Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to the given curve at the given point. $$ x^{2}+y^{2}=169 \text { at }(5,12) $$

Step-by-Step Solution

Answer

The tangent line at (5,12) is \(y = -\frac{5}{12}x + \frac{169}{12}\).

Step 1: Recognize the Equation Type

The given equation \(x^2 + y^2 = 169\) represents a circle centered at the origin with radius 13. The problem asks for the tangent line at the point (5,12).

Step 2: Differentiate the Equation Implicitly

To find the slope of the tangent line at any point on the circle, we need to implicitly differentiate the equation. Differentiating both sides, we get \(2x + 2y\frac{dy}{dx} = 0\).

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

Rearrange the differentiated equation to solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -\frac{x}{y}\).

Step 4: Calculate the Slope at the Given Point

Substitute the given point (5,12) into \(\frac{dy}{dx} = -\frac{x}{y}\) to find the slope of the tangent at that point: \(\frac{dy}{dx} = -\frac{5}{12}\).

Step 5: Use the Point-Slope Form to Write the Equation

With the slope \(-\frac{5}{12}\) and the point (5,12), use the point-slope form \(y - y_1 = m(x - x_1)\). Substituting the values, we get: \(y - 12 = -\frac{5}{12}(x - 5)\).

Step 6: Simplify the Tangent Line Equation

Simplify the equation from the previous step: \(y = -\frac{5}{12}x + \frac{25}{12} + 12\). Converting 12 to a fraction: \(y = -\frac{5}{12}x + \frac{169}{12}\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations where one variable is not explicitly solved in terms of another. For instance, in the equation of a circle like \(x^2 + y^2 = 169\), solving for \(y\) directly as a function of \(x\) can be cumbersome. Implicit differentiation allows us to find the derivative \(\frac{dy}{dx}\) by differentiating both sides of an equation with respect to \(x\), while treating \(y\) as a function of \(x\).
Here's how it works:

• Differentiate the square of \(x\): \(2x\).
• Differentiate the square of \(y\) using the chain rule: \(2y \cdot \frac{dy}{dx}\).
• Combine them as you differentiate the entire equation: \(2x + 2y\frac{dy}{dx} = 0\).

From here, you can solve for \(\frac{dy}{dx}\) to find the slope of the tangent line at any point on the circle.
These steps allow you to handle non-function-like equations smoothly, letting you find tangents and other rates of change efficiently.

Circle Equation

A circle's equation in its standard form is \(x^2 + y^2 = r^2\), where

• \(r\) is the radius.
• The circle is centered at the origin \((0,0)\).

In our example, the equation given is \(x^2 + y^2 = 169\). Recognizing this as a circle, we determine its radius by taking the square root of 169, which is 13.
This means our circle has a radius of 13 units, stretching equally in all directions from the center.
Understanding the circle equation is foundational for identifying its properties and calculating other related aspects, such as tangent lines.

Slope of Tangent Line

Once you have the derivative \(\frac{dy}{dx}\), you can find the slope of the tangent line at a specific point on the curve. For circles, the slope of the tangent at any point \((x, y)\) can be determined as:
\(\frac{dy}{dx} = -\frac{x}{y}\).
In our specific problem, substituting the point \((5, 12)\) results in the slope:

• \(\frac{dy}{dx} = -\frac{5}{12}\).

This calculated slope represents the rate at which the circle's y-value changes relative to its x-value at the point \((5, 12)\).
Knowing how to find and interpret this slope is critical for understanding the nature of tangent lines in geometry and calculus.

Point-Slope Form

The point-slope form is an algebraic method used to determine the equation of a line given a point and the slope. The form is expressed as:
\(y - y_1 = m(x - x_1)\), where:

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

For our tangent line, we use the point \((5, 12)\) and slope \(-\frac{5}{12}\):
Plug these into the formula:
\(y - 12 = -\frac{5}{12}(x - 5)\)
Simplifying this equation provides the final tangent line equation:
\(y = -\frac{5}{12}x + \frac{169}{12}\).
Mastering the point-slope form is essential as it is a quick and effective way to find the equation of a line, particularly useful in calculus and geometry contexts.