Find the angle between the curves `x^(2)-(y^(2))/(3)=a^(2)andax^(3)=c.`

Answer

To find the angle between the curves given by the equations \( x^2 - \frac{y^2}{3} = a^2 \) and \( ax^3 = c \), we will follow these steps:

### Step 1: Differentiate the first curve
The first curve is given by:
\[ x^2 - \frac{y^2}{3} = a^2 \]
To find the slope of the tangent to this curve, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}\left(x^2\right) - \frac{1}{3}\frac{d}{dx}(y^2) = 0 \]
This gives us:
\[ 2x - \frac{2y}{3} \frac{dy}{dx} = 0 \]
Rearranging this, we find:
\[ \frac{dy}{dx} = \frac{3x}{y} \]
Let this slope be \( m_1 = \frac{3x}{y} \).

### Step 2: Differentiate the second curve
The second curve is given by:
\[ ax^3 = c \]
Differentiating both sides with respect to \( x \) using the product rule:
\[ a \cdot 3x^2 = 0 \]
This simplifies to:
\[ 3ax^2 + y^3 \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = -\frac{3ax^2}{y^3} \]
Let this slope be \( m_2 = -\frac{3ax^2}{y^3} \).

### Step 3: Calculate the product of the slopes
Now we calculate the product of the slopes \( m_1 \) and \( m_2 \):
\[ m_1 \cdot m_2 = \left(\frac{3x}{y}\right) \left(-\frac{3ax^2}{y^3}\right) = -\frac{9ax^3}{y^4} \]

### Step 4: Determine the angle between the curves
The angle \( \theta \) between the two curves can be found using the formula:
\[ \tan(\theta) = \left| \frac{m_1 + m_2}{1 - m_1 m_2} \right| \]
Since we have already calculated \( m_1 m_2 = -\frac{9ax^3}{y^4} \), we can substitute this into the formula. However, we note that if \( m_1 m_2 = -1 \), then the curves are perpendicular, and the angle between them is \( \frac{\pi}{2} \).

### Conclusion
In this case, we have:
\[ m_1 \cdot m_2 = -1 \]
Thus, the angle between the curves is:
\[ \theta = \frac{\pi}{2} \]

### Final Answer
The angle between the curves is \( \frac{\pi}{2} \).