Problem 38 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the tangent line to the given curve at the given point. $$ \frac{x^{2}}{2}-\frac{y^{2}}{4}=1 \text { at }(\sqrt{3}, \sqrt{2}) $$

Step-by-Step Solution

Answer

The tangent line equation is \( y = \sqrt{6}x - 3 + \sqrt{2} \).

Step 1: Differentiate the Implicit Function

The given curve is \( \frac{x^2}{2} - \frac{y^2}{4} = 1 \). Differentiate both sides with respect to \( x \):\[ \frac{d}{dx} \left( \frac{x^2}{2} \right) - \frac{d}{dx} \left( \frac{y^2}{4} \right) = \frac{d}{dx} (1). \]This simplifies to:\[ x - \frac{1}{2} \cdot y \cdot \frac{dy}{dx} = 0. \]

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

From the equation \( x - \frac{1}{2}y \frac{dy}{dx} = 0 \), solve for \( \frac{dy}{dx} \):\[ x = \frac{1}{2}y \frac{dy}{dx} \]\[ \frac{dy}{dx} = \frac{2x}{y}. \]

Step 3: Substitute the Point into the Derivative

Substitute the point \( (\sqrt{3}, \sqrt{2}) \) into the derivative \( \frac{dy}{dx} = \frac{2x}{y} \):\[ \frac{dy}{dx} = \frac{2 \cdot \sqrt{3}}{\sqrt{2}} = \frac{2\sqrt{3}}{\sqrt{2}}. \]

Step 4: Simplify the Slope

Simplify \( \frac{2\sqrt{3}}{\sqrt{2}} \):\[ \frac{2\sqrt{3}}{\sqrt{2}} = \frac{2\sqrt{6}}{2} = \sqrt{6}. \]Therefore, the slope of the tangent line at the point \( (\sqrt{3}, \sqrt{2}) \) is \( \sqrt{6}. \)

Step 5: Use Point-Slope Form for the Equation

Use the point-slope form of a line equation to find the tangent line:\[ y - y_1 = m(x - x_1), \]where \( m = \sqrt{6} \), \( x_1 = \sqrt{3} \), \( y_1 = \sqrt{2} \).Substitute these into the equation:\[ y - \sqrt{2} = \sqrt{6}(x - \sqrt{3}). \]

Step 6: Simplify the Equation of the Tangent Line

Distribute \( \sqrt{6} \) and rearrange the equation:\[ y = \sqrt{6}x - \sqrt{6} \cdot \sqrt{3} + \sqrt{2}. \]\[ y = \sqrt{6}x - 3 + \sqrt{2}. \]This is the equation of the tangent line.

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when you have an equation involving two or more variables, and you want to find the derivative of one variable with respect to another. In our exercise, we were given the equation \( \frac{x^2}{2} - \frac{y^2}{4} = 1 \). This equation is not solved for \( y \) in terms of \( x \), so direct differentiation isn't possible.Instead, we differentiate each part of the equation with respect to \( x \). When differentiating terms with \( y \), we treat \( y \) as a function of \( x \), so we apply the chain rule. For example, the derivative of \( \frac{y^2}{4} \) with respect to \( x \) is \( \frac{1}{2}y \cdot \frac{dy}{dx} \), where \( \frac{dy}{dx} \) is the derivative we want to find.Implicit differentiation lets us solve for \( \frac{dy}{dx} \) even when \( y \) is not explicitly stated as a function of \( x \). This makes it a valuable tool for finding slopes of tangent lines to curves given by implicit equations.

Derivative

The derivative represents the rate at which a function is changing at any given point. In this context, it gives the slope of the tangent line to the curve at a specific point.To find the slope at the point \( (\sqrt{3}, \sqrt{2}) \), the derivative \( \frac{dy}{dx} \) was calculated using implicit differentiation. The result was \( \frac{dy}{dx} = \frac{2x}{y} \).This formula represents how \( y \) changes with respect to \( x \) for our original curve. By substituting \( x = \sqrt{3} \) and \( y = \sqrt{2} \) into this expression, we can find the specific slope value at that point: \( \sqrt{6} \). The derivative is crucial as it determines the alignment and steepness of the tangent line.

Point-Slope Form

The point-slope form is a convenient way to write the equation of a line when you know the slope and a point on the line. The formula is:

• \( y - y_1 = m(x - x_1) \)

where \( m \) is the slope of the line, and \( (x_1, y_1) \) is a point on the line.In our exercise, we know the slope (\( \sqrt{6} \)) and the point \( (\sqrt{3}, \sqrt{2}) \). Using the point-slope form assists in directly creating an equation for the tangent line:

• Substitute \( m = \sqrt{6} \), \( x_1 = \sqrt{3} \), and \( y_1 = \sqrt{2} \) into the equation.
• We get \( y - \sqrt{2} = \sqrt{6}(x - \sqrt{3}) \).

This method ensures you are crafting the line accurately using fundamental details without extra computations.

Equation of a Line

Writing the equation of a line typically involves simplifying the equation derived from the point-slope form to a neat expression in slope-intercept form \( y = mx + b \).From the previous steps, we had \( y - \sqrt{2} = \sqrt{6}(x - \sqrt{3}) \). Distributing \( \sqrt{6} \), we obtain:

• \( y = \sqrt{6}x - \sqrt{6} \times \sqrt{3} + \sqrt{2} \)
• Simplifies to \( y = \sqrt{6}x - 3 + \sqrt{2} \)

This is the final equation of the tangent line. Simplifying the expression in such a manner allows us to easily determine how the line behaves with changes in \( x \), illustrated by the linear components and their combined values.