Find the angle at which the two curves `x^(3)-3xy^(2)+2=0and3x^(2)y-y^(3)+3=0` intersect each other.

Answer

To find the angle at which the two curves \( x^3 - 3xy^2 + 2 = 0 \) and \( 3x^2y - y^3 + 3 = 0 \) intersect, we will follow these steps:

### Step 1: Differentiate the first curve
We start with the first curve:
\[ x^3 - 3xy^2 + 2 = 0 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(x^3) - \frac{d}{dx}(3xy^2) + \frac{d}{dx}(2) = 0 \]
Using the product rule for \( 3xy^2 \):
\[ 3x^2 - 3\left(y^2 + 2y\frac{dy}{dx}x\right) = 0 \]
This simplifies to:
\[ 3x^2 - 3y^2 - 6xy\frac{dy}{dx} = 0 \]
Rearranging gives us:
\[ 3x^2 - 3y^2 = 6xy\frac{dy}{dx} \]
Thus, we can express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{x^2 - y^2}{2xy} \]

### Step 2: Differentiate the second curve
Now, we differentiate the second curve:
\[ 3x^2y - y^3 + 3 = 0 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(3x^2y) - \frac{d}{dx}(y^3) + \frac{d}{dx}(3) = 0 \]
Using the product rule for \( 3x^2y \):
\[ 3(2xy + x^2\frac{dy}{dx}) - 3y^2\frac{dy}{dx} = 0 \]
This simplifies to:
\[ 6xy + 3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0 \]
Rearranging gives us:
\[ 6xy = (3y^2 - 3x^2)\frac{dy}{dx} \]
Thus, we can express \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{2xy}{y^2 - x^2} \]

### Step 3: Find the slopes at the point of intersection
Let the point of intersection be \( (x_1, y_1) \). The slopes of the tangents at this point for the two curves are:
1. For the first curve:
\[ m_1 = \frac{x_1^2 - y_1^2}{2x_1y_1} \]
2. For the second curve:
\[ m_2 = \frac{2x_1y_1}{y_1^2 - x_1^2} \]

### Step 4: Find the angle between the two curves
The angle \( \theta \) between the two curves can be found using the formula:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| \]
To find the angle at which the curves intersect, we can also check if the product of the slopes is -1, which indicates that they intersect at right angles.

### Step 5: Check for right angle intersection
Calculating the product of the slopes:
\[ m_1 \cdot m_2 = \left(\frac{x_1^2 - y_1^2}{2x_1y_1}\right) \cdot \left(\frac{2x_1y_1}{y_1^2 - x_1^2}\right) = -1 \]
This implies that the curves intersect at right angles.

### Conclusion
The angle at which the two curves intersect each other is \( 90^\circ \).

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