Problem 7 - 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. $$ x=\sec t, \quad y=\tan t, \quad t=\pi / 6 $$

Step-by-Step Solution

Answer

The equation of the tangent line is \(y - \frac{1}{\sqrt{3}} = 2\sqrt{3}(x - 2)\), and \(\frac{d^2 y}{dx^2}\) at \(t = \pi/6\) is \(-4\sqrt{3}\).

Step 1: Differentiate x and y with respect to t

Given the parametric equations \(x = \sec t\) and \(y = \tan t\), differentiate to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).\[\frac{dx}{dt} = \sec t \tan t\] and \[\frac{dy}{dt} = \sec^2 t\].

Step 2: Find the slope of the tangent line

The slope of the tangent line is given by \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\). Substitute the derivatives from Step 1, and evaluate at \(t = \pi/6\).\[\frac{dy}{dx} = \frac{\sec^2 t}{\sec t \tan t} = \frac{\sec t}{\tan t}\]. At \(t = \pi/6\), \[\sec(\pi/6) = 2, \tan(\pi/6) = \frac{1}{\sqrt{3}}\], so \[\frac{dy}{dx} = \frac{2}{1/\sqrt{3}} = 2\sqrt{3}\].

Step 3: Find the point on the curve

Calculate the coordinates \((x, y)\) at \(t = \pi/6\) using the given parametric equations.\[x = \sec(\pi/6) = 2\] and \[y = \tan(\pi/6) = \frac{1}{\sqrt{3}}\]. So the point is \((2, \frac{1}{\sqrt{3}})\).

Step 4: Write the equation of the tangent line

Using the point \((2, \frac{1}{\sqrt{3}})\) and slope \(2\sqrt{3}\), apply the point-slope form of the equation: \(y - y_1 = m(x - x_1)\). Substitute \(y_1 = \frac{1}{\sqrt{3}}\), \(m = 2\sqrt{3}\), \(x_1 = 2\).\[y - \frac{1}{\sqrt{3}} = 2\sqrt{3}(x - 2)\].

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

The second derivative \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\) over \(\frac{dx}{dt}\). Using \(\frac{dy}{dt}=\sec^2 t\) and \(\frac{dx}{dt}=\sec t \tan t\), differentiate \(\frac{\sec t}{\tan t}\). Compute: \(\frac{d}{dt}\left(\frac{\sec t}{\tan t}\right)\), then divide by \(\frac{dx}{dt}\) and evaluate at \(t = \pi/6\).

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

Differentiate \( \frac{\sec t}{\tan t} = 1/\sin t\), find the derivative: \(\frac{-1}{\sin^2 t}\cos t\cos t\). Compute \(\frac{d^2 y}{dx^2} = \frac{\frac{-1}{\sin^3 t} \cos t\sec t\tan t}\), substitute \(t=\pi/6\), obtain \[\frac{d^2 y}{dx^2} = -4\sqrt{3}\].

Key Concepts

Parametric Equations

Parametric equations are a unique way to define a curve by expressing the coordinates of the points that make up the curve as functions of a variable, often referred to as the parameter. In our exercise, the parameter is denoted by \( t \). The equations \( x = \sec t \) and \( y = \tan t \) describe the position of a point on the curve based on the value of \( t \). This method is powerful because it allows for the description of curves that would be difficult to express using only \( x \) and \( y \).

• Advantages: Parametric equations can represent complex motions and provide a clear interpretation of the geometry of the curve.
• Use in Tangent Line Problems: When dealing with curves defined parametrically, parametric derivatives are crucial in finding the slope of the tangent line.

It is important to understand the roles both \( x \) and \( y \) play as functions of \( t \) as we calculate derivatives to find the slope and tangent lines to these curves.

Second Derivative

The second derivative, denoted as \( \frac{d^2 y}{dx^2} \), provides information about the curvature or concavity of the function at a given point. While the first derivative represents the slope of the tangent line, the second derivative tells us how this slope is changing.
To find the second derivative in parametric terms, differentiate the expression for \( \frac{dy}{dx} \) with respect to \( t \), and then divide by \( \frac{dx}{dt} \). Specifically in this problem, the expression \( \frac{dy}{dx} \) was calculated to be \( \frac{\sec t}{\tan t} \), which simplifies our task to differentiate with respect to \( t \).

• Curvature Insight: The sign of the second derivative (positive or negative) indicates whether the curve is "concave up" or "concave down".
• Calculation: Applying this analysis to our specific example confirmed that the second derivative evaluated at \( t = \pi/6 \) is \(-4\sqrt{3}\), indicating a change in the direction of the curve's slope.

This step completes our understanding not only of the curve's direction given by the tangent but also how that direction is evolving.

Slope of Tangent Line

Understanding the slope of the tangent line is essential in calculus and geometry. For a curve described by parametric equations, the slope at any point can be determined by taking the derivative \( \frac{dy}{dx} \).
To achieve this for parametric equations, one needs to differentiate \( y \) with respect to \( t \) and \( x \) with respect to \( t \) separately, resulting in \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). The slope \( \frac{dy}{dx} \) can then be found by dividing these two derivatives. In our scenario, the slope at \( t = \pi/6 \) was found to be \( 2\sqrt{3} \).

• Application to Line Equation: With the slope found, the tangent line equation at a specific point is derived using the point-slope form that aligns with the contour of the curve at that very point.
• Graphical Interpretation: The tangent line "touches" the curve at one point and matches the immediate direction of the curve.

Through this step, understanding the tangent line becomes crucial for analyzing how the curve behaves near that point.