Question

a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}-2 x\right)^{2}=2\left(x^{2}+y^{2}\right)\) \(\left(x_{0}, y_{0}\right)=(2,2)\) (limaçon of Pascal)

Short answer

Question: Determine the equations of the tangent and normal lines at the point (2, 2) on the curve given by the implicit equation \((x^2 + y^2 - 2x)^2 = 2(x^2 + y^2)\) and plot them on the graph. Answer: The equation of the tangent line is \(y = 2\), and the equation of the normal line is \(x = 2\).

Step-by-step solution

Find the derivative of the implicit equation

Differentiate both sides of the implicit equation \((x^2 + y^2 - 2x)^2 = 2(x^2 + y^2)\) with respect to \(x\) using the chain rule and implicit differentiation: \(2(x^2 + y^2 - 2x)(2x + 2y\frac{dy}{dx} - 2) = 2(2x + 2y\frac{dy}{dx})\)

Solve for dy/dx

Solve the equation from Step 1 for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{x - y}{x + y}\)

Find the slope at the given point

Evaluate the derivative \(\frac{dy}{dx}\) at the given point \((2, 2)\): \(m_t = \frac{dy}{dx}(2, 2) = \frac{2 - 2}{2 + 2} = 0\)

Find the equation of the tangent line

Using the slope \(m_t = 0\) and the point-slope form of a linear equation, find the equation of the tangent line at the point \((2, 2)\): \(y - y_0 = m_t(x - x_0)\) \(y - 2 = 0(x - 2)\) \(y = 2\)

Find the slope of the normal line

Since the tangent line has a slope of 0, the normal line will have an undefined (vertical) slope, as they are perpendicular to each other.

Find the equation of the normal line

Since the normal line is vertical, its equation will be in the form of \(x = c\), where \(c\) is a constant. At the point \((2, 2)\), the equation will be: \(x = 2\)

Graph the tangent and normal lines along with the curve

Finally, plot the curve \((x^2 + y^2 - 2x)^2 = 2(x^2 + y^2)\), the tangent line \(y = 2\), and the normal line \(x = 2\). The tangent line will be a horizontal line passing through \((2, 2)\), while the normal line will be a vertical line passing through the same point.

Key concepts

Implicit Differentiation

In calculus, implicit differentiation is a powerful tool used when dealing with equations that define a relationship between two variables symbolically, rather than explicitly solving one variable in terms of the other. This method is particularly useful for curves and surfaces where separating variables isn't feasible. To apply implicit differentiation, we differentiate both sides of the equation with respect to one variable (usually x), while treating the other variable (usually y) as an implicit function of x. During this process, whenever we differentiate a term with y, we multiply by \( \frac{dy}{dx} \) (the derivative of y with respect to x) as per the chain rule. For example, consider the implicit equation: \((x^2 + y^2 - 2x)^2 = 2(x^2 + y^2)\). Differentiating both sides with respect to x, we get:

• Differentiate each term and apply the chain rule.
• Use algebraic manipulation to isolate \( \frac{dy}{dx} \).

This results in the derivative: \( \frac{dy}{dx} = \frac{x - y}{x + y} \).

Normal Line

The concept of a normal line is fundamental in understanding the geometry of curves. When a tangent line grazes a curve at a point, the normal line is the line that is perpendicular to the tangent at the same point. In our example, after finding the slope of the tangent line at a given point, we use this to identify the slope of the normal line. Given that normal lines and tangent lines are orthogonal, their slopes are negative reciprocals of each other. For instance, if the tangent has a slope \( m_t \), the normal line will have a slope \( -\frac{1}{m_t} \). It is noteworthy that if the slope of the tangent line is zero (horizontal), as found in this problem, then the normal line will have an undefined slope (vertical). Hence, the normal line is represented as a vertical line, such as \( x = 2 \) in our specific situation, passing through the point where the tangent and normal lines intersect the curve.

Slope of Tangents

The slope of the tangent line at a specific point on a curve gives us valuable information about its geometric properties at that juncture. It represents the rate of change or the "steepness" of the curve. Calculating this slope is crucial for finding both the tangent and normal lines to the curve. In the context of the implicit differentiation problem we've been considering, you begin by deriving the equation \( \frac{dy}{dx} = \frac{x - y}{x + y} \). To find the slope at a particular point, substitute the coordinates of that point into the derivative. For the example point \( (2, 2) \), the calculated slope is:

• Substituting into the derivative formula gives \( m_t = \frac{2 - 2}{2 + 2} = 0 \).
• This means the tangent line is horizontal at this point.

Understanding the slope of the tangent line helps us sketch the curve, locate the point of tangency, and visualize the direction of maximum steepness at that point.