Problem 88 - Thomas Calculus

Question

In Exercises \(87-94\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=\cos t, \quad y=\sqrt{3} \cos t, \quad t=2 \pi / 3 $$

Step-by-Step Solution

Answer

The tangent line equation is \( y = \sqrt{3}x + \frac{1}{2} \), and \( \frac{d^2 y}{dx^2} = 0 \) at \( t = \frac{2\pi}{3} \).

Step 1: Parametric Derivatives

To find the tangent line to the curve, we need to calculate the first derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Given \( x = \cos t \) and \( y = \sqrt{3} \cos t \), we find:\[ \frac{dx}{dt} = -\sin t \] \[ \frac{dy}{dt} = -\sqrt{3} \sin t \]

Step 2: Find dy/dx

The slope of the tangent line is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Substitute the derivatives from Step 1:\[ \frac{dy}{dx} = \frac{-\sqrt{3} \sin t}{-\sin t} = \sqrt{3} \]

Step 3: Evaluate Derivatives at t=2π/3

Substitute \( t = \frac{2\pi}{3} \) to find the coordinates and slope:- \( x = \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2} \)- \( y = \sqrt{3} \cos \left( \frac{2\pi}{3} \right) = -\frac{\sqrt{3}}{2} \)- \( \frac{dy}{dx} = \sqrt{3} \)

Step 4: Write Equation of Tangent Line

Use the point-slope form of the equation for the tangent line:\[ y - y_1 = m(x - x_1) \]where \( (x_1, y_1) = \left( -\frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \) and \( m = \sqrt{3} \):\[ y + \frac{\sqrt{3}}{2} = \sqrt{3}(x + \frac{1}{2}) \]Simplifying gives:\[ y = \sqrt{3}x + \frac{1}{2} \]

Step 5: Parametric Second Derivative

To find \( \frac{d^2 y}{dx^2} \), use the formula:\[ \frac{d^2 y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \]where \( \frac{dy}{dx} = \sqrt{3} \) is constant, so \( \frac{d}{dt}\left(\frac{dy}{dx}\right) = 0 \). Thus:\[ \frac{d^2 y}{dx^2} = \frac{0}{-\sin t} = 0 \]

Step 6: Evaluate Second Derivative at t=2π/3

The second derivative \( \frac{d^2 y}{dx^2} \) is zero for any \( t \), including \( t = \frac{2\pi}{3} \), as calculated above.

Key Concepts

Parametric Equations

Parametric equations are a pair of equations that use a third variable, usually denoted as \( t \), to define both \( x \) and \( y \) values in a plane. This way of expressing curves is different from the usual Cartesian equations and allows us to trace the path of a point in a continuous manner.
For example, in the given problem, we have:

• \( x = \cos t \)
• \( y = \sqrt{3} \cos t \)

These equations describe a curve, using \( t \) as the parameter that varies along the curve. As \( t \) changes, different points on the curve are generated by the corresponding \( x \) and \( y \) values.
Parametric equations are particularly useful for describing paths where both horizontal and vertical components need to be expressed in terms of an independent parameter like time.

First Derivative

The first derivative in calculus represents the rate of change of a function. When dealing with parametric equations, the first derivative \( \frac{dy}{dx} \) gives us the slope of the tangent line at a given point on the curve. To find this slope, we need to compute the derivatives of \( x \) and \( y \) with respect to \( t \):

• \( \frac{dx}{dt} = -\sin t \)
• \( \frac{dy}{dt} = -\sqrt{3} \sin t \)

The slope of the tangent line, \( \frac{dy}{dx} \), is given by the ratio of these derivatives:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \sqrt{3}\]This tells us that for our specific curve, the slope of the tangent at \( t = \frac{2\pi}{3} \) is constant at \( \sqrt{3} \), illustrating a steady rate of change.

Second Derivative

The second derivative, denoted \( \frac{d^2y}{dx^2} \), measures the concavity or convexity of the curve at a specific point, which helps us understand how the slope of the tangent line itself changes. For parametric equations, the second derivative can be found using this formula:\[\frac{d^2 y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}\]Since \( \frac{dy}{dx} = \sqrt{3} \) is constant, its derivative with respect to \( t \) is zero:\[\frac{d}{dt}(\frac{dy}{dx}) = 0\]Thus, the second derivative is:\[\frac{d^2 y}{dx^2} = \frac{0}{-\sin t} = 0\]This result indicates that the curve is linear in the vicinity of \( t = \frac{2\pi}{3} \), meaning it does not curve upwards or downwards, showing no acceleration in the path's direction.

Point-Slope Form

The point-slope form of a linear equation is a way to write the equation of a line using the slope and a specific point on the line. This form is particularly useful when we already know these two pieces of information, as in the case of finding a tangent line to a curve.
The equation is written as:\[y - y_1 = m(x - x_1)\]Where:

• \( (x_1, y_1) \) are the coordinates of the point on the line (here, \( x_1 = -\frac{1}{2} \) and \( y_1 = -\frac{\sqrt{3}}{2} \))
• \( m \) is the slope of the line (here, \( m = \sqrt{3} \))

By substituting these values, the tangent line equation becomes:\[y + \frac{\sqrt{3}}{2} = \sqrt{3}(x + \frac{1}{2})\]Simplifying, this leads to:\[y = \sqrt{3}x + \frac{1}{2}\]Using the point-slope form allows us to quickly and accurately find the equation of the tangent line to the curve at the required point with ease.