Problem 18 - 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. $$x=2 t^{3}+3 t^{2}-12 t, \quad y=2 t^{3}+3 t^{2}+1$$

Step-by-Step Solution

Answer

Horizontal tangents at (0, 1) and (13, 2); vertical tangents at (28, 13) and (-7, 6).

Step 1: Calculate Derivatives

First, find the derivatives of \( x \) and \( y \) with respect to \( t \). The derivatives are:\[ \frac{dx}{dt} = 6t^2 + 6t - 12 \]\[ \frac{dy}{dt} = 6t^2 + 6t \]

Step 2: Determine Horizontal Tangents

The tangent is horizontal where \( \frac{dy}{dt} = 0 \) and \( \frac{dx}{dt} eq 0 \). Set \( 6t^2 + 6t = 0 \) and solve for \( t \):1. Factor out 6t: \[ 6t(t + 1) = 0 \]2. Solve for \( t \): \[ t = 0 \quad \text{or} \quad t = -1 \]For horizontal tangents, calculate \( \frac{dx}{dt} \) at these points:- At \( t = 0 \): \( \frac{dx}{dt} = 6(0)^2 + 6(0) - 12 = -12 eq 0 \)- At \( t = -1 \): \( \frac{dx}{dt} = 6(-1)^2 + 6(-1) - 12 = -12 eq 0 \)

Step 3: Calculate Points for Horizontal Tangents

Find the points on the curve corresponding to \( t = 0 \) and \( t = -1 \):- At \( t = 0 \): \( x = 2(0)^3 + 3(0)^2 - 12(0) = 0 \) \( y = 2(0)^3 + 3(0)^2 + 1 = 1 \) So the point is \( (0, 1) \).- At \( t = -1 \): \( x = 2(-1)^3 + 3(-1)^2 - 12(-1) = 13 \) \( y = 2(-1)^3 + 3(-1)^2 + 1 = 2 \) So the point is \( (13, 2) \).

Step 4: Determine Vertical Tangents

The tangent is vertical where \( \frac{dx}{dt} = 0 \) and \( \frac{dy}{dt} eq 0 \). Set \( 6t^2 + 6t - 12 = 0 \) and solve for \( t \):1. Simplify and solve: \( 6(t^2 + t - 2) = 0 \)2. Factor the quadratic: \[ (t + 2)(t - 1) = 0 \]3. Solve for \( t \): \[ t = -2 \quad \text{or} \quad t = 1 \]For vertical tangents, calculate \( \frac{dy}{dt} \) at these points:- At \( t = -2 \): \( \frac{dy}{dt} = 6(-2)^2 + 6(-2) = 12 eq 0 \)- At \( t = 1 \): \( \frac{dy}{dt} = 6(1)^2 + 6(1) = 12 eq 0 \)

Step 5: Calculate Points for Vertical Tangents

Find the points on the curve corresponding to \( t = -2 \) and \( t = 1 \):- At \( t = -2 \): \( x = 2(-2)^3 + 3(-2)^2 - 12(-2) = 28 \) \( y = 2(-2)^3 + 3(-2)^2 + 1 = 13 \) So the point is \( (28, 13) \).- At \( t = 1 \): \( x = 2(1)^3 + 3(1)^2 - 12(1) = -7 \) \( y = 2(1)^3 + 3(1)^2 + 1 = 6 \) So the point is \( (-7, 6) \).

Key Concepts

Horizontal Tangents

In calculus, a horizontal tangent occurs when the slope of a curve is zero at a given point. This means the derivative of the function with respect to the parametric variable is zero. For parametric equations like

• \(x = 2t^3 + 3t^2 - 12t\) and\(y = 2t^3 + 3t^2 + 1\),

the derivative of y with respect to t is set to zero to find these horizontal tangents.First, calculate \(\frac{dy}{dt}\). In this case, \(\frac{dy}{dt} = 6t^2 + 6t\). Set this equal to zero: \[6t^2 + 6t = 0\]Factoring out the common term gives:\[6t(t + 1) = 0\]Solving for \(t\) results in \(t = 0\) or \(t = -1\). After determining \(t\), check if \(\frac{dx}{dt}\) is not zero at these points. Calculation shows:

• At \(t = 0\), \(\frac{dx}{dt} = -12 eq 0\).
• At \(t = -1\), \(\frac{dx}{dt} = -12 eq 0\).

Thus, the points where horizontal tangents occur are valid. For \(t = 0\), the point on the curve is \((0, 1)\). For \(t = -1\), the point is \((13, 2)\). These points are where the curve flattens out in these respective instances.

Vertical Tangents

Vertical tangents occur at points on a curve where the slope is undefined, i.e., \(\frac{dx}{dt} = 0\) but \(\frac{dy}{dt} eq 0\). For the given parametric equations:

• \(x = 2t^3 + 3t^2 - 12t\)
• \(y = 2t^3 + 3t^2 + 1\),

We need to find \(\frac{dx}{dt}\) where it equals zero and \(\frac{dy}{dt}\) is not zero.Calculate \(\frac{dx}{dt}\): \[\frac{dx}{dt} = 6t^2 + 6t - 12\]Set this to zero:\[6t^2 + 6t - 12 = 0\]Solve using factoring:\[6(t^2 + t - 2) = 0\]Further factor the quadratic:\[(t + 2)(t - 1) = 0\]Solving this gives \(t = -2\) or \(t = 1\). Verify by ensuring \(\frac{dy}{dt} eq 0\) at these points,

• At \(t = -2\), \(\frac{dy}{dt} = 12 eq 0\)
• At \(t = 1\), \(\frac{dy}{dt} = 12 eq 0\)

Thus, vertical tangents occur at these points:

• At \(t = -2\), the curve's point is \((28, 13)\).
• At \(t = 1\), the curve's point is \((-7, 6)\).

These points show where the tangent line would run vertically.

Derivatives in Parametric Equations

Derivatives are fundamental in understanding the behavior of curves described by parametric equations. In these equations, both \(x\) and \(y\) are expressed in terms of a third variable \(t\). To find the slope (tangent) at any point, it's crucial to take derivatives.For the equations:

• \(x = 2t^3 + 3t^2 - 12t\)
• \(y = 2t^3 + 3t^2 + 1\),

We find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) as follows:\[ \frac{dx}{dt} = 6t^2 + 6t - 12 \]\[ \frac{dy}{dt} = 6t^2 + 6t \]The slope of the tangent to the curve is given by:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]This calculation involves understanding where \(\frac{dy}{dt} = 0\) (horizontal tangents) and \(\frac{dx}{dt} = 0\) (vertical tangents). Derivatives in parametric equations are a powerful way to analyze curves involving different variables!By knowing these derivatives and their applications, one can accurately find where specific tangents (horizontal vs. vertical) occur. This understanding is essential for graphing and analyzing complex curves mathematically.