If `sqrt((x^(2)+y^(2)))=ae^(tan^-1(y//x)),agt0`, then y''(0) equals

Options

A. `(a)/(2)e^(pi//2)`
B. `ae^(pi//2)`
C. `(-2)/(a)e^(-pi//2)`
D. `(a)/(2)e^(-pi//2)`

Answer

To solve the problem given, we start with the equation:

\[ \sqrt{x^2 + y^2} = a e^{\tan^{-1}(\frac{y}{x})}, \quad a > 0 \]

### Step 1: Square both sides

Squaring both sides gives us:

\[ x^2 + y^2 = a^2 e^{2 \tan^{-1}(\frac{y}{x})} \]

### Step 2: Use the identity for \(e^{2 \tan^{-1}(z)}\)

We can use the identity:

\[ e^{2 \tan^{-1}(z)} = \frac{1 + z^2}{1 - z^2} \]

where \(z = \frac{y}{x}\). Thus, we have:

\[ e^{2 \tan^{-1}(\frac{y}{x})} = \frac{1 + \left(\frac{y}{x}\right)^2}{1 - \left(\frac{y}{x}\right)^2} = \frac{x^2 + y^2}{x^2 - y^2} \]

Substituting this back into our equation gives:

\[ x^2 + y^2 = a^2 \cdot \frac{x^2 + y^2}{x^2 - y^2} \]

### Step 3: Cross-multiply

Cross-multiplying yields:

\[ (x^2 + y^2)(x^2 - y^2) = a^2 (x^2 + y^2) \]

### Step 4: Simplify

Assuming \(x^2 + y^2 \neq 0\), we can divide both sides by \(x^2 + y^2\):

\[ x^2 - y^2 = a^2 \]

### Step 5: Differentiate implicitly

Now we differentiate both sides with respect to \(x\):

\[ \frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}(a^2) \]

This gives:

\[ 2x - 2y \frac{dy}{dx} = 0 \]

Rearranging this, we find:

\[ y' = \frac{x}{y} \]

### Step 6: Find \(y'\) at \(x = 0\)

To find \(y'\) at \(x = 0\), we substitute \(x = 0\) into our earlier equation:

\[ 0^2 - y^2 = a^2 \implies -y^2 = a^2 \implies y = \pm a \]

Taking \(y = a\) (since \(a > 0\)) gives:

\[ y'(0) = \frac{0}{a} = 0 \]

### Step 7: Differentiate again to find \(y''\)

Now we differentiate \(y' = \frac{x}{y}\) again with respect to \(x\):

Using the quotient rule:

\[ y'' = \frac{y \cdot 1 - x \cdot y'}{y^2} \]

Substituting \(y' = \frac{x}{y}\):

\[ y'' = \frac{y - x \cdot \frac{x}{y}}{y^2} = \frac{y^2 - x^2}{y^3} \]

### Step 8: Evaluate \(y''(0)\)

At \(x = 0\), we have:

\[ y''(0) = \frac{y^2 - 0^2}{y^3} = \frac{y^2}{y^3} = \frac{1}{y} \]

Substituting \(y = a\):

\[ y''(0) = \frac{1}{a} \]

### Final Answer

Thus, the value of \(y''(0)\) is:

\[ y''(0) = \frac{1}{a} \]