Problem 35 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to the given curve at the given point. \(\frac{x^{2}}{27}+\frac{y^{2}}{9}=1\) at \((3, \sqrt{6})\)

Step-by-Step Solution

Answer

The equation of the tangent line is \(y = -\frac{1}{\sqrt{6}}x + \frac{9}{\sqrt{6}}\).

Step 1: Understand the Problem

The problem requires us to find the tangent line to the ellipse \(\frac{x^{2}}{27}+\frac{y^{2}}{9}=1\) at a specific point \((3, \sqrt{6})\). This involves finding the slope of the tangent line and using point-slope form of the equation of a line.

Step 2: Differentiate the Equation Implicitly

Differentiate both sides of the equation \(\frac{x^{2}}{27}+\frac{y^{2}}{9}=1\) with respect to \(x\). This will give us the derivative in terms of both \(x\) and \(y\):\[\frac{d}{dx}\left(\frac{x^2}{27} + \frac{y^2}{9}\right) = \frac{d}{dx}(1)\]This results in:\[\frac{2x}{27} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0\]

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

Rearrange the equation \(\frac{2x}{27} + \frac{2y}{9} \cdot \frac{dy}{dx} = 0\) to solve for \(\frac{dy}{dx}\):\[\frac{2y}{9} \cdot \frac{dy}{dx} = -\frac{2x}{27}\]Then:\[\frac{dy}{dx} = -\frac{2x}{27} \cdot \frac{9}{2y} = -\frac{x}{3y}\]

Step 4: Plug in the Point \((3, \sqrt{6})\) to Find the Slope

Substitute \(x = 3\) and \(y = \sqrt{6}\) into the derivative \(\frac{dy}{dx} = -\frac{x}{3y}\):\[\left.\frac{dy}{dx} \right|_{(3, \sqrt{6})} = -\frac{3}{3\sqrt{6}} = -\frac{1}{\sqrt{6}}\]

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

Use the point-slope form of the linear equation \(y - y_1 = m(x - x_1)\) with the point \((3, \sqrt{6})\) and slope \(-\frac{1}{\sqrt{6}}\):\[y - \sqrt{6} = -\frac{1}{\sqrt{6}}(x - 3)\]Simplify to get the equation of the tangent line:\[y = -\frac{1}{\sqrt{6}}x + \frac{3}{\sqrt{6}} + \sqrt{6}\]

Step 6: Simplify the Tangent Line Equation

Combine and simplify the constant terms in the equation:\[y = -\frac{1}{\sqrt{6}}x + \frac{3}{\sqrt{6}} + \sqrt{6}\]The constants simplify to:\[y = -\frac{1}{\sqrt{6}}x + \frac{3 + 6}{\sqrt{6}}\]\[y = -\frac{1}{\sqrt{6}}x + \frac{9}{\sqrt{6}}\]

Key Concepts

Ellipse

An ellipse is a fascinating geometric shape that resembles an elongated circle. It is defined as the set of all points where the sum of the distances from two fixed points, called foci, is constant. In the case of the equation given in the exercise, \(\frac{x^{2}}{27}+\frac{y^{2}}{9}=1\), the ellipse is centered at the origin, \((0,0)\). Here, the denominators 27 and 9 help to determine the lengths of the semi-major and semi-minor axes.

- The semi-major axis is the longer radius of the ellipse, along which the ellipse stretches further. In our equation, it is \(\sqrt{27}\).- The semi-minor axis is the shorter radius, which is \(\sqrt{9}\) in our case.Ellipses are important in various fields such as astronomy, physics, and engineering because of their unique properties, like reflecting angles. Understanding their properties helps when working with problems involving ellipses, ensuring accurate calculations when finding tangent lines or assessing symmetry.

Implicit Differentiation

Implicit differentiation is a technique used in calculus to find the derivative of an equation not solved for one variable in terms of another. This is especially useful when dealing with curves like ellipses or circles, where the equation includes both \(x\) and \(y\), as in \(\frac{x^{2}}{27}+\frac{y^{2}}{9}=1\). When performing implicit differentiation, we differentiate all terms with respect to a single variable, typically \(x\), while considering \(y\) as a function of \(x\) (i.e., \(y = y(x)\)).

- Each occurrence of \(y\) in the equation will require us to apply the chain rule, which states that \(\frac{d}{dx}[y^n] = n\cdot y^{n-1} \cdot \frac{dy}{dx}\). This introduces \(\frac{dy}{dx}\), the derivative of \(y\) with respect to \(x\).- In this exercise, implicit differentiation helps find the slope of the tangent at a particular point \((3, \sqrt{6})\), which is essential for constructing the tangent line equation.Understanding this concept allows you to tackle various problems involving non-linear equations, making implicit differentiation a valuable tool in calculus.

Point-Slope Form

The point-slope form is a useful way to write the equation of a line when you know a point on the line and the slope. This form is given by the equation: \[y - y_1 = m(x - x_1)\]where \((x_1, y_1)\) is the point on the line and \(m\) is the slope. This form emphasizes the idea of how a line’s slope, indicating its steepness, determines the line’s exact path starting from the given point.

- It is particularly advantageous in situations, like in our exercise, where you are directly asked to find a tangent line. After finding the slope \(-\frac{1}{\sqrt{6}}\) through implicit differentiation, you can directly plug into this form along with the given point \((3, \sqrt{6})\).- Simplifying the resulting equation gives the exact linear equation you need.Using the point-slope form ensures that the line accurately reflects both the calculated slope and the position of the known point. This concept broadens your capability to express linear relationships in mathematics accurately.