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. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right)\) \(\left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli)

Short answer

Answer: The equation of the tangent line at the point \((2, -1)\) is \(y = -\frac{3}{5}x + \frac{1}{5} + 1\), and the equation of the normal line is \(y = \frac{5}{3}x - \frac{10}{3} - 1\).

Step-by-step solution

Find the first derivative of the curve equation with respect to x

To do this, we first rewrite the given equation as an implicit function of x and y. Then apply the chain rule to differentiate the equation with respect to x using implicit differentiation. Original Equation: \((x^{2}+y^{2})^{2}=\frac{25}{3}(x^2 -y^2)\) Implicit differentiation: \(4(x^{2}+y^{2})(2x+2yy')=\frac{25}{3}(2x - 2yy')\)

Solve for y'

We now want to isolate y' in the equation above to find the first derivative: \(4(x^{2}+y^{2})(2x+2yy') - \frac{25}{3}(2x - 2yy') = 0\) Now, we can factor out \(2y'\): \(2y'\left( 4(x^{2}+y^{2}) + \frac{25}{3}\right)= 4(x^2 + y^2)(2x) - \frac{25}{3}(2x)\) Next, divide both sides by the expression in parentheses: \(y'= \frac{4(x^2 + y^2)(2x) - \frac{25}{3}(2x)}{2\left( 4(x^{2}+y^{2}) + \frac{25}{3}\right)}\)

Evaluate y' at \((x_0, y_0) = (2, -1)\)

Plug the coordinates \((x_0, y_0)\) into the expression for y' to determine the slope of the tangent line at the given point \(y'_{(2, -1)}\): \(y'_{(2, -1)} = \frac{4((2)^2 + (-1)^2)(2(2)) - \frac{25}{3}(2(2))}{2\left( 4((2)^2+(-1)^2) + \frac{25}{3}\right)}\) After simplifying: \(y'_{(2, -1)} = \frac{-10}{\frac{50}{3}} = \frac{-30}{50} = -\frac{3}{5}\)

Determine the equation for the tangent line

Using the slope, \(m_t = -\frac{3}{5}\), and the point \((x_0, y_0) = (2, -1)\), we can now find the equation of the tangent line using the point-slope form of a line equation: \(tangent: y - (-1) = -\frac{3}{5}(x - 2)\) Simplifying: \(y = -\frac{3}{5}x + \frac{1}{5} + 1\)

Determine the equation for the normal line

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent line's slope: \(m_n = -\frac{1}{-\frac{3}{5}} = \frac{5}{3}\) Using this slope and the point \((2,-1)\), we find the equation of the normal line: \(normal: y - (-1) = \frac{5}{3}(x - 2)\) Simplifying: \(y = \frac{5}{3}x - \frac{10}{3} - 1\)

Graph the tangent and normal lines

Since the exercise request to graph the tangent and normal lines, the student will need to graph the given lemniscate of Bernoulli curve, and the tangent and normal lines based on the equations we found in previous steps. Here are the equations for each line: - Lemniscate of Bernoulli (implicit function): \((x^{2}+y^{2})^{2}=\frac{25}{3}(x^2 -y^2)\) - Tangent line: \(y = -\frac{3}{5}x + \frac{1}{5} + 1\) - Normal line: \(y = \frac{5}{3}x - \frac{10}{3} - 1\) The student can use graphing tools, such as a graphing calculator or graphing software (e.g. GeoGebra, Desmos, etc.) to plot these equations and observe their relationships.

Key concepts

Tangent Line Equation

The equation of a tangent line is a fundamental concept in calculus, describing a line that touches a curve at exactly one point without crossing it. When dealing with complex curves like the lemniscate of Bernoulli, finding the tangent line involves using implicit differentiation to calculate the derivative at the point of tangency.

Consider a point \(x_0, y_0\) on the curve. To find the equation of the tangent line at that point, we begin by finding the derivative \(y'\) of the curve's equation with respect to \(x\). This derivative represents the slope \(m_t\) of the tangent line. With the slope and the given point, we can use the point-slope form: \(y - y_0 = m_t(x - x_0)\).

The resulting equation provides a linear function that represents the tangent at the specified point on the curve. This process is crucial because it reveals how the curve behaves locally and helps in finding the rate of change at any point along the curve.

Normal Line Equation

A normal line at a given point on a curve is a line that is perpendicular to the tangent line at that point. The slope of the normal line, \(m_n\), is the negative reciprocal of the slope of the tangent line, \(m_t\).

If the slope of the tangent line is \(m_t\), then the slope of the normal line would be \(m_n = -1/m_t\). Subsequently, we use the point-slope equation with the slope of the normal line and the point of tangency \(x_0, y_0\) to get the normal line equation: \(y - y_0 = m_n(x - x_0)\).

Creating the equation for the normal line allows students to better understand orthogonal relationships in geometry and aids in problems that require constructing perpendiculars from a curve.

Lemniscate of Bernoulli

The lemniscate of Bernoulli is a fascinating figure-eight or infinity symbol-shaped curve and is a special case of the Cassini oval. The standard equation for the lemniscate is \( (x^2 + y^2)^2 = a^2 (x^2 - y^2) \), where \(a\) is a constant that determines the size of the lemniscate.

The equation represents a relationship between the coordinates \(x\) and \(y\) for each point on the curve. Unlike a circle or an ellipse, the lemniscate consists of two loops that intersect at the origin. Graphically, the curve can be plotted using the implicit function provided, which requires tools capable of handling non-function graphs since it doesn't pass the vertical line test.

Understanding the lemniscate of Bernoulli not only embodies aesthetic appeal but also introduces students to geometries beyond the elementary shapes and to the intriguing properties of complex curves.

First Derivative Calculus

First derivative calculus is an essential tool in mathematics, particularly in the study of the behavior of functions. It involves finding the derivative of a function, which represents the rate at which the function is changing at any given point. The first derivative is denoted as \(y'\) or \(f'(x)\) and is obtained by applying the derivative rules, such as the power rule, chain rule, and product rule.

In the context of implicit differentiation, such as with the lemniscate of Bernoulli, we treat \(y\) as a function of \(x\) and differentiate both sides of the equation with respect to \(x\). This allows us to solve for \(y'\) even when the equation cannot be written explicitly as \(y=f(x)\).

The first derivative is highly significant; it is used to draw conclusions about the increasing or decreasing behavior of the function, determine the slope of tangent lines, and locate extrema—points where the function reaches local minimums or maximums.