Problem 3 - 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. $$ x=4 \sin t, \quad y=2 \cos t, \quad t=\pi / 4 $$

Step-by-Step Solution

Answer

The tangent line is \(y = -\frac{1}{2}x + 2\sqrt{2}\) and \(\frac{d^2y}{dx^2} = -\frac{\sqrt{2}}{4}\) at \(t=\frac{\pi}{4}\).

Step 1: Differentiate Parametric Equations

To find the tangent line, first find \( \frac{dy}{dx} \) by differentiating both \(x\) and \(y\) with respect to \(t\):\[\frac{dx}{dt} = 4 \cos t \quad \text{and} \quad \frac{dy}{dt} = -2 \sin t\]Then, apply the chain rule:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2 \sin t}{4 \cos t} = -\frac{1}{2} \tan t\]

Step 2: Evaluate the Derivative at t = \( \frac{\pi}{4} \)

Substitute \( t = \frac{\pi}{4} \) into the expression for \( \frac{dy}{dx} \):\[\frac{dy}{dx} \bigg|_{t=\frac{\pi}{4}} = -\frac{1}{2} \tan\left(\frac{\pi}{4}\right) = -\frac{1}{2} \times 1 = -\frac{1}{2}\]

Step 3: Find the Coordinates at t = \( \frac{\pi}{4} \)

Find the coordinates of the point by substituting \( t = \frac{\pi}{4} \) into the parametric equations:\[x = 4 \sin\left(\frac{\pi}{4}\right) = 4 \times \frac{\sqrt{2}}{2} = 2\sqrt{2}, \quad y = 2 \cos\left(\frac{\pi}{4}\right) = 2 \times \frac{\sqrt{2}}{2} = \sqrt{2}\]Thus, the point is \((2\sqrt{2}, \sqrt{2})\).

Step 4: Write the Equation of the Tangent Line

Use the point-slope form \(y - y_1 = m(x - x_1)\) where \((x_1, y_1) = (2\sqrt{2}, \sqrt{2})\) and \(m = -\frac{1}{2}\):\[y - \sqrt{2} = -\frac{1}{2}(x - 2\sqrt{2})\]Simplifying gives:\[y = -\frac{1}{2}x + \sqrt{2} + \sqrt{2} = -\frac{1}{2}x + 2\sqrt{2}\]

Step 5: Calculate \( \frac{d^2y}{dx^2} \)

The second derivative \( \frac{d^2y}{dx^2} \) involves differentiating \( \frac{dy}{dx} \) with respect to \(t\) and then dividing by \( \frac{dx}{dt} \):\[\frac{d^2y}{dx^2} = \frac{d}{dt} \left( -\frac{1}{2} \tan t \right) \div \frac{dx}{dt}\]Calculate the derivative:\[-\frac{1}{2} \sec^2 t \quad \text{and divide by} \quad 4 \cos t\]Thus:\[\frac{d^2y}{dx^2} = \frac{-\frac{1}{2} \sec^2 t}{4 \cos t}\]

Step 6: Evaluate \( \frac{d^2y}{dx^2} \) at t = \( \frac{\pi}{4} \)

Substitute \( t = \frac{\pi}{4} \) into the expression for \( \frac{d^2y}{dx^2} \):\[\frac{d^2y}{dx^2} \bigg|_{t=\frac{\pi}{4}} = \frac{-\frac{1}{2} \sec^2\left(\frac{\pi}{4}\right)}{4 \cdot \frac{\sqrt{2}}{2}}\]Calculate \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \):\[\frac{-\frac{1}{2} \cdot 2}{4 \cdot \frac{\sqrt{2}}{2}} = \frac{-1}{4 \cdot \frac{1}{\sqrt{2}}} = \frac{-\sqrt{2}}{4}\]

Key Concepts

parametric equations

Parametric equations provide a way to define a curve using parameters, usually denoted as \( t \). Instead of expressing \( y \) directly in terms of \( x \), both \( x \) and \( y \) are written as separate functions of the parameter \( t \). This allows for a flexible representation of more complex curves that can't easily be achieved with standard Cartesian equations.

In the given exercise, the parametric equations are:

• \( x = 4 \sin t \)
• \( y = 2 \cos t \)

By substituting different values of \( t \), you can find the corresponding values of \( x \) and \( y \) to plot the curve on a graph. Parametric equations are particularly useful for modeling paths and motions in physics and engineering.

derivative

The derivative is a fundamental concept used to understand how a function changes at any point. In the context of parametric equations, since both \( x \) and \( y \) are functions of \( t \), we find their derivatives with respect to \( t \).

The derivatives are evaluated as follows:

• \( \frac{dx}{dt} = 4 \cos t \)
• \( \frac{dy}{dt} = -2 \sin t \)

By applying the chain rule, we find the derivative of \( y \) with respect to \( x \), noted as \( \frac{dy}{dx} \):

\( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -\frac{1}{2} \tan t \)

This represents the slope of the tangent line to the curve at any given \( t \). Evaluating this derivative at \( t = \frac{\pi}{4} \), we find that the slope at this point is \(-\frac{1}{2}\).

second derivative

While the first derivative \( \frac{dy}{dx} \) gives us the slope or rate of change at any point, the second derivative \( \frac{d^2y}{dx^2} \) provides information about the curvature or concavity of the function. It helps us understand how the curve itself changes, not just its slope.

To find the second derivative for parametric equations, you must first differentiate \( \frac{dy}{dx} \) with respect to \( t \), then divide the result by \( \frac{dx}{dt} \). For this exercise, it is calculated as:

\( \frac{d^2y}{dx^2} = \frac{-\frac{1}{2} \sec^2 t}{4 \cos t} \)

By evaluating \( \frac{d^2y}{dx^2} \) at \( t = \frac{\pi}{4} \), you find the value \( \frac{-\sqrt{2}}{4} \), which indicates the curvature at that specific point on the curve.

chain rule

The chain rule is a technique in calculus for differentiating composite functions. It is especially useful in parametric equations because it allows us to find derivatives with respect to one variable, while both components are expressed as functions of a third variable \( t \).

For a function \( y = f(g(t)) \), the chain rule states that \( \frac{dy}{dt} = \frac{df}{dg} \cdot \frac{dg}{dt} \).

In our example, the chain rule is applied to determine \( \frac{dy}{dx} \) as \( \frac{dy}{dt} \div \frac{dx}{dt} \). This calculated slope is essential for understanding the behavior of the curve generated by the parametric equations and is used to find the tangent line and the second derivative.

• Useful for dealing with derivatives in complex situations.
• Simplifies the differentiation process for parametric forms.

By mastering the chain rule, you can tackle a variety of problems involving interconnected functions and parameters. It is crucial for working confidently with parametric curves and understanding their geometric properties.