Problem 4 - Thomas Calculus

Question

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

Tangent line: \( y = \sqrt{3}x \), \( \frac{d^2y}{dx^2} = 0 \) at \( t = \frac{2\pi}{3} \).

Step 1: Parameterize the curve and find coordinates

First, substitute the given value of \( t \) into the parametric equations to find the point where the tangent line touches the curve. Here, \( x = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \) and \( y = \sqrt{3}\cos\left(\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2} \), so the point is \((-\frac{1}{2}, -\frac{\sqrt{3}}{2})\).

Step 2: Compute first derivatives

Calculate the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = \cos t \), \( \frac{dx}{dt} = -\sin t \). For \( y = \sqrt{3}\cos t \), \( \frac{dy}{dt} = -\sqrt{3}\sin t \). At \( t = \frac{2\pi}{3} \), \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \), so \( \frac{dx}{dt} = -\frac{\sqrt{3}}{2} \) and \( \frac{dy}{dt} = -\frac{3}{2} \).

Step 3: Find slope \( \frac{dy}{dx} \)

The slope of the tangent line \( \frac{dy}{dx} \) is calculated using \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Substitute the derivatives: \( \frac{dy}{dx} = \frac{-\frac{3}{2}}{-\frac{\sqrt{3}}{2}} = \frac{3}{\sqrt{3}} = \sqrt{3} \).

Step 4: Equation of the tangent line

Use the point-slope form of the line equation, \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (-\frac{1}{2}, -\frac{\sqrt{3}}{2}) \) and the slope \( m = \sqrt{3} \). The equation is: \( y + \frac{\sqrt{3}}{2} = \sqrt{3}(x + \frac{1}{2}) \). Simplify this to: \( y = \sqrt{3}x \).

Step 5: Compute second derivative \( \frac{d^2y}{dx^2} \)

To find the second derivative, use the formula: \[ \frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) / \frac{dx}{dt} \]. First, differentiate \( \frac{dy}{dx} = \sqrt{3} \) with respect to \( t \): \( \frac{d}{dt}(\sqrt{3}) = 0 \). Since \( \frac{dx}{dt} = -\frac{\sqrt{3}}{2} \) at \( t = \frac{2\pi}{3} \), \( \frac{d^2y}{dx^2} = \frac{0}{-\frac{\sqrt{3}}{2}} = 0 \).

Key Concepts

Parametric Equations

Parametric equations are a powerful way to describe curves in the plane. Instead of using a single equation like "y = f(x)", parametric equations express both x and y as functions of a third variable, often called \(t\), which represents a parameter. This approach allows us to describe more complex curves, such as ellipses or spirals, by varying \(t\) over an interval.
In our specific context, the parametric equations are given as \(x = \cos t\) and \(y = \sqrt{3} \cos t\). Here, \(t\) is an angle parameter. As \(t\) varies, these equations trace out a path on the plane. For the exercise, the given value of \(t\) is \(\frac{2\pi}{3}\), which sits at a specific point on the curve. By substituting this value into the parametric equations, we easily find the point \((-\frac{1}{2}, -\frac{\sqrt{3}}{2})\), indicating where the tangent line touches the curve.

First Derivative

The first derivative is a fundamental concept when discussing tangent lines. In parametric equations, we find the first derivative, \(\frac{dy}{dx}\), by using the chain rule. This involves taking the derivative of \(y\) with respect to \(t\) and \(x\) with respect to \(t\), and then forming the derivative \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).
From our example, \( \frac{dx}{dt} = -\sin t \) and \( \frac{dy}{dt} = -\sqrt{3}\sin t \) are the derivatives of \(x\) and \(y\) respectively. At \( t = \frac{2\pi}{3} \), \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \), leading to \( \frac{dx}{dt} = -\frac{\sqrt{3}}{2} \) and \( \frac{dy}{dt} = -\frac{3}{2} \). With these, the slope of the tangent line at the given point becomes \( \sqrt{3} \). This slope is crucial for determining the tangent line equation using the point-slope form of a line.

Second Derivative

The second derivative, \(\frac{d^2y}{dx^2}\), provides information about the curvature of the curve at a point, indicating how the direction of the tangent line changes. It gives us insight into the concavity of the curve.
To calculate it in parametric form, we use the formula: \[ \frac{d^2y}{dx^2} = \frac{d}{dt}(\frac{dy}{dx}) / \frac{dx}{dt} \]This involves differentiating the first derivative \(\frac{dy}{dx}\) with respect to \(t\). In our example, since \(\frac{dy}{dx} = \sqrt{3}\), its derivative with respect to \(t\) is zero, making the second derivative \( \frac{d^2y}{dx^2} = 0 \). This result indicates that near the point \((-\frac{1}{2}, -\frac{\sqrt{3}}{2})\), the curve does not exhibit concavity, and the direction of the tangent line doesn't change.