of the other near a specific point.
  • This is pivotal in our example, where the curve is defined implicitly by \( y^2 + x^2 = y^4 - 2x \), meaning \( y \) isn't isolated on one side.
  • The theorem guarantees the differentiability of such functions, giving us the confidence to find \( \frac{dy}{dx} \) using implicit differentiation.
In essence, the Implicit Function Theorem is our reliable support, ensuring that implicit differentiation is not only possible but mathematically sound.

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Most popular questions from this chapter

Diagonals If \(x, y,\) and \(z\) are lengths of the edges of a rectangular box, the common length of the box's diagonals is \(s=\) \(\sqrt{x^{2}+y^{2}+z^{2}}\) a. Assuming that \(x, y,\) and \(z\) are differentiable functions of \(t\) how is \(d s / d t\) related to \(d x / d t, d y / d t,\) and \(d z / d t\) ? b. How is \(d s / d t\) related to \(d y / d t\) and \(d z / d t\) if \(x\) is constant? c. How are \(d x / d t, d y / d t,\) and \(d z / d t\) related if \(s\) is constant?

Assume that \(y=5 x\) and \(d x / d t=2 .\) Find \(d y / d t.\)

$$If r=\sin (f(t)), f(0)=\pi / 3, and f^{\prime}(0)=4, then what is d r / d t at t=0 ?$$

Use a CAS to perform the following steps in Exercises \(53-60 .\) \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x^{3}-x y+y^{3}=7, \quad P(2,1) \end{equation}

The edge \(x\) of a cube is measured with an error of at most 0.5\(\%\) . What is the maximum corresponding percentage error in computing the cube's $$\text{a. surface area?} \quad \text{b. volume?}$$
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