Problem 310 - Calculus Volume 1

Question

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 tangent line equation is \(y = -5x - 6\).

Step 1: Find the Derivative

Since the equation is given as an implicit function, we need to differentiate both sides with respect to \(x\). Use implicit differentiation on the equation \(x^4y - xy^3 = -2\). For each term, apply the product rule and differentiate.

Step 2: Apply Implicit Differentiation

Differentiate \(x^4y\) to get \(4x^3y + x^4\frac{dy}{dx}\). Differentiate \(xy^3\) to get \(y^3 + 3xy^2\frac{dy}{dx}\). Set the equation as: \(4x^3y + x^4\frac{dy}{dx} - y^3 - 3xy^2\frac{dy}{dx} = 0\).

Step 3: Solve for \(\frac{dy}{dx}\)

Rearrange the terms to solve for \(\frac{dy}{dx}\): \(x^4\frac{dy}{dx} - 3xy^2\frac{dy}{dx} = y^3 - 4x^3y\). Therefore, \(\frac{dy}{dx} = \frac{y^3 - 4x^3y}{x^4 - 3xy^2}\).

Step 4: Evaluate the Derivative at the Point

Substitute \(x = -1\) and \(y = -1\) into the derived expression for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{(-1)^3 - 4(-1)^3(-1)}{(-1)^4 - 3(-1)(-1)^2}\). Simplify to get the slope \(-5\).

Step 5: Write the Equation of the Tangent Line

Use the point-slope form of a line equation \(y - y_1 = m(x - x_1)\), with \(m = -5\), \((x_1, y_1) = (-1, -1)\). The equation becomes \(y + 1 = -5(x + 1)\). Simplify to get \(y = -5x - 6\).

Step 6: Graph the Function and Tangent Line

Use a calculator or computer software to plot the original function \(x^4y - xy^3 = -2\) and the tangent line \(y = -5x - 6\) on the same graph to visualize their intersection at \((-1, -1)\).

Key Concepts

Tangent Line Equation

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. It describes the behavior of the curve nearly at a specific location and can be represented mathematically by an equation. To find this equation, we need two pieces of information:

• The slope of the tangent line, which is given by the derivative of the function at that point.
• The specific coordinates of the point on the curve.

Using these, we can apply the point-slope form of the linear equation: \[ y - y_1 = m(x - x_1) \] where \(m\) is the slope and \((x_1, y_1)\) represents the coordinates of the point. In the exercise, with a slope \(m = -5\) and point \(x_1, y_1 = (-1, -1)\), we derive the tangent line equation \(y = -5x - 6\). This equation gives us a line that merely touches the curve at \((-1, -1)\), illustrating immediate local behavior.

Implicit Function

An implicit function is a function where the dependent variable (usually \(y\)) isn't isolated on one side of the equation. Instead, \(x\) and \(y\) are mingled together, like in the equation \[ x^4y - xy^3 = -2 \].
Unlike explicit functions where \(y\) is neatly expressed as \(y=f(x)\), implicit functions require more creativity to extract useful calculus information, such as derivatives. Often, implicit functions describe complex relationships that aren't easily separable or simplified. Solving problems involving implicit functions frequently utilizes techniques like implicit differentiation, allowing us to find gradients without needing to first solve for one variable in terms of the other, which can be quite complex in practice.

Product Rule

The product rule is a critical tool in calculus used to differentiate the product of two functions. When dealing with implicit differentiation, especially with complex forms such as \(x^4y\) or \(xy^3\), the product rule becomes indispensable.
Mathematically, if you have two functions, say \(u(x)\) and \(v(x)\), their product's derivative is expressed as: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \].
Applying the product rule in our example:

• For \(x^4y\), consider \(u(x) = x^4\) and \(v(x) = y\). Differentiating gives \(4x^3y + x^4 \frac{dy}{dx}\).
• For \(xy^3\), consider \(u(x) = x\) and \(v(x) = y^3\). Differentiation results in \(y^3 + 3xy^2 \frac{dy}{dx}\).

Using the product rule facilitates managing these differentiations systematically and accurately even in challenging implicit contexts.

Derivative Evaluation

Derivative evaluation is the process of calculating the slope of a curve at a specific point by substituting the point's coordinates into the derivative. In calculus, this is crucial for determining rates of change and can be particularly revealing in implicit functions.
Once we have the derivative expression from implicit differentiation, such as \[ \frac{dy}{dx} = \frac{y^3 - 4x^3y}{x^4 - 3xy^2} \], we proceed by evaluating this expression at the point of interest.

• Plug \(x = -1\) and \(y = -1\) into the derivative.
• Calculations yield: \[ \frac{(-1)^3 - 4(-1)^3(-1)}{(-1)^4 - 3(-1)(-1)^2} = -5 \].

This value, \-5\, is the slope of the tangent line at the specific point, which is used in the equation to describe the tangent's precise direction and angle on the graph.