Problem 310 - Calculus Volume 1

Question

For the following exercises, find the equation of the tangent line to the graph of the given equation at the indicated point. Use a calculator or computer software to graph the function and the tangent line. $$ x^{4} y-x y^{3}=-2,(-1,-1) $$

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = \frac{5}{2}x + \frac{3}{2}\).

Step 1: Implicit Differentiation

We start by differentiating both sides of the equation \(x^4 y - x y^3 = -2\) with respect to \(x\). This gives us: \(d/dx (x^4 y) - d/dx (x y^3) = d/dx (-2)\). Using the product rule, \(d/dx (x^4 y) = x^4 (dy/dx) + 4x^3 y\) and \(d/dx (x y^3) = y^3 + x 3y^2 (dy/dx)\). Thus, the differentiated equation becomes: \(x^4 (dy/dx) + 4x^3 y - y^3 - 3xy^2 (dy/dx) = 0\).

Step 2: Solve for \(dy/dx\)

Rearrange the terms in the differentiated equation to solve for \(dy/dx\). We have: \((x^4 - 3xy^2)(dy/dx) = y^3 - 4x^3 y\). Solving for \(dy/dx\), we get: \(dy/dx = \frac{y^3 - 4x^3 y}{x^4 - 3xy^2}\).

Step 3: Evaluate the Derivative at the Point

Substitute \(x = -1\) and \(y = -1\) into \(dy/dx = \frac{y^3 - 4x^3 y}{x^4 - 3xy^2}\). This gives \(dy/dx = \frac{(-1)^3 - 4(-1)^3(-1)}{(-1)^4 - 3(-1)(-1)^2}\). Simplifying, we find \(dy/dx = \frac{-1 - 4}{1 - 3} = -\frac{5}{-2} = \frac{5}{2}\).

Step 4: Equation of the Tangent Line

The formula for the tangent line is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope \(\frac{5}{2}\) and \((x_1, y_1) = (-1, -1)\). Substitute these values to get \(y + 1 = \frac{5}{2}(x + 1)\). Simplifying, the equation becomes \(y = \frac{5}{2}x + \frac{3}{2}\).

Step 5: Graph the Function and Tangent Line

Use a graphing calculator or computer software to graph the original function with its implicit form, \(x^4 y - x y^3 = -2\), and the tangent line equation, \(y = \frac{5}{2}x + \frac{3}{2}\). Verify the point \((-1, -1)\) is where the tangent line touches the curve.

Key Concepts

Tangent Line Equation

The tangent line equation is a crucial concept in calculus, often deriving from finding the slope at a particular point on a curve. A tangent line touches a curve at exactly one point, just like drawing a straight line that kisses the edge of a circle. To express this mathematically, we use the formula:

• \( y - y_1 = m(x - x_1) \)

In this equation:

• \( m \) stands for the slope of the tangent line.
• \((x_1, y_1)\) is the point where the tangent touches the curve.

For our problem, we found the slope \( m \) to be \( \frac{5}{2} \) at the point \((x_1, y_1) = (-1, -1)\). Substituting these into the formula, the tangent line equation becomes:

• \( y = \frac{5}{2}x + \frac{3}{2}\)

Slope of a Tangent Line

Understanding the slope of a tangent line is akin to understanding the steepness of a hill; it's a measure of how much \( y \) changes with a change in \( x \). In implicit differentiation, calculating the slope involves finding the derivative \( \frac{dy}{dx} \).
In this exercise, we've worked through implicit differentiation to find the slope of the tangent line at the point \((-1, -1)\), which turned out to be \( \frac{5}{2} \). So, with every two steps in the \( x \)-direction, the \( y \) value climbs by five steps upwards.

• At the heart, \( \frac{dy}{dx} = \frac{y^3 - 4x^3 y}{x^4 - 3xy^2} \) is the expression we evaluated.

By substituting the point values \((x, y) = (-1, -1)\), we simplified the expression to get our specific slope value.This calculated slope tells us how sharp or gentle the tangent line inclines at a given point on the curve.

Product Rule

The product rule is an essential rule in calculus used for differentiating products of two functions. Simply put, it helps find the derivative when two variables are multiplied together. If you have a function \( u(x) \times v(x) \), the product rule states:

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

In our problem, we encountered terms like \(x^4 y\) and \(xy^3\), which are products of two functions. Applying the product rule:

• For \( x^4 y \), we differentiated by: \( x^4(dy/dx) + 4x^3 y \).
• For \( xy^3 \), it became: \( y^3 + x \, 3y^2(dy/dx) \).

These steps allowed us to express the differentiated equation, ultimately necessary to solve for \(dy/dx\). The product rule is a fundamental tool for dealing with complex expressions that aren't simple sums or differences.

Implicit Function

An implicit function is where dependent and independent variables are intermingled in an equation, instead of being expressed in the traditional y=f(x) form. With implicit differentiation, we can find derivatives to help analyze such equations.
In this problem, we deal with an implicit expression: \( x^4 y - x y^3 = -2 \). Here, \( x \) and \( y \) are not neatly separated, yet we can still differentiate both sides with respect to \( x \). The process involves treating \( y \) as a function of \( x \) and using rules like the product rule to differentiate each part of the equation.

• By doing so, we obtained a derivative equation that could be solved for \( dy/dx \).

Implicit functions are common in complex curves or systems where direct separation isn't possible but analysis is crucial, particularly for finding characteristics like slopes of tangent lines.