The angle of intersection of the curves `y=x^(2), 6y=7-x^(3)` at (1, 1), is

Options

A. `pi//4`
B. `pi//3`
C. `pi//2`
D. none of these

Answer

To find the angle of intersection of the curves \( y = x^2 \) and \( 6y = 7 - x^3 \) at the point \( (1, 1) \), we will follow these steps:

### Step 1: Differentiate the first curve
The first curve is given by:
\[ y = x^2 \]
Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = 2x \]

### Step 2: Evaluate the slope of the first curve at the point (1, 1)
Substituting \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{(1, 1)} = 2 \cdot 1 = 2 \]
Thus, the slope \( M_1 \) of the first curve at the point \( (1, 1) \) is:
\[ M_1 = 2 \]

### Step 3: Differentiate the second curve
The second curve is given by:
\[ 6y = 7 - x^3 \]
Rearranging gives:
\[ y = \frac{7 - x^3}{6} \]
Now, differentiating with respect to \( x \):
\[ \frac{dy}{dx} = \frac{-3x^2}{6} = -\frac{x^2}{2} \]

### Step 4: Evaluate the slope of the second curve at the point (1, 1)
Substituting \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{(1, 1)} = -\frac{1^2}{2} = -\frac{1}{2} \]
Thus, the slope \( M_2 \) of the second curve at the point \( (1, 1) \) is:
\[ M_2 = -\frac{1}{2} \]

### Step 5: Use the formula for the angle of intersection
The angle \( \theta \) between the two curves can be found using the formula:
\[ \tan \theta = \left| \frac{M_1 - M_2}{1 + M_1 M_2} \right| \]
Substituting the values of \( M_1 \) and \( M_2 \):
\[ \tan \theta = \left| \frac{2 - (-\frac{1}{2})}{1 + 2 \cdot (-\frac{1}{2})} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{2 + \frac{1}{2}}{1 - 1} \right| = \left| \frac{\frac{5}{2}}{0} \right| \]
Since the denominator is zero, this indicates that \( \tan \theta \) approaches infinity, which means:
\[ \theta = \frac{\pi}{2} \]

### Conclusion
The angle of intersection of the curves at the point \( (1, 1) \) is:
\[ \theta = \frac{\pi}{2} \text{ radians} \]