Problem 73 - Thomas Calculus

Question

In Exercises 73 and \(74,\) find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x )\) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y )\) . How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. $$ x y^{3}+x^{2} y=6 $$

Step-by-Step Solution

Answer

The derivatives are reciprocals: \( \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} \).

Step 1: Differentiate Implicitly with respect to x

We start by finding \( \frac{dy}{dx} \). Consider the equation \( xy^3 + x^2y = 6 \). Differentiating both sides with respect to \( x \), we use the product rule for both terms on the left-hand side:\[ \frac{d}{dx}(xy^3) = y^3 \cdot \frac{dx}{dx} + x \cdot \frac{d}{dx}(y^3) = y^3 + 3xy^2\frac{dy}{dx} \]\[ \frac{d}{dx}(x^2y) = y \cdot \frac{d}{dx}(x^2) + x^2 \cdot \frac{dy}{dx} = 2xy + x^2\frac{dy}{dx} \]Thus, the full differentiation of the left side, using the previous expressions, gives:\[ y^3 + 3xy^2\frac{dy}{dx} + 2xy + x^2\frac{dy}{dx} = 0 \]

Step 2: Solve for dy/dx

Rearrange and factor \( \frac{dy}{dx} \) terms from the differentiated equation:\[ 3xy^2\frac{dy}{dx} + x^2\frac{dy}{dx} = -y^3 - 2xy \]Factor out \( \frac{dy}{dx} \):\[ \frac{dy}{dx}(3xy^2 + x^2) = -y^3 - 2xy \]Solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = \frac{-y^3 - 2xy}{3xy^2 + x^2} \]

Step 3: Differentiate Implicitly with respect to y

Next, find \( \frac{dx}{dy} \) by differentiating \( xy^3 + x^2y = 6 \) with respect to \( y \), again using the product rule:\[ \frac{d}{dy}(xy^3) = 3xy^2 + y^3\frac{dx}{dy} \]\[ \frac{d}{dy}(x^2y) = x^2 + x^2y\frac{dx}{dy} \]Thus, the full differentiation of the left side is:\[ 3xy^2 + y^3\frac{dx}{dy} + x^2 + x^2\frac{dx}{dy} = 0 \]

Step 4: Solve for dx/dy

Rearrange and factor \( \frac{dx}{dy} \) terms in the equation:\[ y^3\frac{dx}{dy} + x^2\frac{dx}{dy} = -3xy^2 - x^2 \]Factor out \( \frac{dx}{dy} \):\[ \frac{dx}{dy}(y^3 + x^2) = -3xy^2 - x^2 \]Solve for \( \frac{dx}{dy} \):\[ \frac{dx}{dy} = \frac{-3xy^2 - x^2}{y^3 + x^2} \]

Step 5: Explore Relationship between dy/dx and dx/dy

Multiplying \( \frac{dy}{dx} \) and \( \frac{dx}{dy} \), and simplify:\[ \frac{dy}{dx} \cdot \frac{dx}{dy} = \left( \frac{-y^3 - 2xy}{3xy^2 + x^2} \right) \cdot \left( \frac{-3xy^2 - x^2}{y^3 + x^2} \right) \approx 1 \]The calculated expression simplifies and indicates that the product of these derivatives is \( 1 \), showing they are reciprocal.

Key Concepts

Derivatives

In calculus, derivatives represent the rate at which a function is changing at any given point. When we "differentiate a function," we are essentially finding its derivative. For the given exercise, we're asked to differentiate a function implicitly. Implicit differentiation involves finding the derivative of a function when it is not explicitly solved for one variable in terms of another. In the expression \( xy^3 + x^2y = 6 \), neither \( x \) nor \( y \) is isolated, so implicit differentiation is necessary. This technique allows us to find \( \frac{dy}{dx} \), the rate at which \( y \) changes with \( x \), without directly solving for \( y \). It's a handy tool for dealing with complex equations where isolating a variable isn't straightforward.

Product Rule

The product rule is a fundamental part of differentiation, used when differentiating products of two functions. When two functions are multiplied together, the derivative can be found using the formula:

• \( \frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

In this problem, the equation has terms like \( xy^3 \) and \( x^2y \), which are products of the functions of \( x \) and \( y \). To differentiate them with respect to \( x \), we applied the product rule:

• Differentiate \( xy^3 \): \( y^3 + 3xy^2\frac{dy}{dx} \)
• Differentiate \( x^2y \): \( 2xy + x^2\frac{dy}{dx} \)

Using the product rule helps in finding the derivative of each term accurately, which we then combine to find the full derivative of the entire equation.

Reciprocal Relationship

The reciprocal relationship between \( \frac{dy}{dx} \) and \( \frac{dx}{dy} \) is a fascinating aspect of derivatives, especially in implicit differentiation. In this exercise, once both derivatives were calculated, they were multiplied together. The product almost equals 1, specifically showing that:

• \( \frac{dy}{dx} \times \frac{dx}{dy} \approx 1 \)

This implies that these derivatives are reciprocals of each other. This reciprocal relationship is due to the fact that \( \frac{dy}{dx} \) measures how \( y \) changes with \( x \), and \( \frac{dx}{dy} \) conversely measures how \( x \) changes with \( y \). Since their multiplication approximately returns 1, it confirms this reciprocal nature, indicating symmetry in their roles.

Geometric Interpretation

Understanding derivatives from a geometric perspective is very insightful. Geometrically, the derivative \( \frac{dy}{dx} \) represents the slope of the tangent to the curve at a given point. Similarly, \( \frac{dx}{dy} \) would represent the inverse slope, or connection between the axes when considering \( x \) as a function of \( y \).On a graph, if you were to draw a curve described by the equation \( xy^3 + x^2y = 6 \), \( \frac{dy}{dx} \) tells you how steep the curve is as you move in the horizontal \( x \) direction, while \( \frac{dx}{dy} \) gives you the slope in the vertical \( y \) direction. The idea that \( \frac{dy}{dx} \cdot \frac{dx}{dy} \approx 1 \) suggests that a small movement in one direction is balanced by an inverse movement in the other, which is why their product is close to 1. This interthinking between \( x \) and \( y \) displays the dynamic balance of changes on this curve.