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Chapter 2: Problem 5

Differentiate implicitly to find \(d y / d x .\) Then find the slope of the curve at the given point. $$ x^{2}+y^{2}=1 ; \quad\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) $$

Short Answer

Expert verified
The slope of the curve at \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) is \(-\frac{\sqrt{3}}{3}\).

Step by step solution

01

Implicit Differentiation Setup

Start with the equation of the curve: \( x^2 + y^2 = 1 \). We need to differentiate both sides with respect to \( x \). This means we will treat \( y \) as a function of \( x \) and use the chain rule when differentiating terms involving \( y \).
02

Differentiate Each Term

Differentiate \(x^2\) to get \(2x\). Differentiate \(y^2\) using the chain rule to get \(2y \cdot \frac{dy}{dx}\). The right side of the equation, which is a constant (1), differentiates to 0. So, the differentiated equation is: \(2x + 2y \cdot \frac{dy}{dx} = 0\).
03

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

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\): \(2y \cdot \frac{dy}{dx} = -2x\). Divide both sides by \(2y\) to isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -\frac{x}{y}\).
04

Substitute the Point into the Derivative

To find the slope of the curve at the given point \(\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)\), substitute \(x = \frac{1}{2}\) and \(y = \frac{\sqrt{3}}{2}\) into \(\frac{dy}{dx} = -\frac{x}{y}\). This gives: \(\frac{dy}{dx} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}}\).
05

Simplify the Derivative Value

The slope of the curve at the given point is \(-\frac{1}{\sqrt{3}}\). To rationalize the denominator, multiply by \(\frac{\sqrt{3}}{\sqrt{3}}\), which yields \(-\frac{\sqrt{3}}{3}\) as the simplified slope.

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Key Concepts

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

Chain Rule
Implicit differentiation requires us to find derivatives when one or more variables are intertwined in an equation. When dealing with equations like this, the Chain Rule is crucial. The Chain Rule is used to differentiate composite functions, which are functions of functions. In our example, we have the equation \[ x^2 + y^2 = 1.\] Here, both x and y are entangled in the equation, and y is a function of x. When we differentiate terms involving y, we need to apply the Chain Rule.
  • Differentiate x虏 normally to get 2x.
  • For y虏, treat y as a function of x. The derivative of y虏 w.r.t. y is 2y. Then, multiply by the derivative of y w.r.t. x, which is \( \frac{dy}{dx} \), giving us \( 2y \cdot \frac{dy}{dx} \).
Thus, the Chain Rule helps to correctly account for the dependence of y on x, allowing us to differentiate the whole equation accurately.
Slope of a Curve
The slope of a curve at a specific point tells you how steep the curve is at that point. Geometrically, it is the tangent to the curve, which is the line that just 'touches' the curve at that specific point without cutting through it. This slope can be found by deriving the function to get \( \frac{dy}{dx} \), and then substituting the given point into this derivative to find the actual numeric value of the slope.In our example:
  • After implicit differentiation, we find \( \frac{dy}{dx} = -\frac{x}{y} \).
  • For the point \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \), substitute x and y into the derivative to find the slope at this particular point:
  • \( \frac{dy}{dx} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \) (after rationalization).
Therefore, the slope of the curve at \( \left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right) \) is \(-\frac{\sqrt{3}}{3}\), indicating the curve is sloping downwards at that point.
Differentiation Techniques
Differentiation is a powerful tool used to find rates of changes in functions. To handle various types of functions, certain techniques are employed. When dealing with relationships between x and y that can't be easily separated, implicit differentiation comes into play.For the equation given, \( x^2 + y^2 = 1 \), we use several techniques:
  • Implicit Differentiation: This allows us to find the derivative \( \frac{dy}{dx} \) when y is not isolated. By differentiating each term with respect to x and applying the chain rule, we handle the implicit function.
  • Algebraic Manipulation: After differentiating, rearrange the resulting expression to solve for \( \frac{dy}{dx} \). This involves simple algebraic steps like moving terms and dividing both sides to isolate \( \frac{dy}{dx} \).
  • Substitution: To find the specific slope at a given point, substitute the coordinates of the point into your expression for \( \frac{dy}{dx} \).
These techniques together enable us to navigate through complex equations and uncover the hidden behaviors of curves, providing insight into their slope and general behavior at any point.

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