Problem 77 - 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. $$ \begin{aligned} &\varphi_{1}(t)=1-3 t^{2}, \varphi_{2}(t)=t-3 t^{3}\\\ &P_{0}=(1 / 4,-1 / 8)(t=-1 / 2) \quad \text { (Cubic of Tschirnhaus) } \end{aligned} $$

Step-by-Step Solution

Answer

The tangent line at \( P_0 \) is \( y = -\frac{7}{12}x + \frac{1}{8} \).

Step 1: Parameterize the Curve

The given parametric equations are \( x(t) = 1 - 3t^2 \) and \( y(t) = t - 3t^3 \). These functions describe how the x and y coordinates change as t changes.

Step 2: Confirm Point on Curve

Substitute \( t = -\frac{1}{2} \) into the parametric equations to confirm that \( P_0 \) is on the curve. Calculate \( x(-\frac{1}{2}) = 1 - 3(-\frac{1}{2})^2 = \frac{1}{4} \) and \( y(-\frac{1}{2}) = -\frac{1}{2} - 3(-\frac{1}{2})^3 = -\frac{1}{8} \), confirming \( P_0 = (\frac{1}{4}, -\frac{1}{8}) \).

Step 3: Calculate Derivatives

To find the tangent line, calculate the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). \[ \frac{dx}{dt} = -6t \; , \quad \frac{dy}{dt} = 1 - 9t^2 \].

Step 4: Evaluate Derivatives at \( t = -\frac{1}{2} \)

Substitute \(t = -\frac{1}{2}\) to find \( \frac{dx}{dt} = -6(-\frac{1}{2}) = 3 \) and \( \frac{dy}{dt} = 1 - 9(-\frac{1}{2})^2 = \frac{-7}{4} \).

Step 5: Determine Slope of Tangent Line

The slope \( m \) of the tangent line at \( (\frac{1}{4}, -\frac{1}{8}) \) is given by the ratio of derivatives: \( m = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-7/4}{3} = -\frac{7}{12} \).

Step 6: Equation of Tangent Line

Using point-slope form \( y - y_1 = m (x - x_1) \), the equation of the tangent line is \( y + \frac{1}{8} = -\frac{7}{12} (x - \frac{1}{4}) \). Simplify to get \( y = -\frac{7}{12}x + \frac{1}{8} \).

Step 7: Plot the Curve and Tangent Line

Plot the curve by evaluating \( x(t) \) and \( y(t) \) over a suitable range of \( t \) that includes \( t = -\frac{1}{2} \). Add the line \( y = -\frac{7}{12}x + \frac{1}{8} \) to the plot, passing through \( P_0 = (\frac{1}{4}, -\frac{1}{8}) \).

Key Concepts

Tangent Line

In mathematics, a tangent line is a straight line that just touches a curve at a given point, without crossing it. Think of it as the line that best represents the direction in which the curve is heading at that precise spot. Its slope is crucial because it describes the instantaneous rate at which the curve is changing at that point.

To find the equation of a tangent line to a parametric curve at a specific point, we need two things: the derivatives of the parametric equations and the coordinates of the point of tangency. The derivative helps us determine the slope of the tangent line, while the coordinates allow us to position the line accurately.

When the curve is given in parametric form, like in the exercise, we derive these slopes from the parametric equations to calculate the line's equation. We then use point-slope form, a handy tool in calculus, which is expressed as:

• \(y - y_1 = m(x - x_1)\),

where \(m\) is the slope, and \((x_1, y_1)\) is the point of tangency on the curve.

Cubic of Tschirnhaus

The Cubic of Tschirnhaus refers to a specific type of parametric curve. Named after the German mathematician Ehrenfried Walther von Tschirnhaus, this curve is defined by polynomial expressions, particularly cubic polynomials, in both x and y directions.

In our case, the curve is parameterized using:

• \(x(t) = 1 - 3t^2\)
• \(y(t) = t - 3t^3\)

Cubic curves are commonplace in calculus, offering substantial insights into curve behaviors due to their inherent flexibility and smooth transitions. They can generate loops or inflection points and exhibit diverse geometric shapes, making them essential in mathematical visuals and real-world applications, such as computer graphics and animations. Understanding the Cubic of Tschirnhaus is fundamental for grasping how complex curves can behave under different conditions.

Derivatives Calculation

Derivatives are at the heart of calculus and help us understand how things change. In parametric equations, calculating derivatives involves treating both x and y as functions of another variable, often denoted as \(t\). This derivative tells us how quickly each coordinate changes as \(t\) changes.

To find the slope of the tangent line to a parametric curve, we calculate:

• The derivative of x with respect to t: \( \frac{dx}{dt} \)
• The derivative of y with respect to t: \( \frac{dy}{dt} \)

For the given parametric curve, we determined:

• \( \frac{dx}{dt} = -6t \)
• \( \frac{dy}{dt} = 1 - 9t^2 \)

By substituting \( t = -\frac{1}{2} \) into these derivatives, we find the rate of change of each coordinate precisely at that point. This is how derivatives give us the tool to explore the slope and thus the tangent.

Slope of a Curve

The slope of a curve at any point is a measure of its steepness at that point. It's essentially the angle the curve makes with the horizontal direction. In the context of calculus, the slope of a curve at a particular point is the same as the slope of the tangent line at that point.

To find the slope of the tangent line for parametric equations, we use:

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

In our exercise, it simplified to \( m = -\frac{7}{12} \), indicating the rate at which y changes with respect to x as we move along the curve.

Understanding this slope helps in visualizing how the curve aligns or deviates from a straight path. A positive slope indicates an upward movement, a negative denotes downward, and a zero slope means the curve is momentarily flat. These concepts are integral to not only mathematics but also fields like physics, where they interpret real-world phenomena.