Problem 74 - Calculus Single Variable

Question

Plot the given parametric curve \(x=\varphi_{1}\) \((t), y=\varphi_{2}(t)\) in a viewing rectangle that contains the given point \(P_{0}\). Find the equation of the tangent line at \(P_{0}\). Add the tangent line to your plot. $$ \begin{array}{l} \varphi_{1}(t)=100 t /\left(1+t^{2}\right)^{2}, \varphi_{2}(t)=100 t^{2} /\left(1+t^{2}\right)^{2} \\ P_{0}=(32,16)(\mathrm{t}=1 / 2) \quad \text { (Regular Bifolium) } \end{array} $$

Step-by-Step Solution

Answer

The equation of the tangent line at \( P_0 = (32, 16) \) is \( y = x - 16 \).

Step 1: Understand the Given Parametric Equations

We have the parametric equations for the curve: \( x = 100 \frac{t}{(1+t^2)^2} \) and \( y = 100 \frac{t^2}{(1+t^2)^2} \). These equations describe the shape and behavior of a curve in the plane as \( t \) varies.

Step 2: Confirm Point on the Curve

The point \( P_0 = (32, 16) \) is given as a known point on the curve when \( t = \frac{1}{2} \). Substitute \( t = \frac{1}{2} \) into the equations: \( x = 100 \frac{0.5}{(1+0.25)^2} = 32 \) and \( y = 100 \frac{0.5^2}{(1+0.25)^2} = 16 \). Thus, \( P_0 \) is indeed on the curve.

Step 3: Calculate Derivatives for Slope of Tangent Line

To find the tangent line at \( P_0 \), we need \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).\( \frac{dx}{dt} = \frac{d}{dt} \left( \frac{100t}{(1+t^2)^2} \right) = \frac{100(1-t^2)}{(1+t^2)^3} \).\( \frac{dy}{dt} = \frac{d}{dt} \left( \frac{100t^2}{(1+t^2)^2} \right) = \frac{200t(1-t^2)}{(1+t^2)^3} \).

Step 4: Compute Derivative Values at \( t = \frac{1}{2} \)

Evaluate the derivatives at \( t = \frac{1}{2} \):\( \frac{dx}{dt} \bigg|_{t=0.5} = \frac{100(1-(0.5)^2)}{(1+(0.5)^2)^3} = \frac{96}{9.375} = 10.24 \).\( \frac{dy}{dt} \bigg|_{t=0.5} = \frac{200(0.5)(1-(0.5)^2)}{(1+(0.5)^2)^3} = \frac{96}{9.375} = 10.24 \).

Step 5: Slope and Equation of the Tangent Line

The slope \( m \) of the tangent line is the ratio of these derivatives: \( m = \frac{dy/dt}{dx/dt} = \frac{10.24}{10.24} = 1 \).Using point-slope form, \( y - y_1 = m(x - x_1) \), with \( x_1 = 32 \), \( y_1 = 16 \), gives: \( y - 16 = 1(x - 32) \), simplifying to \( y = x - 16 \).

Step 6: Plot the Curve and Tangent Line

Now plot the parametric equations for the curve in a suitable domain around \( t \), and include the tangent line \( y = x - 16 \) on the same graph. Ensure that the viewing window includes \( P_0 = (32, 16) \). Use graphing technology to confirm accuracy.

Key Concepts

Tangent Line

Let's talk about tangent lines. Imagine you have a curve on a graph. The tangent line is a straight line that just touches the curve at one particular point. It doesn't cross the curve at that point but rather brushes against it, moving in the same direction the curve is heading at that very spot.
To find the equation of a tangent line at a given point on a curve, like point \(P_0 = (32, 16)\) in our exercise, you need to know the slope of the curve at that point. Once you have the slope, you can use the point-slope form of a linear equation: \(y - y_1 = m(x - x_1)\), where \(m\) represents the slope, and \(x_1\) and \(y_1\) are coordinates of the point. In this case, after calculation, the tangent line equation becomes \(y = x - 16\).
This equation shows how the line behaves around the point \(P_0\), reflecting the instant rate of change of the curve at that location.

Derivatives

Derivatives are a fundamental idea in calculus, representing the rate of change. When dealing with parametric equations, derivatives can be calculated with respect to the parameter, in this case, \(t\). For the parametric curve given by equations \(x(t)\) and \(y(t)\), we find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

• \(\frac{dx}{dt}\) measures how much \(x\) changes with a small change in \(t\).
• \(\frac{dy}{dt}\) measures how much \(y\) changes with a small change in \(t\).

To find the slope of the tangent line, we use the ratio \(\frac{dy/dt}{dx/dt}\). In this problem, both derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are evaluated at \(t = \frac{1}{2}\), and the result is a slope of 1. Derivatives allow us to understand not just where the curve is but how it moves or changes.

Bifolium Curve

A bifolium curve is a special kind of parametric curve characterized by its elegant shape. It's crucial to grasp its structure when analyzing how it behaves, especially at specific points like \(P_0 = (32, 16)\).
The equations \(x(t) = 100 \frac{t}{(1+t^2)^2}\) and \(y(t) = 100 \frac{t^2}{(1+t^2)^2}\) describe this type of curve. As the parameter \(t\) changes, it traces out the path of the bifolium curve, detailing a path that resembles two loops.

• The curve has symmetry, which means it looks the same on one side as it does on the other.
• It's interesting to examine because its shape provides insights into the beauty and complexity of geometry and parametrization.

Having an understanding of this curve type helps to predict its behavior and characteristics, especially in mathematical and graphical analyses.

Calculus

Calculus is like the language of change, focusing on how things move and evolve. It uses tools like derivatives and integrals to delve into these changes.
In our exercise, calculus helps us understand how the parametric equations describe the bifolium curve. By differentiating these parametric equations with respect to \(t\), we are using calculus to analyze the instantaneous rate of change of each component of the curve.

• Calculus allows us to model real-world phenomena where variables are continuously changing.
• By understanding the derivatives, you can paint a picture of how quantities like speed, position, and more behave continuously over time.

In mathematical problems, calculus serves as a bridge to transform graphical and numerical data into analytical insights, providing a deeper comprehension of the shapes and paths depicted by equations.