Problem 311 - 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. $$ x^{2} y^{2}+5 x y=14,(2,1) $$

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = -\frac{1}{2}x + 2\).

Step 1: Find the Gradient of the Implicit Function

First, we need to differentiate the given implicit equation \(x^2 y^2 + 5xy = 14\) with respect to \(x\). This involves implicit differentiation. Apply the product rule and chain rule where needed:\[\frac{d}{dx}(x^2 y^2) + \frac{d}{dx}(5xy) = 0.\]The derivative of \(x^2 y^2\) is \(2xy^2 + 2x^2 y \frac{dy}{dx}\) and the derivative of \(5xy\) is \(5y + 5x \frac{dy}{dx}\). Put these together:\[2xy^2 + 2x^2 y \frac{dy}{dx} + 5y + 5x \frac{dy}{dx} = 0.\]

Step 2: Solve for \( \frac{dy}{dx} \)

Rearrange the differentiated equation to solve for \( \frac{dy}{dx} \):\[2x^2 y \frac{dy}{dx} + 5x \frac{dy}{dx} = -2xy^2 - 5y.\]Factor out \( \frac{dy}{dx} \):\[\frac{dy}{dx}(2x^2 y + 5x) = -2xy^2 - 5y.\]Thus,\[\frac{dy}{dx} = \frac{-2xy^2 - 5y}{2x^2 y + 5x}.\]

Step 3: Calculate the Slope at the Given Point

Substitute the given point \((2, 1)\) into the expression for \( \frac{dy}{dx} \) to find the slope of the tangent line:\[\frac{dy}{dx} = \frac{-2(2)(1)^2 - 5(1)}{2(2)^2(1) + 5(2)} = \frac{-2(2) - 5}{8 + 10}.\]Simplify both the numerator and denominator,\[\frac{dy}{dx} = \frac{-4 - 5}{18} = \frac{-9}{18} = -\frac{1}{2}.\]

Step 4: Write the Equation of the Tangent Line

Using the slope \(-\frac{1}{2}\) and point \((2, 1)\), use the point-slope form of the equation of a line:\[y - y_1 = m(x - x_1),\]where \(m = -\frac{1}{2}\) and \((x_1, y_1) = (2,1)\). Substitute these values:\[y - 1 = -\frac{1}{2}(x - 2).\]Simplify:\[y = -\frac{1}{2}x + 1 + 1,\]\[y = -\frac{1}{2}x + 2.\]

Step 5: Graph the Function and the Tangent Line

Use a graphing calculator or computer software to plot the implicit function \(x^2 y^2 + 5xy = 14\) and the tangent line equation \(y = -\frac{1}{2}x + 2\). Make sure the tangent line intersects the curve at the point (2,1).

Key Concepts

Tangent Line

Understanding a tangent line is crucial in calculus and helps us grasp the behavior of curves. A tangent line touches a curve at a single point and provides the slope or direction of the curve at that point.
For instance, imagine a curve as a winding road and the tangent line as a straight road connecting only at one point. It indicates the instant direction we would take if we continued at that point. In the context of implicit differentiation, determining the tangent line involves finding the slope at a specific point on the curve. The tangent line equation is generally derived in three major steps:

• Firstly, finding the derivative, which represents the rate of change or the slope of the curve.
• Secondly, substituting the given point into the derivative to find the specific slope at that point.
• Lastly, inserting the slope and point into the point-slope form equation to establish the tangent line equation.

By understanding these aspects, solving for the tangent line becomes a direct task of applying calculus concepts effectively.

Product Rule

In calculus, the product rule is a fundamental principle used when differentiating the product of two functions. It can be represented as: For two functions, say, \(u(x)\) and \(v(x)\), their derivative is given by:

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

This rule is vital for implicit differentiation as it deals with functions that are products of each other, just like in our original problem. This is why you encounter it repeatedly in calculus when handling complex functions composed of multiplicative terms. To apply the product rule, always:

• Differentiating the first function while keeping the second one constant.
• Then differentiate the second function while maintaining the first as is.

Combining these results gives you the derivative of the product of the two functions. The product rule thus bridges the gap for differentiating products effectively.

Chain Rule

The chain rule is another pivotal tool in calculus that helps differentiate composite functions. When encountering implicit functions, it becomes especially useful as these are often composed of two or more related functions.Imagine two functions related in a nested manner. The chain rule helps peel these layers:

• If there are two functions \(f(g(x))\), its derivative is \(f'(g(x))g'(x)\).

In the context of implicit differentiation, this means that if a term like \(x^2y^2\) appears, the chain rule allows one to recognize it as a composite function of \(x\) and \(y(x)\) and differentiate accordingly. By explicitly distinguishing and differentiating the inner and outer functions, the chain rule ensures accuracy in handling complex derivations. This rule is hence crucial for breaking down multifaceted expressions often encountered in calculus problems.

Point-Slope Form

The point-slope form of a line is one of the most straightforward ways to write the equation of a tangent line in calculus. It's particularly convenient when you know the slope of a line and a point through which it passes.The general format of the point-slope form is:

• \(y - y_1 = m(x - x_1)\)

Where:

• \((x_1, y_1)\) is the known point.
• \(m\) is the slope of the line.

This form directly uses the slope and point to construct the linear equation, simplifying the process of finding the tangent line. It eliminates unnecessary complexity by neatly tying the slope and point data together into a clear expression. Thus, utilizing the point-slope form makes translating calculus results into tangible, graphical line equations efficient and straightforward.