2.4 Q-3

Question Statement

Find $\frac{dy}{dx}$ for the given parametric equations:

1. $x = \theta + \frac{1}{\theta}, \quad y = \theta + 1$
2. $x = \frac{a(1-t)}{1+t^2}, \quad y = \frac{2bt}{1+t^2}$

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Background and Explanation

To find the derivative $\frac{dy}{dx}$ for parametric equations, we use the chain rule. The chain rule for parametric equations is:

$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$$

This requires calculating the derivatives of $x$ and $y$ with respect to the parameter (e.g., $\theta$ or $t$).

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Solution

Part 1: $x = \theta + \frac{1}{\theta}, \quad y = \theta + 1$

1. Find $\frac{dx}{d\theta}$:

$$\frac{dx}{d\theta} = \frac{d}{d\theta}\left(\theta + \frac{1}{\theta}\right) = 1 - \frac{1}{\theta^2}.$$

Combine under a common denominator:

$$\frac{dx}{d\theta} = \frac{\theta^2 - 1}{\theta^2}.$$

2. Find $\frac{dy}{d\theta}$:

$$y = \theta + 1 \implies \frac{dy}{d\theta} = 1.$$

3. Find $\frac{dy}{dx}$: Using the chain rule:

$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{1}{\frac{\theta^2 - 1}{\theta^2}} = \frac{\theta^2}{\theta^2 - 1}.$$

Answer:

$$\frac{dy}{dx} = \frac{\theta^2}{\theta^2 - 1}.$$

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Part 2: $x = \frac{a(1-t)}{1+t^2}, \quad y = \frac{2bt}{1+t^2}$

1. Find $\frac{dx}{dt}$: Using the quotient rule:

$$x = \frac{a(1-t)}{1+t^2} \implies \frac{dx}{dt} = \frac{(1+t^2)(-a) - a(1-t)(2t)}{(1+t^2)^2}.$$

Simplify the numerator:

$$\frac{dx}{dt} = \frac{-a(1+t^2) - 2at(1-t)}{(1+t^2)^2}.$$

Expand terms:

$$\frac{dx}{dt} = \frac{-a - at^2 - 2at + 2at^2}{(1+t^2)^2} = \frac{-a - 2at + at^2}{(1+t^2)^2}.$$

2. Find $\frac{dy}{dt}$: Using the quotient rule:

$$y = \frac{2bt}{1+t^2} \implies \frac{dy}{dt} = \frac{(1+t^2)(2b) - 2bt(2t)}{(1+t^2)^2}.$$

Simplify the numerator:

$$\frac{dy}{dt} = \frac{2b(1+t^2) - 4bt^2}{(1+t^2)^2}.$$

Combine terms:

$$\frac{dy}{dt} = \frac{2b - 2bt^2}{(1+t^2)^2}.$$

3. Find $\frac{dy}{dx}$: Using the chain rule:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2b(1-t^2)}{(1+t^2)^2}}{\frac{-a(1+t^2) - 2at}{(1+t^2)^2}}.$$

Cancel common terms $(1+t^2)^2$:

$$\frac{dy}{dx} = \frac{2b(1-t^2)}{-a(1+t^2) - 2at}.$$

Factor out common terms:

$$\frac{dy}{dx} = -\frac{b(1-t^2)}{a(1+t^2) + 2at}.$$

Answer:

$$\frac{dy}{dx} = -\frac{b(1-t^2)}{a(1+t^2) + 2at}.$$

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Key Formulas or Methods Used

1. Derivative of parametric equations:

$$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}.$$

2. Quotient Rule:

$$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.$$

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Summary of Steps

Part 1

1. Compute $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$.
2. Use the chain rule to find $\frac{dy}{dx}$.
3. Simplify the result: $\frac{dy}{dx} = \frac{\theta^2}{\theta^2 - 1}$.

Part 2

1. Compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ using the quotient rule.
2. Use the chain rule: $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.
3. Simplify the result: $-\frac{b(1-t^2)}{a(1+t^2) + 2at}$.