Problem 17 - Calculus Early Transcendentals

Question

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work. $$x=10-t^{2}, \quad y=t^{3}-12 t$$

Step-by-Step Solution

Answer

Horizontal tangents at (6, -16) and (6, 16), vertical tangent at (10, 0).

Step 1: Understand the Problem

We are given the parametric equations \(x = 10 - t^2\) and \(y = t^3 - 12t\). We need to find points where the tangent to the curve is horizontal and vertical.

Step 2: Find Derivatives

First, calculate the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). These represent the change in \(x\) and \(y\) with respect to \(t\). For \(x\), \(\frac{dx}{dt} = -2t\). For \(y\), \(\frac{dy}{dt} = 3t^2 - 12\).

Step 3: Conditions for Horizontal Tangents

Horizontal tangents occur when \(\frac{dy}{dt} = 0\). Set \(3t^2 - 12 = 0\) and solve for \(t\). \[ 3t^2 = 12 \] \[ t^2 = 4 \] \[ t = \pm 2 \].

Step 4: Conditions for Vertical Tangents

Vertical tangents occur when \(\frac{dx}{dt} = 0\). Set \(-2t = 0\) and solve for \(t\). Thus, \(t = 0\).

Step 5: Find Corresponding Points on the Curve

Substitute \(t = \pm 2\) into the parametric equations to find the points for horizontal tangents. For \(t = 2\): \(x = 10 - 4 = 6\), \(y = 8 - 24 = -16\), point is (6, -16). For \(t = -2\): \(x = 10 - 4 = 6\), \(y = -8 + 24 = 16\), point is (6, 16). For vertical tangent at \(t = 0\): \(x = 10\), \(y = 0\), point is (10, 0).

Step 6: Check the Results with a Graph

To verify our solutions, use a graphing tool to plot the parametric curve given by \(x = 10 - t^2\) and \(y = t^3 - 12t\). Ensure the graph shows horizontal tangents at points (6, -16) and (6, 16), and a vertical tangent at (10, 0).

Key Concepts

Horizontal Tangents

When dealing with parametric equations, horizontal tangents occur where the slope of the tangent is zero. This means that the derivative of the y-component with respect to the parameter, here denoted as \(t\), must equal zero.
For the parametric equations given by \(x = 10 - t^2\) and \(y = t^3 - 12t\), we calculate the derivative \(\frac{dy}{dt} = 3t^2 - 12\).
To find where the horizontal tangent occurs, we set \(\frac{dy}{dt} = 0\):

• \(3t^2 - 12 = 0\)
• This simplifies to \(t^2 = 4\)
• Solving for \(t\), we get \(t = \pm 2\)

Substitute these values back into the parametric equations to find the corresponding points:

• For \(t = 2\), the point is \((6, -16)\)
• For \(t = -2\), the point is \((6, 16)\)

These points are where the curve has horizontal tangents, meaning the slope at these points is zero, and the tangent lines are flat.

Vertical Tangents

Vertical tangents occur in parametric equations where the slope of the tangent is undefined. This happens when the derivative of the x-component with respect to the parameter, \(t\), equals zero.
Given the parametric equation \(x = 10 - t^2\), we find the derivative \(\frac{dx}{dt} = -2t\).
To locate where the vertical tangent appears, set \(\frac{dx}{dt} = 0\):

• \(-2t = 0\)
• Solving yields \(t = 0\)

Now, substitute \(t = 0\) into the parametric equations to find the vertical tangent's corresponding point:

• At \(t = 0\), the point is \((10, 0)\)

This means that at this point, the slope is undefined, suggesting the tangent line is vertical.

Derivatives

Derivatives play a crucial role in analyzing parametric curves. They help determine the nature of the slope of the tangent at any point on the curve.
In parametric equations, separate derivatives are calculated with respect to the parameter \(t\):

• For the x-component, \(\frac{dx}{dt} = -2t\)
• For the y-component, \(\frac{dy}{dt} = 3t^2 - 12\)

The derivative \(\frac{dy}{dx}\) can then be found by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\).
This quotient gives the actual slope of the tangent line:

• \(\frac{dy}{dx} = \frac{3t^2 - 12}{-2t}\)

By studying these derivatives, you can glean where the tangent slopes to infinite values (vertical) or zero (horizontal).
Learning to compute these derivatives efficiently is key to mastering parametric curve graphing.

Graphing Parametric Curves

Graphing parametric curves allows for visualization of equations defined by parameters. These curves don't follow the typical \(y = f(x)\) form but instead rely on pairwise defined equations like \(x(t)\) and \(y(t)\).
To graph the curve described by \(x = 10 - t^2\) and \(y = t^3 - 12t\), start by noting key points where derivative conditions (horizontal and vertical tangents) provide vital features.

• At \( t = 2\) and \( t = -2\), the curve has horizontal tangents at \((6, -16)\) and \((6, 16)\)
• At \( t = 0\), the curve displays a vertical tangent at point \((10, 0)\)

Using these points as guides along with a graphing calculator or software, sketch the curve to visualize these tangents.
Graphing provides a powerful way to confirm analytical results and to observe the overall behavior of the curve formed by the parametric equations.