Problem 45 - Thomas Calculus

Question

In Exercises 45 and \(46,\) find the slope of the curve at the given points. $$ y^{2}+x^{2}=y^{4}-2 x \text { at }(-2,1) \text { and }(-2,-1) $$

Step-by-Step Solution

Answer

The slopes are -1 at (-2,1) and 1 at (-2,-1).

Step 1: Differentiate Implicitly

First, differentiate the given equation implicitly with respect to \(x\):\[ \frac{d}{dx}(y^2 + x^2) = \frac{d}{dx}(y^4 - 2x) \] This gives us: \[ 2y \frac{dy}{dx} + 2x = 4y^3 \frac{dy}{dx} - 2 \] because when differentiating \(y^2\) or \(y^4\) we must apply the chain rule.

Step 2: Rearrange the Equation

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

Step 3: Evaluate at the Point (-2, 1)

Substitute \(x = -2\) and \(y = 1\) into the expression for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{-2 - 2(-2)}{2(1) - 4(1)^3} = \frac{-2 + 4}{2 - 4} = \frac{2}{-2} = -1 \] The slope at the point \((-2, 1)\) is \(-1\).

Step 4: Evaluate at the Point (-2, -1)

Substitute \(x = -2\) and \(y = -1\) into the expression for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = \frac{-2 - 2(-2)}{2(-1) - 4(-1)^3} = \frac{-2 + 4}{-2 + 4} = \frac{2}{2} = 1 \] The slope at the point \((-2, -1)\) is \(1\).

Key Concepts

Slope of Curve

The slope of a curve at a given point indicates how steep or flat the curve is at that location. It is essentially the rate at which the curve is rising or falling.
For a curve defined by an equation involving two variables like \(y\) and \(x\), finding the slope involves determining the derivative, typically denoted as \(\frac{dy}{dx}\).
The derivative gives us the slope of the tangent line to the curve at any point.

• The slope might be positive (curve rising), negative (curve falling), or zero (flat or peak/valley).
• To find the slope at specific points, you substitute the coordinate values into the derivative.

Differentiation Techniques

Differentiation is a technique used to find the rate at which a function changes. In this scenario, since the equation \(y^2 + x^2 = y^4 - 2x\) involves both \(x\) and \(y\), implicit differentiation is a useful technique.
Implicit differentiation allows for differentiation when \(y\) is not isolated on one side of the equation. Here's how:

• Differentiating each term with respect to \(x\), even those involving \(y\), by applying the derivative rules.
• Every time you differentiate something with \(y\), multiply by \(\frac{dy}{dx}\), because \(y\) is a function of \(x\).


• This method allows you to derive \(\frac{dy}{dx}\) without explicitly solving for \(y\) first, making it very powerful for complex equations.

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of compositions of functions.
When differentiating implicitly, as seen with the given equation, the chain rule plays a crucial role.
For example, in differentiating \(y^2\) with respect to \(x\), we think of it as a composition function, where \(y\) is a function of \(x\).

• The formula for the chain rule is: if \(u(x)\) is a function of \(x\) and \(y = f(u)\), then \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\).
• Applying this, the derivative of \(y^2\) is \(2y \cdot \frac{dy}{dx}\), illustrating how the chain rule connects \(dy\) and \(dx\).
• Similarly, for \(y^4\), the derivative becomes \(4y^3 \cdot \frac{dy}{dx}\).

This approach systematically handles variations in \(y\) by recognizing dependencies in \(y\) on \(x\).