Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to the given curve at the given point. \(\frac{x^{2}}{24}+\frac{y^{2}}{16}=1\) at \((3 \sqrt{2},-2)\)

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = \sqrt{2}x - 8\).

Step 1: Identify the Curve and Known Point

The curve given is an ellipse described by the equation \(\frac{x^2}{24} + \frac{y^2}{16} = 1\). The point provided is \((3\sqrt{2}, -2)\). This point lies on the curve which is the first thing we check to ensure that we are finding the tangent correctly.

Step 2: Differentiate the Curve Implicitly

To find the slope of the tangent line, differentiate the equation \(\frac{x^2}{24} + \frac{y^2}{16} = 1\) with respect to \(x\). This gives: \(\frac{d}{dx}(\frac{x^2}{24}) + \frac{d}{dx}(\frac{y^2}{16}) = 0\). Applying the power rule and implicit differentiation, we get: \(\frac{x}{12} + \frac{y}{8} \cdot \frac{dy}{dx} = 0\).

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

Rearrange the differentiated equation to solve for the slope \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -\frac{8}{12} \cdot \frac{x}{y} = -\frac{2}{3} \cdot \frac{x}{y}\).

Step 4: Substitute Known Values to Find the Slope

Plug in the point \((3\sqrt{2}, -2)\) into the equation \(\frac{dy}{dx} = -\frac{2}{3} \cdot \frac{x}{y}\) to find the slope at that specific point: \(\frac{dy}{dx} = -\frac{2}{3} \cdot \frac{3\sqrt{2}}{-2} = \sqrt{2}\).

Step 5: Use Point-Slope Form to Find Tangent Line Equation

The point-slope form of the equation of a line is: \(y - y_1 = m(x - x_1)\). Substitute the known point \((3\sqrt{2}, -2)\) and the slope \(\sqrt{2}\) into this formula: \(y - (-2) = \sqrt{2}(x - 3\sqrt{2})\). Simplify to find the equation of the tangent line: \(y + 2 = \sqrt{2}x - 6\). Then, rearrange to standard form if necessary: \( y = \sqrt{2}x - 8 \).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of functions that are not explicitly solved for one variable in terms of another. When a relationship between variables is not expressed as a function, such as an ellipse or a circle, it can be tricky to differentiate directly.

For instance, with the ellipse equation \( \frac{x^2}{24} + \frac{y^2}{16} = 1 \), both \(x\) and \(y\) contribute to the shape, making it necessary to use implicit differentiation.

• Differentiate both sides of the equation with respect to \(x\).
• Apply the chain rule when differentiating terms involving \(y\), treating \(y\) as a function of \(x\), so you add the term \( \frac{dy}{dx} \).
• Finally, solve the new equation for \( \frac{dy}{dx} \).

In this way, you can find the rate at which \(y\) changes with respect to \(x\) along the ellipse.

Tangent Line

A tangent line is a straight line that just "touches" a curve at a given point, without crossing it. The purpose of finding a tangent line is to understand the behavior of the curve at a particular location. In this exercise, the objective is to find the tangent line to the given ellipse at a specific point.

When you find the tangent line, you obtain its slope by differentiating the curve equation. This slope, along with a specific point on the curve, allows you to write the equation of the tangent line using the point-slope form:

• Point-slope form: \( y - y_1 = m(x - x_1) \).
• Ensure to substitute the slope \(m\) and the coordinates of the point \((x_1, y_1)\) into this formula.

This equation describes how the line behaves at the exact point where it touches the curve.

Ellipse Tangent

Ellipses are unique because they are not defined by a simple function. An ellipse, such as \( \frac{x^2}{24} + \frac{y^2}{16} = 1 \), requires using implicit differentiation to find tangents because it is not a function of \(x\) or \(y\) alone.

To find the tangent at a particular point on an ellipse, follow these steps:

• Verify that the point lies on the ellipse. Substitute coordinates into the ellipse equation to ensure they satisfy it.
• Differentiating implicitly gives you a formula for the slope of the tangent at any point.
• Substitute the particular \(x\) and \(y\) into this slope equation to find the slope of the tangent at that specific point.

This process leads to an understanding of how the ellipse's curve behaves around that point.

Slope of Tangent Line

Calculating the slope of a tangent line is crucial for determining the tangent's direction. It's found by determining \( \frac{dy}{dx} \), which represents how the curve function \(y\) changes as \(x\) changes.

For our ellipse \( \frac{x^2}{24} + \frac{y^2}{16} = 1 \), we use implicit differentiation to find \( \frac{dy}{dx} \), which in this case is \(-\frac{2}{3} \cdot \frac{x}{y} \). Following this, plug in the specific \((x, y)\) value to calculate the actual slope at the point of interest:

• Use \(x = 3\sqrt{2}\) and \(y = -2\) to find \( \frac{dy}{dx} = \sqrt{2} \).

This slope signifies the steepness and direction of the tangent as it touches the ellipse precisely at the point \((3\sqrt{2}, -2)\). Understanding this slope helps in sketching or analyzing the curve's behavior at that point.