Problem 94 - 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. $$ x=\sec ^{2} t-1, \quad y=\tan t, \quad t=-\pi / 4 $$

Step-by-Step Solution

Answer

Tangent line: \(y = -\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}} - 1\); \(\frac{d^2y}{dx^2} = -\frac{\sqrt{2}}{16}\).

Step 1: Find the coordinates at the given t-value

To find the coordinates of the point on the curve where \(t = -\frac{\pi}{4}\), substitute \(t = -\frac{\pi}{4}\) into the parametric equations.For \(x\):\[x = \sec^2\left(-\frac{\pi}{4}\right) - 1 = (\sqrt{2})^2 - 1 = 2 - 1 = 1\]For \(y\):\[y = \tan\left(-\frac{\pi}{4}\right) = -1\]So the point is \((1, -1)\).

Step 2: Determine the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

Calculate \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) for the given parametric equations.For \(x = \sec^2 t - 1\):\[\frac{dx}{dt} = \frac{d}{dt}(\sec^2 t) = 2 \sec t \tan t\]For \(y = \tan t\):\[\frac{dy}{dt} = \sec^2 t\]

Step 3: Calculate \(\frac{dy}{dx}\) using \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

The derivative \(\frac{dy}{dx}\) is given by the quotient \(\frac{dy/dt}{dx/dt}\).\[\frac{dy}{dx} = \frac{\sec^2 t}{2 \sec t \tan t} = \frac{\sec t}{2 \tan t} = \frac{1}{2 \sin t}\]

Step 4: Evaluate \(\frac{dy}{dx}\) at \(t = -\frac{\pi}{4}\)

Substitute \(t = -\frac{\pi}{4}\) into \(\frac{dy}{dx}\):\[\frac{dy}{dx} = \frac{1}{2 \sin\left(-\frac{\pi}{4}\right)} = \frac{1}{2(-\frac{\sqrt{2}}{2})} = -\frac{1}{\sqrt{2}}\]

Step 5: Write the equation of the tangent line

The equation of the tangent line is given by the formula \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point.Substituting \((1, -1)\) and \(m = -\frac{1}{\sqrt{2}}\), we get:\[y + 1 = -\frac{1}{\sqrt{2}}(x - 1)\]Simplifying further:\[y = -\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}} - 1\]

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

First, find \(\frac{d^2y}{dt^2}\) and substitute into \[\frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right)/\frac{dx}{dt}\].Since \(\frac{dy}{dx} = \frac{\sec t}{2 \tan t}\), differentiate to get \(\frac{d}{dt}\left(\frac{1}{2\sin t}\right)= -\frac{1}{2\cos^2 t}\).Substituting \(\frac{dx}{dt} = 2\sec t \tan t\):\[\frac{d^2y}{dx^2} = \frac{-\frac{1}{2\cos^2 t}}{2\sec t \tan t}\approx -\frac{\cos^3 t}{4\sin^2 t}\],Then evaluate at \(t = -\frac{\pi}{4}\): \[\frac{d^2y}{dx^2} = -\frac{(\sqrt{2}/2)^3}{4(\sqrt{2}/2)^2} = -\frac{\sqrt{2}/8}{2} = -\frac{\sqrt{2}}{16}\].

Key Concepts

Parametric Equations

Parametric equations allow us to define a set of equations where the coordinates of the points

• \(x(t)\) describes the horizontal position as a function of \(t\)
• \(y(t)\) describes the vertical position as a function of \(t\)

In our exercise, the parametric equations given are:\[ x = \sec^2 t - 1 \] \[ y = \tan t \]We are tasked with understanding these equations to analyze a curve more effectively.These are useful because they express complex curves or paths in terms of a parameter \(t\), rather than relying on a static function of \(x\) and \(y\).

This allows us to model motion and transformation more naturally by adjusting \(t\). Evaluating these at \(t = -\frac{\pi}{4}\) gives us the point (1, -1).

Second Derivative

The second derivative in calculus helps us to understand the curvature or concavity of a function or curve.While the first derivative gives the slope of the tangent line, the second derivative

• Indicates the rate of change of the slope.
• Determines if a curve is concave up or concave down at a given point.

In the given context, the expression for the second derivative \( \frac{d^2y}{dx^2} \) uses:\[\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}\]Here, we found through evaluation at \(t = -\frac{\pi}{4}\) that

\[\frac{d^2y}{dx^2} = -\frac{\sqrt{2}}{16}\]This result conveys that at that particular \(t\) value, the curve is bending in such a way, and we understand how the rate of change of the slope is behaving.

Calculus Exercises

Working through calculus exercises, like finding tangent lines and derivatives, develops strong foundational skills.

Such exercises bolster:

• Conceptual understanding of derivatives and second derivatives.
• Ability to manipulate and differentiate equations, whether parametric or not.

By practicing problems involving parametric equations and deriving important features like tangent lines, students improve their problem-solving skills in calculus.
Each exercise might seem challenging, but breaking it down into smaller steps, as shown, can make the process more approachable. Grasping these concepts is essential for further study in calculus and higher-level mathematics.

Tangent Line Equation

The equation of the tangent line is a fundamental concept that provides a linear approximation to a curve at a designated point.In order to find the equation,

• We start with finding the slope of the tangent line using the derivative \( \frac{dy}{dx} \).
• The slope is then used in the linear equation form \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point on the curve.

For our parametric equations, we found \( \frac{dy}{dx} = -\frac{1}{\sqrt{2}} \) at \(t = -\frac{\pi}{4}\), giving us the tangent line:\[y = -\frac{1}{\sqrt{2}}x + \frac{1}{\sqrt{2}} - 1\]This equation describes a straight line that just touches the curve at the point (1, -1) and represents the best linear approximation of the curve's behavior near that point.
Understanding tangent lines is crucial in calculus as they help in approximations and can provide insights into the behavior of a complex curve.