Problem 235 - Calculus 3

Question

For the following exerises, find \(\frac{d y}{d x}\) using parial derivatives. $$ x^{2 / 3}+y^{2 / 3}=a^{2 / 3} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3} \)

Step 1: Implicit Differentiation Setup

We start by differentiating both sides of the given equation with respect to the variable \( x \). Our equation is \( x^{2/3} + y^{2/3} = a^{2/3} \). Since \( a \) is a constant, its derivative is zero.

Step 2: Differentiate Each Term

Differentiate \( x^{2/3} \) with respect to \( x \), which gives \( \frac{2}{3}x^{-1/3} \). For \( y^{2/3} \), since \( y \) is a function of \( x \), use the chain rule: differentiate to get \( \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} \).

Step 3: Write the Derivative Equation

Combine the differentiated terms into an equation: \( \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = 0 \).

Step 4: Solve for \( \frac{dy}{dx} \)

To isolate \( \frac{dy}{dx} \), move the first term to the right side: \( \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = -\frac{2}{3}x^{-1/3} \). Simplify by dividing both sides by \( \frac{2}{3}y^{-1/3} \): \( \frac{dy}{dx} = -\left(\frac{x}{y}\right)^{-1/3} \).

Step 5: Rewrite the Final Expression

Rearrange the expression to clear the negative exponent: \( \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3} \).

Key Concepts

Partial Derivatives

Partial derivatives are essential in dealing with functions of multiple variables. In the context of our exercise, where the equation is given implicitly as \( x^{2/3} + y^{2/3} = a^{2/3} \), partial derivatives help break down the change of functions in relation to each independent variable. When you take the partial derivative of a function with respect to one of its variables, you treat the other variables as constants. This concept is crucial when working with equations involving multiple variables, as seen in our example with \( x \) and \( y \). To compute \( \frac{\partial}{\partial x} \) of \( x^{2/3} \), we keep all other terms constant, resulting in \( \frac{2}{3}x^{-1/3} \). Whereas for \( y^{2/3} \), since \( y \) is a function of \( x \), it is differentiated using additional techniques, specifically the chain rule.

Chain Rule

The chain rule is a fundamental derivative rule used when differentiating compositions of functions. It plays a pivotal role in implicit differentiation, which is pertinent in our exercise. In the equation \( x^{2/3} + y^{2/3} = a^{2/3} \), the term \( y^{2/3} \) requires the application of the chain rule because \( y \) itself depends on \( x \). The idea is to first differentiate the outer function (raising to the power of \( 2/3 \)) with respect to its inner function \( y \), and then multiply by the derivative of \( y \) with respect to \( x \), giving us \( \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} \). This chain rule application efficiently manages the multilayered nature of composite functions, ensuring a correct differentiation process for expressions where a variable is nested within another.

Implicit Functions

Implicit functions become apparent when we face situations where \( y \) is defined in relation to \( x \) through an equation, and not as an explicit function of \( x \). In our case, the equation \( x^{2/3} + y^{2/3} = a^{2/3} \) does not solve for \( y \) directly in terms of \( x \), thus requiring implicit differentiation. Given the implicit nature, the relationship between \( x \) and \( y \) is understood by differentiating implicitly. We differentiate each term on their own, while the derivative of terms linked to \( y \) involve \( \frac{dy}{dx} \), an indicator of \( y's \) dependency on \( x \). By arranging the resulting differentiated equation, you are able to isolate \( \frac{dy}{dx} \), offering insight on how \( y \) changes with respect to \( x \) implicitly without having a solved explicit function at hand.

Calculus Problem Solving

Calculus problem solving is the broader context wherein methodologies like partial derivatives and the chain rule are applied to tackle complex mathematical expressions. The step-wise process of implicit differentiation highlighted in our exercise underlines a systematic approach often used in calculus. To effectively approach these problems:

• First, understand the requirement: differentiating given expressions implicitly.
• Apply the correct rules: the chain rule and concept of partial derivatives as dictated by the problem setup.
• Use algebraic manipulation where necessary, as seen in isolating \( \frac{dy}{dx} \).
• Verify and rearrange your final results for clarity and simplicity.

This organized methodology allows for gaining a deeper understanding of the relationship between variables, particularly in more elaborate calculus problems involving implicit functions.