Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Determine the derivative of the curve

To find the slope of the tangent line, first determine the derivative of the curve given by the equation \(x^2 - y^2 = -1\). Use implicit differentiation. Differentiating both sides with respect to \(x\) gives: \(2x - 2y \frac{dy}{dx} = 0\).

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

From the equation \(2x - 2y \frac{dy}{dx} = 0\), solve for \(\frac{dy}{dx}\) to find the slope of the tangent line. Rearranging gives \(2x = 2y \frac{dy}{dx}\). Therefore, \(\frac{dy}{dx} = \frac{x}{y}\).

Step 3: Substitute the point into the derivative

Substitute the given point \((\sqrt{2}, \sqrt{3})\) into the derivative \(\frac{dy}{dx} = \frac{x}{y}\) to find the slope of the tangent line at that point. This gives \(\frac{dy}{dx} = \frac{\sqrt{2}}{\sqrt{3}}\).

Step 4: Write the equation of the tangent line

Use the point-slope form of a line \(y - y_1 = m(x - x_1)\) to write the equation of the tangent line, where \(m = \frac{\sqrt{2}}{\sqrt{3}}\) and \((x_1, y_1) = (\sqrt{2}, \sqrt{3})\). Substituting these values, the equation becomes: \(y - \sqrt{3} = \frac{\sqrt{2}}{\sqrt{3}}(x - \sqrt{2})\).

Step 5: Simplify the equation of the tangent line

Distribute and simplify the equation from the previous step to get it in a more standard form. After simplification: \(y = \frac{\sqrt{2}}{\sqrt{3}}x - \frac{2}{\sqrt{3}} + \sqrt{3}\).

Key Concepts

Implicit Differentiation

When faced with an equation that isn't solved for either variable, like \( x^{2} - y^{2} = -1 \), we use implicit differentiation. Unlike explicit differentiation, where we differentiate \( y \) with respect to \( x \) when \( y \) is expressed as a function of \( x \), implicit differentiation allows us to handle equations where \( y \) isn’t isolated.

The key idea is to differentiate both sides of the equation with respect to \( x \), treating \( y \) as an implicit function of \( x \). Each time we differentiate \( y \), we apply the chain rule. In our given equation, differentiating \( x^2 - y^2 = -1 \) gives us:

• Derive \( x^2 \) to get \( 2x \).
• Derive \( -y^2 \) using the chain rule: \(-2y \cdot \frac{dy}{dx}\).

The right side of the equation, \(-1\), which is a constant, becomes 0 after differentiation, leading to: \( 2x - 2y \cdot \frac{dy}{dx} = 0 \).

Implicit differentiation helps find the slope (\( \frac{dy}{dx} \)) for equations where \( y \) is entangled with \( x \) in complex ways, facilitating calculations of tangent lines where an explicit formula for \( y \) doesn't exist.

Derivative Calculation

After differentiating implicitly, the next step is solving for \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \). This derivative represents the slope of the tangent line at any given point on the curve.

From the differentiated equation \( 2x - 2y \cdot \frac{dy}{dx} = 0 \), rearrange to isolate \( \frac{dy}{dx} \):

• Add \( 2y \cdot \frac{dy}{dx} \) to both sides: \( 2x = 2y \cdot \frac{dy}{dx} \).
• Divide both sides by \( 2y \): \( \frac{dy}{dx} = \frac{x}{y} \).

This equation \( \frac{dy}{dx} = \frac{x}{y} \) is crucial because it tells us the slope at any point \( (x, y) \) on the curve, as long as \( y eq 0 \).

To find the specific slope at the point \( (\sqrt{2}, \sqrt{3}) \), substitute these values into the derivative \( \frac{dy}{dx} \): \( \frac{dy}{dx} = \frac{\sqrt{2}}{\sqrt{3}} \).

The calculation of the derivative via implicit differentiation gives us an elegant way to handle non-linear functions, simplifying the process of finding slopes needed for tangent line equations.

Point-Slope Form

Once you know the slope of the tangent line from your derivative calculation, the point-slope form helps to quickly write the equation of the tangent line. The point-slope form is given by:

• \( y - y_1 = m(x - x_1) \), where \( m \) is the slope.
• \((x_1, y_1)\) is the point on the line, here \((\sqrt{2}, \sqrt{3})\).

Substituting into the point-slope equation:

• Slope \( m = \frac{\sqrt{2}}{\sqrt{3}} \)
• Point \((x_1, y_1) = (\sqrt{2}, \sqrt{3}) \)

gives \( y - \sqrt{3} = \frac{\sqrt{2}}{\sqrt{3}}(x - \sqrt{2}) \).

This provides the equation of the tangent line in a straight-forward manner. To express the tangent line in a more conventional form, you can distribute and simplify, resulting in: \( y = \frac{\sqrt{2}}{\sqrt{3}}x - \frac{2}{\sqrt{3}} + \sqrt{3} \).

The point-slope form is a powerful and simple way to find the equation of a line when you know a point on the line and the slope, fostering a swift completion of tangent lines in calculus.