Problem 44 - Calculus Single Variable

Question

Find the tangent line to the parametric curve \(x=\varphi_{1}(t), y=\varphi_{2}(t)\) at the point corresponding to the given value \(t_{0}\) of the parameter. $$ \varphi_{1}(t)=1 /\left(1+t^{2}\right), \varphi_{2}(t)=t(1+\ln (t)) \quad t_{0}=1 $$

Step-by-Step Solution

Answer

The tangent line at \( t_0 = 1 \) is \( y = -4x + 3 \).

Step 1: Determine the coordinates of the point

To find the coordinates corresponding to the parameter value \( t_0 = 1 \), substitute \( t = 1 \) into \( \varphi_1(t) \) and \( \varphi_2(t) \). We get \( x = \varphi_1(1) = \frac{1}{1 + 1^2} = \frac{1}{2} \) and \( y = \varphi_2(1) = 1(1 + \ln(1)) = 1 \cdot 1 = 1 \). So, the point is \( \left( \frac{1}{2}, 1 \right) \).

Step 2: Calculate the derivatives

We need the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) to find the slope of the tangent line. Calculate: \( \frac{dx}{dt} = -\frac{2t}{(1+t^2)^2} \). For \( \frac{dy}{dt} \), use the product rule: \( \frac{dy}{dt} = (1 + \ln(t)) + \frac{t}{t} = 1 + \ln(t) + 1 = 1 + \ln(t) + 1 = 2 + \ln(t) \).

Step 3: Evaluate the derivatives at \( t_0 \)

Substitute \( t = 1 \) into the derivatives: \( \frac{dx}{dt} \big|_{t=1} = -\frac{2 \cdot 1}{(1+1^2)^2} = -\frac{2}{4} = -\frac{1}{2} \) and \( \frac{dy}{dt} \big|_{t=1} = 2 + \ln(1) = 2 \).

Step 4: Find the slope of the tangent line

The slope of the tangent line is \( m = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{2}{-\frac{1}{2}} = -4 \).

Step 5: Write the equation of the tangent line

Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \), with point \( \left( \frac{1}{2}, 1 \right) \) and slope \( -4 \), we get \( y - 1 = -4 \left(x - \frac{1}{2}\right) \). Simplify to: \( y = -4x + 3 \).

Key Concepts

Parametric Equations

Parametric equations are a mathematical way of expressing wide-ranging and effective relationships between variables. Instead of describing a curve with a single equation in terms of x and y, parametric equations use a third variable, known as the parameter, to express the equations for x and y separately. This technique is especially useful for plotting curves on the coordinate plane where standard y = f(x) forms can become complex.

In this exercise, we used parametric equations to describe a curve in the plane:

• For x, we had: \( \varphi_1(t) = \frac{1}{1 + t^2} \)
• For y, we used: \( \varphi_2(t) = t (1 + \ln (t)) \)

To find the tangent line at a specific point, the value of the parameter \( t_0 = 1 \) was substituted into both equations, giving the point on the curve as \( \left( \frac{1}{2}, 1 \right) \).

Understanding parametric equations allows us to study complex curves that simple direct equations may not adequately describe.

Derivatives

Derivatives are crucial in calculus as they help us understand how functions change. In simple terms, a derivative tells us about the slope or rate of change at any given point on a function's graph. When dealing with parametric equations, we look for the derivatives of both x and y concerning the parameter t:

The derivatives for our specific functions are:

• \( \frac{dx}{dt} = -\frac{2t}{(1 + t^2)^2} \)
• \( \frac{dy}{dt} = 2 + \ln(t) \)

These derivative expressions allow us to determine the tangent line's slope at any point specified by t.

This process involves using various differentiation techniques, such as the product rule when needed, especially for more complicated expressions like \( \varphi_2(t) \). Calculating these derivatives is pivotal for finding how x and y change with t, which directly influences the behavior of the curve.

Tangent Line Equation

The equation of a tangent line can be obtained using the derivatives of the parametric equations. The tangent line is effectively a straight line that just grazes the curve at a specific point, providing an accurate approximation of the curve's behavior at that point.

To write the tangent line's equation, we first need to find the slope, \( m \), of the line using:

• \( m = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } \)

For our exercise, substituting the evaluated derivatives at \( t_0 = 1 \), we found:

• \( m = \frac{2}{-\frac{1}{2}} = -4 \)

Then, using the point-slope form of a line, \( y - y_1 = m (x - x_1) \), with the point \( \left( \frac{1}{2}, 1 \right) \) and the slope \( -4 \), the tangent line equation becomes:

• \( y = -4x + 3 \)

Understanding how to derive this equation is crucial for analyzing the behavior of parametric curves.

Calculus

Calculus combines the concepts of limits, derivatives, and integrals, giving us powerful tools to analyze the changes and behaviors of functions. It has two main branches: differential calculus and integral calculus. Here, we focus on differential calculus, which deals primarily with derivatives and their applications.

In solving problems involving parametric curves, calculus allows us to:

• Understand the rate of change of functions through derivatives.
• Examine concavity and convexity of curves by inspecting second derivatives in more advanced applications.
• Investigate the behavior of curves at particular points by determining their tangents.

Such skills are essential for studying movements, trends, and predicting future values in natural sciences, econometrics, and various engineering fields.

By mastering calculus, students gain deeper insight into both theoretical and practical concerns, aiding their comprehension of complex systems described by mathematical models.