Find the angle between the curve `y=lnx` and `y=(lnx)^(2)` at their point of intersections.

Answer

To find the angle between the curves \( y = \ln x \) and \( y = (\ln x)^2 \) at their points of intersection, we can follow these steps:

### Step 1: Find the Points of Intersection
We need to find the points where the two curves intersect, which means we set them equal to each other:

\[ \ln x = (\ln x)^2 \]

Rearranging gives:

\[ (\ln x)^2 - \ln x = 0 \]

Factoring out \( \ln x \):

\[ \ln x (\ln x - 1) = 0 \]

This gives us two cases:

1. \( \ln x = 0 \) which implies \( x = e^0 = 1 \)
2. \( \ln x = 1 \) which implies \( x = e^1 = e \)

Thus, the points of intersection are \( x = 1 \) and \( x = e \).

### Step 2: Find the Slopes of the Tangents
Next, we differentiate both functions to find the slopes at the points of intersection.

For \( y_1 = \ln x \):

\[ \frac{dy_1}{dx} = \frac{1}{x} \]

For \( y_2 = (\ln x)^2 \), we use the chain rule:

\[ \frac{dy_2}{dx} = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x} \]

### Step 3: Calculate the Slopes at \( x = 1 \)
Substituting \( x = 1 \):

For \( y_1 \):

\[ m_1 = \frac{1}{1} = 1 \]

For \( y_2 \):

\[ m_2 = \frac{2 \ln 1}{1} = \frac{2 \cdot 0}{1} = 0 \]

### Step 4: Find the Angle Between the Curves at \( x = 1 \)
The formula for the tangent of the angle \( \theta \) between two curves is given by:

\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} \]

Substituting the values:

\[ \tan \theta = \frac{1 - 0}{1 + (1)(0)} = \frac{1}{1} = 1 \]

Thus,

\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Step 5: Calculate the Slopes at \( x = e \)
Now, we find the slopes at \( x = e \):

For \( y_1 \):

\[ m_1 = \frac{1}{e} \]

For \( y_2 \):

\[ m_2 = \frac{2 \ln e}{e} = \frac{2 \cdot 1}{e} = \frac{2}{e} \]

### Step 6: Find the Angle Between the Curves at \( x = e \)
Using the same formula:

\[ \tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2} = \frac{\frac{1}{e} - \frac{2}{e}}{1 + \left(\frac{1}{e}\right)\left(\frac{2}{e}\right)} = \frac{-\frac{1}{e}}{1 + \frac{2}{e^2}} = \frac{-1}{e(1 + \frac{2}{e^2})} = \frac{-1}{\frac{e^2 + 2}{e}} = \frac{-e}{e^2 + 2} \]

Thus,

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

### Summary of Results
- The angle between the curves at \( x = 1 \) is \( 45^\circ \).
- The angle between the curves at \( x = e \) is \( \tan^{-1}\left(\frac{-e}{e^2 + 2}\right) \).