Problem 28 - Single Variable Calculus

Question

Find the rate of change of \(y\) with respect to \(x\) at the given values of \(x\) and \(y\). $$ x^{2 / 3}+y^{2 / 3}=5 ; \quad x=1, y=8 $$

Step-by-Step Solution

Answer

The rate of change of \(y\) with respect to \(x\) at the point (\(x=1\), \(y=8\)) is \(-3\).

Step 1: Understand Implicit Differentiation

Implicit differentiation is used when it's difficult or impossible to explicitly solve for one variable in terms of another. Since we have both \(x\) and \(y\) in the equation \(x^{\frac{2}{3}} + y^{\frac{2}{3}} = 5\), it's a great candidate for implicit differentiation. We will differentiate both sides of the equation with respect to \(x\), and whenever we differentiate a term with \(y\), we'll multiply by \(\frac{dy}{dx}\) by chain rule.

Step 2: Differentiate the equation implicitly

Differentiate both sides of the equation \(x^{\frac{2}{3}} + y^{\frac{2}{3}} = 5\) with respect to \(x\). Using the chain rule, we find: \[\frac{d}{dx}(x^{\frac{2}{3}}) + \frac{d}{dx}(y^{\frac{2}{3}}) = \frac{d}{dx}(5)\] Differentiate each term, remembering to multiply by \(\frac{dy}{dx}\) for the term with \(y\): \[\frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}} \frac{dy}{dx} = 0\]

Step 3: Solve for dy/dx

Now, we can solve for \(\frac{dy}{dx}\) from the equation obtained in step 2: \[\frac{2}{3}y^{-\frac{1}{3}} \frac{dy}{dx} = -\frac{2}{3}x^{-\frac{1}{3}}\] Multiply both sides by \(\frac{3}{2}\) to simplify: \[\frac{dy}{dx} = -\frac{3}{2}x^{-\frac{1}{3}}y^{\frac{1}{3}}\]

Step 4: Plug in the given point

Now, we can plug in the given point (\(x=1\), \(y=8\)) into our formula for \(\frac{dy}{dx}\) to find the rate of change of \(y\) with respect to \(x\) at this point: \[\frac{dy}{dx} = -\frac{3}{2}(1)^{-\frac{1}{3}}(8)^{\frac{1}{3}}\]

Step 5: Simplify and find the result

Finally, simplify the expression to find the rate of change of \(y\) with respect to \(x\) at the given point: \[\frac{dy}{dx} = -\frac{3}{2}(1)(2)\] \[\frac{dy}{dx} = -3\] So, the rate of change of \(y\) with respect to \(x\) at the point (\(x=1\), \(y=8\)) is \(-3\).

Key Concepts

Rate of Change

The rate of change refers to how quickly one quantity changes in relation to another. In this exercise, we are interested in the rate at which the variable \( y \) changes with respect to \( x \), represented mathematically as \( \frac{dy}{dx} \). This gives us the rate of change of \( y \) for a tiny movement in \( x \). This concept is very useful in various fields, such as physics and economics, for understanding how one variable affects another.
Understanding the rate of change is like looking at the slope of a tangent line to the curve described by our equation. It's the value telling us whether \( y \) increases or decreases as \( x \) changes. Here, the rate of change at the point \( (x=1, y=8) \) was found to be \(-3\), meaning that \( y \) decreases by 3 units for every one unit increase in \( x \).

• The negative sign shows that the relationship is inverse: as \( x \) increases, \( y \) decreases.
• If the result had been positive, \( y \) would increase as \( x \) increases.

Chain Rule

The chain rule is a fundamental tool in calculus used for differentiating composite functions. It helps to find the derivative of a function when one variable depends on another, and that second variable also depends on a third. In our exercise, we saw this in action while differentiating the term \( y^{\frac{2}{3}} \) with respect to \( x \).
When we apply the chain rule, we differentiate the outer function (\( y^{\frac{2}{3}} \)) and multiply by the derivative of the inner function (\( y \) with respect to \( x \), or \( \frac{dy}{dx} \)). Mathematically, it looks like this: \[\frac{d}{dx}(y^{\frac{2}{3}}) = \frac{2}{3}y^{-\frac{1}{3}} \cdot \frac{dy}{dx}\]
This technique ensures that we accurately capture the effect of changing \( x \) on \( y \).

• By using the chain rule, we recognize that \( y \) isn't just a constant but varies with \( x \).
• This careful attention to variables dependent on one another is crucial for exact calculations.

Derivatives

Derivatives are a core concept in calculus that measure the rate of change of a function. They provide us a way to determine how a function changes at any point. In the exercise given, we find the derivative \( \frac{dy}{dx} \) which represents the rate at which \( y \) changes with respect to \( x \).
The derivative of \( x^{\frac{2}{3}} \) was found using power rule, which is a simple rule for finding derivatives of functions in the form of \( x^n \). For the term \( y^{\frac{2}{3}} \), implicit differentiation was necessary due to the dependency between \( x \) and \( y \), resulting in the chain rule application.

• Derivatives express the concept of instantaneous rate of change, akin to the velocity of an object at a precise moment.
• Practically, they are used in curve sketching, optimization problems, and in determining local maxima and minima.
• Understanding derivatives allows us to model and predict natural phenomena, such as graphing the path of a moving object.