Problem 54 Find all values of $t$ at whic... [FREE SOLUTION]

Chapter 10: Problem 54

Find all values of $t$ at which the parametric curve has (a) a horizontal tangent line and (b) a vertical tangent line. $$ x=2 t^{3}-15 t^{2}+24 t+7, y=t^{2}+t+1 $$

Short Answer

Expert verified

Horizontal at $t = -\frac{1}{2}$, vertical at $t = 2$.

Step by step solution

Find Derivatives for Parametric Equations

First, we need to find the derivatives $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $. These derivatives will be used to determine the slope $ \frac{dy}{dx} $ of the tangent line at any point on the curve. For $ x = 2t^3 - 15t^2 + 24t + 7 $, $ \frac{dx}{dt} = 6t^2 - 30t + 24 $. For $ y = t^2 + t + 1 $, $ \frac{dy}{dt} = 2t + 1 $.

Determine Horizontal Tangent (dy/dx = 0)

A tangent line is horizontal when the slope is zero, meaning $ \frac{dy}{dx} = 0 $. Compute $ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $. Set $ \frac{dy}{dt} = 2t + 1 = 0 $. Solving for $ t $, we find $ t = -\frac{1}{2} $ is when $ \frac{dy}{dx} = 0 $, meaning the tangent is horizontal.

Determine Vertical Tangent (dx/dt = 0)

A tangent line is vertical when $ \frac{dx}{dt} = 0 $, since $ \frac{1}{0} $ would make $ \frac{dy}{dx} $ undefined. Set $ \frac{dx}{dt} = 6t^2 - 30t + 24 = 0 $. To solve, use the quadratic formula: $ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $ with $ a = 6 $, $ b = -30 $, and $ c = 24 $. The solutions are $ t = 2 $ and $ t = 2 $.

List Values of t with Horizontal and Vertical Tangents

Combine the values found in steps 2 and 3. For a horizontal tangent, $ t = -\frac{1}{2} $. For a vertical tangent, $ t = 2 $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Tangent Line

In parametric equations, a horizontal tangent line occurs when the derivative of the parameterized curve in the y-direction is zero. This makes the slope of the tangent line zero, producing a horizontal line. To find this, we need to compute the slope in terms of parametric derivatives. When dealing with equations like

$ x = 2t^3 - 15t^2 + 24t + 7 $ and
$ y = t^2 + t + 1 $,

you start by finding

$ \frac{dy}{dt} = 2t + 1 $

For horizontal tangents, set $ \frac{dy}{dt} $ to zero. Solving the equation $ 2t + 1 = 0 $ gives us $ t = -\frac{1}{2} $. This means at this parameter value, the slope becomes $ \frac{dy}{dx} = 0 $, confirming a horizontal tangent.

Vertical Tangent Line

A vertical tangent line in parametric equations occurs when the derivative of the curve with respect to the parameter in the x-direction is zero. This makes the slope $ \frac{dy}{dx} $ undefined, as you cannot divide by zero. For the equations

$ x = 2t^3 - 15t^2 + 24t + 7 $,
find $ \frac{dx}{dt} = 6t^2 - 30t + 24 $

and set this equal to zero to find when tangents become vertical. Use the quadratic formula

$ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $
with coefficients $ a = 6 $, $ b = -30 $, and $ c = 24 $.

Solving this, we find $ t = 2 $. This result indicates that when $ t = 2 $, the tangent line to the curve becomes vertical.

Derivatives of Parametric Equations

The concept of finding derivatives of parametric equations involves differentiating both the x and y components with respect to the parameter $ t $. For parametric curves defined by

$ x(t) $ and $ y(t) $, compute the derivatives
$ \frac{dx}{dt} $ and $ \frac{dy}{dt} $.

For our example, we derived

$ \frac{dx}{dt} = 6t^2 - 30t + 24 $
$ \frac{dy}{dt} = 2t + 1 $.

These derivatives help determine how the coordinates of a point on the curve change with respect to each other as $ t $ changes. These derivatives are essential because they facilitate the calculation of the slope $ \frac{dy}{dx} $.

Slope of Tangent Line

The slope of a tangent line to a parametric curve quantifies how steep the line is at a particular point on the curve. To calculate this slope, use the chain rule for derivatives, resulting in

$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $.

This formula enables us to evaluate the change in y with respect to a change in x. For the curve given by

$ x = 2t^3 - 15t^2 + 24t + 7 $,
$ y = t^2 + t + 1 $,

we previously found

$ \frac{dy}{dt} = 2t + 1 $
$ \frac{dx}{dt} = 6t^2 - 30t + 24 $.

Substituting these into the slope equation $ \frac{dy}{dx} $ helps identify key points like horizontal and vertical tangents, providing insights into the curve's geometry.

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91影视

Find all values of \(t\) at which the parametric curve has (a) a horizontal tangent line and (b) a vertical tangent line. $$ x=2 t^{3}-15 t^{2}+24 t+7, y=t^{2}+t+1 $$

Short Answer

Step by step solution

Find Derivatives for Parametric Equations

Determine Horizontal Tangent (dy/dx = 0)

Determine Vertical Tangent (dx/dt = 0)

List Values of t with Horizontal and Vertical Tangents

Key Concepts

Horizontal Tangent Line

Vertical Tangent Line

Derivatives of Parametric Equations

Slope of Tangent Line

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