Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find \(\frac{d y}{d x}\) for the curve \(x^{3}+3 y+y^{2}=6\). What is the slope at \((2,-1)\) ? At what points is the slope zero?

Step-by-Step Solution

Answer

The slope at \((2, -1)\) is \(-2\). The slope is zero at the points \((0, 1)\) and \((0, -3)\).

Step 1: Differentiate Both Sides of the Equation

Differentiate both sides of the equation with respect to \(x\). Use the chain rule when differentiating \(y\) (and \(y^2\)) treated as a function of \(x\). You get: \(3x^2 + 3\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\).

Step 2: Solve for \(\frac{dy}{dx}\)

Rearrange the differentiated equation to isolate \(\frac{dy}{dx}\) on one side: \(\frac{dy}{dx} = \frac{-3x^2}{3 + 2y}\).

Step 3: Find the Slope at \((2,-1)\) by Substituting

To find the slope at the point \((2, -1)\), substitute \(x = 2\) and \(y = -1\) into the derivative equation from step 2: \(\frac{dy}{dx} = \frac{-3 \cdot (2)^2}{3 + 2\cdot (-1)} = -2\). So the slope at \((2, -1)\) is \(-2\).

Step 4: Find the Points Where the Slope is Zero

To find the points where the slope is zero, set the derivative (from step 2) equal to zero and solve for \(x\). You'll get the equation \(0 = \frac{-3x^2}{3 + 2y}\). This only happens when the numerator \( -3x^2 = 0\) and solving for \(x\) gives \(x = 0\). Substitute \(x = 0\) back into the original equation to get the \(y\)-coordinate, which is \(y = 1\) or \(y = -3\). So the points where the slope is zero are \((0, 1)\) and \((0, -3)\).

Key Concepts

Chain Rule

When we differentiate equations that involve more than one variable, like in the equation \(x^3 + 3y + y^2 = 6\), we often use a method called implicit differentiation. Implicit differentiation is particularly useful when we can't easily solve for \(y\) in terms of \(x\). This is where the chain rule comes into play.

The chain rule is a fundamental rule in calculus that allows us to differentiate composite functions. In this problem, when you differentiate terms like \(3y\) and \(y^2\), you treat \(y\) as a function of \(x\). The chain rule tells us how to differentiate these terms correctly:

• For \(3y\), the derivative with respect to \(x\) is \(3\frac{dy}{dx}\).
• For \(y^2\), the derivative is \(2y\frac{dy}{dx}\).

The chain rule is crucial here because it gives us the correct differentiation of terms involving \(y\), which can't be ignored, as they contribute to the overall rate of change of the curve.

Slope of a Curve

The slope of a curve at any given point reflects how steep the curve is at that particular point. The slope is essentially the same as the derivative \(\frac{dy}{dx}\).

In our example, once we differentiated the original equation using the chain rule, we obtained \(3x^2 + 3\frac{dy}{dx} + 2y\frac{dy}{dx} = 0\). Solving for \(\frac{dy}{dx}\) gave us \(\frac{dy}{dx} = \frac{-3x^2}{3 + 2y}\).

• At the point \((2, -1)\), by substituting \(x = 2\) and \(y = -1\) into this derivative, we discovered the slope to be \(-2\).
• This means the curve is decreasing at this point with a slope of \(-2\), indicating a fairly steep decline.

The slope effectively tells us the direction and steepness of the curve at the point, which is why it's such an important concept in calculus.

Equation of a Curve

Understanding the equation of a curve allows us to sketch its graph and optimize functions. In the problem \(x^3 + 3y + y^2 = 6\), this equation describes a relationship between \(x\) and \(y\) that forms a particular shape in a graph, which can be complex due to the presence of both \(x\) and \(y\) on the same side.

Using implicit differentiation and finding the slope at certain points—like we did at \((2, -1)\) and when finding where the slope is zero—helps us understand the nature of the curve's graph, including:

• Using \(x = 0\) to find points like \((0, 1)\) and \((0, -3)\) where the slope is zero helps identify where the curve has a horizontal tangent.
• These horizontal tangents can indicate minima or maxima points or changes in the direction of the curve.

These calculations are crucial for analyzing and understanding the behavior and shape of the curve dictated by the equation.