Problem 5 (a) If $y=x^{2}+1 / x^{2}$, fi... [FREE SOLUTION]

Chapter 8: Problem 5

(a) If $y=x^{2}+1 / x^{2}$, find $\mathrm{d} y / \mathrm{d} x$ and $\mathrm{d}^{2} y / \mathrm{d} x^{2}$. Hence show that $$ x^{2} \frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+4 x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 y=12 x^{2} $$ (b) If $x=\tan t$ and $y=\cot t$, show that $\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}+2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=0$

Short Answer

Expert verified

(a) For part (a), the derivatives confirm the identity $ x^2 \frac{d^2 y}{dx^2} + 4x \frac{dy}{dx} + 2y = 12x^2 $. (b) For part (b), the transformation verifies $ \frac{d^2 y}{dx^2} + 2y \frac{dy}{dx} = 0 $.

Step by step solution

Find the first derivative of y with respect to x

Given that $ y = x^2 + \frac{1}{x^2} $, we need to differentiate it using the power rule and the derivative of a reciprocal function. First, differentiate $ x^2 $:\[ \frac{d}{dx}(x^2) = 2x \]Now, differentiate $ \frac{1}{x^2} $:\[ \frac{d}{dx}\left(\frac{1}{x^2}\right) = \frac{d}{dx}(x^{-2}) = -2x^{-3} = -\frac{2}{x^3} \]Therefore, the first derivative is:\[ \frac{dy}{dx} = 2x - \frac{2}{x^3} \]

Find the second derivative of y with respect to x

Differentiate the expression for $ \frac{dy}{dx} $ obtained in Step 1:$ \frac{dy}{dx} = 2x - \frac{2}{x^3} $.Differentiate further:- $ \frac{d}{dx}(2x) = 2 $.- $ \frac{d}{dx}\left(-\frac{2}{x^3}\right) = -2 \cdot (-3) x^{-4} = \frac{6}{x^4} $.Combine the results:\[ \frac{d^2y}{dx^2} = 2 + \frac{6}{x^4} \]

Verify the given equation involving second derivatives

The task is to show:\[ x^2 \frac{d^2 y}{dx^2} + 4x \frac{dy}{dx} + 2y = 12x^2 \]Substitute $ \frac{dy}{dx} = 2x - \frac{2}{x^3} $ and $ \frac{d^2y}{dx^2} = 2 + \frac{6}{x^4} $ into the equation:First term: \[ x^2 \left( 2 + \frac{6}{x^4} \right) = 2x^2 + \frac{6}{x^2} \]Second term:\[ 4x \left( 2x - \frac{2}{x^3} \right) = 8x^2 - \frac{8}{x^2} \]Third term ($ 2y = 2(x^2 + \frac{1}{x^2}) $):\[ 2(x^2 + \frac{1}{x^2}) = 2x^2 + \frac{2}{x^2} \]Combine and simplify the terms:\[ (2x^2 + \frac{6}{x^2}) + (8x^2 - \frac{8}{x^2}) + (2x^2 + \frac{2}{x^2}) = 12x^2 \]

Transform variables for part (b)

Given $ x=\tan t $ and $ y=\cot t $, we need to compute $ \frac{d^2 y}{dx^2} $ and prove the given differential equation.Find $ \frac{dx}{dt} $:\[ \frac{dx}{dt} = \sec^2 t \]Find $ \frac{dy}{dt} $:\[ \frac{dy}{dt} = -\csc^2 t \]

Calculate the derivatives with respect to x

Using the chain rule:First derivative $ \frac{dy}{dx} $:\[ \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \left(-\csc^2 t\right) \left(\frac{1}{\sec^2 t}\right) = -\cot^2 t \]Second derivative $ \frac{d^2y}{dx^2} $:\[ \frac{d^2y}{dx^2} = \frac{d}{dt}\left(-\cot^2 t\right) \times \frac{dt}{dx} = -2\cot t \left(-\csc^2 t \right) \times \frac{1}{\sec^2 t} = 2\cot^3 t \]

Conclude the verification for part (b)

