If `x^(2)+y^(2)=a^(2)" and "k=1//a` then k is equal to

Options

A. `(y'')/(sqrt(1+y'))`
B. (1)/y(sqrt(1+yy'')
C. `(2y'')/(sqrt(1+y'^(2)))`
D. `(y'')/(2sqrt((1+y'^(2))^(3)))`

Answer

To solve the problem step by step, we start with the given equation and the definition of \( k \).

### Step 1: Start with the given equation
We have the equation:
\[ x^2 + y^2 = a^2 \]

### Step 2: Differentiate both sides with respect to \( x \)
Differentiating both sides with respect to \( x \) gives:
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(a^2) \]
Using the chain rule on \( y^2 \), we get:
\[ 2x + 2y \frac{dy}{dx} = 0 \]
This simplifies to:
\[ x + y \frac{dy}{dx} = 0 \]

### Step 3: Solve for \( \frac{dy}{dx} \)
From the equation \( x + y \frac{dy}{dx} = 0 \), we can isolate \( \frac{dy}{dx} \):
\[ y \frac{dy}{dx} = -x \implies \frac{dy}{dx} = -\frac{x}{y} \]

### Step 4: Differentiate again to find \( \frac{d^2y}{dx^2} \)
Now we differentiate \( \frac{dy}{dx} = -\frac{x}{y} \) with respect to \( x \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) \]
Using the quotient rule, we have:
\[ \frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} \]
Substituting \( \frac{dy}{dx} = -\frac{x}{y} \) into the equation:
\[ \frac{d^2y}{dx^2} = -\frac{y + x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y - \frac{x^2}{y}}{y^2} \]
This simplifies to:
\[ \frac{d^2y}{dx^2} = -\frac{y^2 - x^2}{y^3} \]

### Step 5: Relate \( k \) to \( a \)
Given that \( k = \frac{1}{a} \) and from the original equation \( x^2 + y^2 = a^2 \), we can express \( a \):
\[ a = \sqrt{x^2 + y^2} \]
Thus, substituting this into the expression for \( k \):
\[ k = \frac{1}{\sqrt{x^2 + y^2}} \]

### Final Result
Therefore, the value of \( k \) is:
\[ k = \frac{1}{\sqrt{x^2 + y^2}} \]