Question

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. 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) (Graph cant copy)

Short answer

Answer: The equation of the tangent line is \(y=-x+1\), and the equation of the normal line is \(y=x-3\).

Step-by-step solution

Find the derivative dy/dx

First, find the derivative of the equation \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right)\) with respect to x by differentiating both sides. Use the Chain Rule and Implicit Differentiation: Differentiate both sides with respect to \(x\): \(2\left(x^{2}+y^{2}\right)(2x+2yy') = \frac{25}{3}(2x-2yy')\) Now, we'll solve for \(y'\): \(4\left(x^{2}+y^{2}\right)(x+yy') = \frac{50}{3}(x-y'y)\)

Find the slope of the tangent at the given point

Plug in the given point \((x_0, y_0) = (2, -1)\) into the equation we just found for \(y'\): \(4\left(2^2+(-1)^2\right)(2+(-1)y') = \frac{50}{3}(2-(-1)y)\) This equation simplifies to: \(20(2-y') = \frac{50}{3}(3+y)\) Now, solve for \(y'\): \(40-20y' = 50+50y/3\) \(120-60y'=150+50y\) \(-30y'=30\) \(y'=-1\) So, the slope of the tangent line at point \((2, -1)\) is \(-1\).

Find the equation of the tangent line

Now we have a point and the slope, we will use point-slope form to find the equation of the tangent line: \(y - y_0 = m(x - x_0)\) Where \(m=-1\), \(x_0=2\), and \(y_0=-1\): \(y - (-1) = -1(x - 2)\) Simplifying, we get: \(y+1 = -x + 2\) The equation of the tangent line is \(y=-x+1\).

Find the slope and equation of the normal line

The normal line is perpendicular to the tangent line, so their slopes are negative reciprocals. Since the slope of the tangent line is \(-1\), the slope of the normal line is \(1\). Now, we'll find the equation of the normal line using point-slope form: \(y - y_0 = m(x - x_0)\) Where \(m=1\), \(x_0=2\), and \(y_0=-1\): \(y - (-1) = 1(x - 2)\) Simplifying, we get: \(y+1 = x - 2\) The equation of the normal line is \(y=x-3\).

Graph the tangent and normal lines on the given graph

Without the exact graph we can't visualize it for you. However, you can now plot the curve \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right)\), the tangent line \(y=-x+1\), the normal line \(y=x-3\), and the given point \((2, -1)\) on the same graph to visualize the result.

Key concepts

Tangent line

When studying calculus, the tangent line is a crucial concept. It represents the line that just "touches" a curve at a given point, without crossing it. Imagine you have a curve, and you're zooming in at a particular point. The tangent line is what you'd see if you magnified that section to the limit.This tangent line tells us about the behavior of the curve at that specific point, and its slope is essentially the derivative of the curve at that point. In practice, we use point-slope form to write the equation of the tangent line:\[ y - y_0 = m(x - x_0) \]Here, \(m\) is the slope derived from differentiating the curve, and \((x_0, y_0)\) is the point of tangency. In our exercise, we found the tangent line at the point \((2, -1)\) on a lemniscate of Bernoulli to be \(y = -x + 1\). This gave us a precise linear representation of the curve at that point.

Normal line

The normal line exists perpendicular to the tangent line at the same point on the curve. In essence, while the tangent line might tell us how steep a curve is at a point, the normal line acts like a cross-section.The slope of the normal line is the negative reciprocal of the tangent line's slope. For instance, if the slope of the tangent line is \(-1\), the slope of the normal will be \(1\). That's because when two lines are perpendicular, their slopes multiply to \(-1\).Using the point-slope form once more:\[ y - y_0 = m(x - x_0) \]With our derived slope \(1\) and point \((2, -1)\), the normal line becomes \(y = x - 3\). This perpendicular relationship is essential in various applications, especially in orthogonal trajectories and optimizing curves.

Implicit differentiation

Implicit differentiation is a method used in calculus when dealing with equations not easily solved for a single variable. Instead of rearranging to \(y = f(x)\), implicit differentiation allows us to take the derivative directly, respecting both \(x\) and \(y\) as variables.In our exercise, the given curve was \((x^2 + y^2)^2 = \frac{25}{3}(x^2 - y^2)\). Solving explicitly for \(y\) is difficult, if not impossible. Implicit differentiation involves the chain rule, treating \(y\) as a function of \(x\) during differentiation. This allows us to find \(\frac{dy}{dx}\) in terms of both \(x\) and \(y\). The step-by-step differentiation and solving process uncovered the slope of the tangent line at \((2, -1)\). Implicit differentiation makes it manageable to find these rates of change in complex equations.

Slope

The slope in calculus is the measure of steepness or inclination of a line or curve at a point. Calculating slope is one of the most frequent tasks because it gives valuable information about the function's behavior and trend.In this context, we used the slope to determine the tangent and normal lines for a lemniscate of Bernoulli. For the line at any given point \((x_0, y_0)\), the slope \(m\) of the tangent line is determined by finding the derivative \(\frac{dy}{dx}\) through implicit differentiation.After performing the necessary calculations, we found the tangent line's slope was \(-1\). This slope was vital for writing the tangent's equation, \(y = -x + 1\), and is equally important in calculating the normal line's slope, turning it into \(y = x - 3\). Calculating these slopes efficiently connects rates of change to the geometry of the curve, arming us with both understanding and predictive power.

Lemniscate of Bernoulli

The lemniscate of Bernoulli is a specific type of plane curve known for its figure-eight or infinity-like shape. In our exercise, the equation given \[ (x^2 + y^2)^2 = \frac{25}{3}(x^2 - y^2) \]is a form of lemniscate. The lemniscate is interesting in calculus due to its symmetry and how it can demonstrate complex concepts like implicit differentiation.Working with this curve, we focused on finding tangent and normal lines. These lines help depict how the curve behaves around any given point, such as \((2, -1)\). Lemniscates can visualize a range of mathematical phenomena, making them relevant in both theoretical exploration and practical graphical applications.By analyzing these structures, students can better appreciate how calculus translates to dynamic visual structures, offering a meld of analysis with geometry.