Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=\frac{2}{1+t^{2}}, y=\frac{2}{t\left(1+t^{2}\right)} ; t \neq 0 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1 + 3t^2}{2t^3} \), \( \frac{d^2y}{dx^2} = \frac{3(1+t^2)}{8t^5} \).

Step 1: Find dx/dt

Given the parameterized function \( x = \frac{2}{1+t^2} \), we need to differentiate \( x \) with respect to \( t \). Applying the quotient rule, \( \frac{d}{dt}\left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = 2 \) and \( v = 1 + t^2 \):\[ \frac{dx}{dt} = \frac{0 \cdot (1+t^2) - 2 \cdot 2t}{(1+t^2)^2} = \frac{-4t}{(1+t^2)^2} \]

Step 2: Find dy/dt

Now, differentiate \( y = \frac{2}{t(1+t^2)} \) with respect to \( t \):Here, \( u = 2 \) and \( v = t(1 + t^2) \).Using quotient rule:\[ \frac{d}{dt}(t(1+t^2)) = (1+t^2) + t \cdot 2t = 1 + 3t^2 \]Thus:\[ \frac{dy}{dt} = \frac{0 \cdot t(1+t^2) - 2(1 + 3t^2)}{t^2(1+t^2)^2} = \frac{-2(1 + 3t^2)}{t^2(1+t^2)^2} \]

Step 3: Determine dy/dx

To find \( \frac{dy}{dx} \) using parametric equations, use the formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \):\[ \frac{dy}{dx} = \frac{-2(1 + 3t^2)}{t^2(1+t^2)^2} \times \frac{(1+t^2)^2}{-4t} \]After cancelling and simplifying:\[ \frac{dy}{dx} = \frac{2(1 + 3t^2)}{4t^3} = \frac{1 + 3t^2}{2t^3} \]

Step 4: Differentiate dy/dx with respect to t

To find \( \frac{d^2y}{dx^2} \), we first differentiate \( \frac{dy}{dx} = \frac{1 + 3t^2}{2t^3} \) with respect to \( t \):Use the quotient rule again:Let \( u = 1 + 3t^2 \) and \( v = 2t^3 \).\[ \frac{d}{dt}(2t^3) = 6t^2 \]\[ \frac{d}{dt}(1 + 3t^2) = 6t \]Using quotient rule:\[ \frac{d}{dt}\left( \frac{1 + 3t^2}{2t^3} \right) = \frac{6t(2t^3) - (1 + 3t^2)6t^2}{(2t^3)^2} \]Simplifying:\[ = \frac{12t^4 - 6t^2 - 18t^4}{4t^6} = \frac{-6t^2 - 6t^4}{4t^6} = \frac{-6(1 + t^2)}{4t^4} \]\[ = \frac{-3(1 + t^2)}{2t^4} \]

Step 5: Compute d^2y/dx^2

Now, substitute back the derivative \( \frac{dx}{dt} \) to get the second derivative with respect to \( x \):\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \bigg/ \frac{dx}{dt} \]\[ \frac{d^2y}{dx^2} = \frac{-3(1+t^2)}{2t^4} \times \frac{(1+t^2)^2}{-4t} \]After cancellation and simplification:\[ = \frac{3(1+t^2)}{8t^5} \]

Key Concepts

Parametric Equations

Parametric equations allow for the expression of a set of related quantities as functions of an independent variable, known as a parameter. Rather than describing a curve using a direct relation between two variables, say \(x\) and \(y\), parametric equations use an intermediary variable \(t\) to define \(x\) and \(y\) separately. This can make it easier to describe complex curves.
In this example, we have:

• \(x = \frac{2}{1+t^2}\)
• \(y = \frac{2}{t(1+t^2)}\)

The parameter \(t\) is used to express both \(x\) and \(y\), allowing one to explore the relationship between the two by varying \(t\). It provides flexibility to express curves that cannot be represented by simple functions in Cartesian coordinates. This approach can also simplify the process of differentiation and integration, as seen in calculus problems involving these types of equations.

Quotient Rule

The quotient rule is a technique for differentiating expressions that involve division of one function by another. It states that if you have a function in the form \(\frac{u}{v}\), its derivative is given by:\[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
In the parametric context of this exercise:

• When differentiating \(x = \frac{2}{1+t^2}\), \(u = 2\) and \(v = 1 + t^2\).
• For \(y = \frac{2}{t(1+t^2)}\), \(u = 2\) and \(v = t(1 + t^2)\).

This rule is crucial for finding the rate of change in parametric equations, especially when both \(x\) and \(y\) are expressed in terms of \(t\). By applying the quotient rule, we can effectively derive \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) which are necessary for further calculations like finding the derivatives of \(x\) and \(y\) with respect to \(t\).

Derivative

The derivative represents the rate of change of a function. For parametric equations, the derivative \(\frac{dy}{dx}\) reveals how \(y\) changes with respect to \(x\). To find this, we use the chain rule of calculus in the form:\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]
This equation allows us to calculate the slope of the tangent line to a parametric curve without eliminating the parameter \(t\). From this exercise:

• \(\frac{dx}{dt} = \frac{-4t}{(1+t^2)^2}\)
• \(\frac{dy}{dt} = \frac{-2(1 + 3t^2)}{t^2(1+t^2)^2}\)

Substituting these into the formula, we get \(\frac{dy}{dx} = \frac{1 + 3t^2}{2t^3}\), showing how \(y\) changes with \(x\) for each value of \(t\). This is a fundamental concept in calculus, enabling us to understand and explore the behavior of functions at any given point.

Second Derivative

The second derivative gives insight into the curvature and acceleration of a curve, by determining how the rate of change itself changes. It's found by differentiating the first derivative \(\frac{dy}{dx}\) with respect to \(t\) once more:\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{dy}{dx}\right) \bigg/ \frac{dx}{dt} \]
In the context of the exercise, to find \(\frac{d^2y}{dx^2}\), we first differentiate \(\frac{dy}{dx} = \frac{1 + 3t^2}{2t^3}\) with respect to \(t\). Applying the quotient rule again leads to:

• The simplified derivative \(\frac{-3(1 + t^2)}{2t^4}\)

Finally, divide by \(\frac{dx}{dt}\) to find the second derivative in terms of \(t\). After simplification, the second derivative is \(\frac{3(1+t^2)}{8t^5}\). This tells us how the slope of the curve changes as \(x\) varies, offering deeper insight into the behavior and characteristics of the curve as traced by \(t\).