The differential equation required to be proven is:\[ \frac{d^2 y}{dx^2} + 2y \frac{dy}{dx} = 0 \]Substitute in the values:\[ 2\cot^3 t + 2\left(\cot t\right)\left(-\cot^2 t\right) = 2\cot^3 t - 2\cot^3 t = 0 \]This proves the identity is correct, showing $ \frac{d^2 y}{dx^2} + 2y \frac{dy}{dx} = 0 $.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Derivative

In differential calculus, finding the first derivative of a function with respect to a variable is crucial. It indicates the rate of change or slope of the function. Given the function $ y = x^2 + \frac{1}{x^2} $, we want to compute $ \frac{dy}{dx} $. This involves differentiating each term separately.

The derivative of $ x^2 $ is straightforward, using the power rule, which gives $ 2x $.
The term $ \frac{1}{x^2} $ is a bit more complex, treated as $ x^{-2} $, resulting in $-2x^{-3} = -\frac{2}{x^3} $.

Hence, the first derivative is: \[ \frac{dy}{dx} = 2x - \frac{2}{x^3} \]Understanding how to calculate the first derivative is foundational for solving many calculus problems, especially when identifying specific behaviors like increasing or decreasing trends.

Second Derivative

To explore further changes in curves, such as concavity or points of inflection, we calculate the second derivative. Using the first derivative from before: $ \frac{dy}{dx} = 2x - \frac{2}{x^3} $, we differentiate again relative to $ x $ to find the second derivative.

The derivative of $ 2x $ is simply $ 2 $.
The derivative of $ -\frac{2}{x^3} $ requires the power rule once more, resulting in $ \frac{6}{x^4} $.

Thus, the second derivative is: \[ \frac{d^2y}{dx^2} = 2 + \frac{6}{x^4} \]This helps us understand more profound behavior changes in the function, revealing features of the graph related to curvature and even stability in systems.

Trigonometric Functions

Trigonometric functions are vital in calculus and various scientific applications. These include functions like sine, cosine, tangent, and cotangent. In the given problem: $ x = \tan t $ and $ y = \cot t $, we're handling derivatives of trigonometric functions.

The derivative of $ \tan t $ with respect to $ t $ is $ \sec^2 t $.
The derivative of $ \cot t $ with respect to $ t $ is $ -\csc^2 t $.

These derivatives are crucial because trig functions often convert complex relationships into simpler ones, easing the analysis and solution of differential equations and other calculus problems.

Chain Rule

The chain rule in calculus is indispensable when dealing with composite functions. It allows us to differentiate a function with respect to an outer variable, especially critical when transformations are involved. We utilize the chain rule to find derivatives with respect to $ x $ indirectly through $ t $, where $ x = \tan t $ and $ y = \cot t $.

First, compute $ \frac{dy}{dt} $ and $ \frac{dx}{dt} $.
Then relate them back to $ x $ using $ \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} $.

Here, it helped us form the expression for $ \frac{dy}{dx} $ as $ -\cot^2 t $, integrating this method is vital in calculus to bridge relationships between different variable transformations.

Differential Equations

Differential equations play a central role in modeling real-world phenomena. Often, these equations define relationships involving rates of change. In the problem, one equation to verify is: \[ x^2 \frac{d^2 y}{dx^2} + 4x \frac{dy}{dx} + 2y = 12x^2 \]We use the derivatives found to validate this equation, substitute $ \frac{dy}{dx} = 2x - \frac{2}{x^3} $ and $ \frac{d^2y}{dx^2} = 2 + \frac{6}{x^4} $. Simplifying these into the equation confirms its validity.
The process demonstrates how differential equations are solved and verified, laying the groundwork for practical applications in engineering, physics, and beyond.

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Short Answer

Step by step solution

Find the first derivative of y with respect to x

Find the second derivative of y with respect to x

Verify the given equation involving second derivatives

Transform variables for part (b)

Calculate the derivatives with respect to x

Conclude the verification for part (b)

Key Concepts

First Derivative

Second Derivative

Trigonometric Functions

Chain Rule

Differential Equations

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