2.7 Q-3

Question Statement

Find $y_2$ for the following equations:

i. $x^2 + y^2 = a^2$

ii. $x^3 - y^3 = a^3$

iii. $x = a \cos \phi$, $y = a \sin \phi$

iv. $x = a t^2$, $y = b t^4$

v. $x^2 + y^2 + 2gx + 2fy + c = 0$

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Background and Explanation

To solve these types of problems, the key concept is to differentiate the given equations with respect to $x$ to find the first derivative $y_1$. Then, differentiate again to find the second derivative $y_2$. The general approach involves applying implicit differentiation and algebraic manipulation to solve for $y_2$.

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Solution

i. $x^2 + y^2 = a^2$

1. First Differentiation: Differentiating both sides with respect to $x$, we get:

$$2x + 2y \frac{dy}{dx} = 0$$

Simplifying:

$$x + y \frac{dy}{dx} = 0 \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x}{y} \quad \text{(Equation 1)}$$

2. Second Differentiation: Differentiating Equation 1 with respect to $x$:

$$y_2 = \frac{-y^2 - x^2}{y^2} = \frac{a^2}{y^2}$$

Answer: $y_2 = \frac{a^2}{y^2}$

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ii. $x^3 - y^3 = a^3$

1. First Differentiation: Differentiating both sides with respect to $x$:

$$3x^2 - 3y^2 \frac{dy}{dx} = 0$$

Simplifying:

$$y^2 \frac{dy}{dx} = x^2 \quad \Rightarrow \quad \frac{dy}{dx} = \frac{x^2}{y^2} \quad \text{(Equation 1)}$$

2. Second Differentiation: Differentiating Equation 1 with respect to $x$:

$$y_2 = \frac{2xy^2 - 2x^4 / y}{y^4} = \frac{2x(x^3 - y^3)}{y^5} = \frac{2a^3x}{y^5}$$

Answer: $y_2 = \frac{2a^3x}{y^5}$

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iii. $x = a \cos \phi$, $y = a \sin \phi$

1. First Differentiation:

$$\frac{dx}{d\phi} = -a \sin \phi, \quad \frac{dy}{d\phi} = a \cos \phi$$

Using the chain rule:

$$\frac{dy}{dx} = \frac{a \cos \phi}{-a \sin \phi} = -\cot \phi \quad \text{(Equation 1)}$$

2. Second Differentiation: Differentiating Equation 1:

$$y_2 = \csc^2 \phi \times \left( -\frac{1}{a \sin \phi} \right) = -\frac{1}{a \sin^3 \phi}$$

Answer: $y_2 = -\frac{1}{a \sin^3 \phi}$

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iv. $x = a t^2$, $y = b t^4$

1. First Differentiation:

$$\frac{dx}{dt} = 2at, \quad \frac{dy}{dt} = 4bt^3$$

Using the chain rule:

$$\frac{dy}{dx} = \frac{4bt^3}{2at} = \frac{2bt^2}{a} \quad \text{(Equation 1)}$$

2. Second Differentiation: Differentiating Equation 1:

$$y_2 = \frac{2b}{a} \left( 2t \cdot \frac{1}{2at} \right) = \frac{2b}{a^2}$$

Answer: $y_2 = \frac{2b}{a^2}$

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v. $x^2 + y^2 + 2gx + 2fy + c = 0$

1. First Differentiation: Differentiating both sides with respect to $x$:

$$2x + 2y \frac{dy}{dx} + 2g + 2f \frac{dy}{dx} = 0$$

Simplifying:

$$(y + f) \frac{dy}{dx} = -(x + g) \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{x + g}{y + f} \quad \text{(Equation 1)}$$

2. Second Differentiation: Differentiating Equation 1:

$$y_2 = \frac{(y + f)(1) - (x + g)\left(-\frac{x + g}{y + f}\right)}{(y + f)^2}$$

Simplifying:

$$y_2 = \frac{(y + f)^2 + (x + g)^2}{(y + f)^3}$$

Answer: $y_2 = \frac{c - f^2 - g^2}{(y + f)^3}$

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Key Formulas or Methods Used

• Implicit Differentiation: Used to differentiate equations that involve both $x$ and $y$ with respect to $x$.
• Chain Rule: Used in problems where $x$ and $y$ are functions of a third variable, such as $t$ or $\phi$.

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Summary of Steps

1. Differentiate the given equation with respect to $x$ to find the first derivative $y_1$.
2. Differentiate $y_1$ with respect to $x$ to find the second derivative $y_2$.
3. Substitute the values from the previous step into the formula for $y_2$ to find the final result.