calculate the equation of the normal line of the polar equation r=e^theta at theta = 0

To find the equation of the normal line to the polar curve \( r = e^\theta \) at \( \theta = 0 \), follow these steps:

### 1. Convert the polar equation to Cartesian form
The polar coordinates \((r, \theta)\) relate to Cartesian coordinates \((x, y)\) through the following equations:

\[
x = r \cos \theta
\]
\[
y = r \sin \theta
\]

Given \( r = e^\theta \), we can express \( x \) and \( y \) as:
\[
x = e^\theta \cos \theta
\]
\[
y = e^\theta \sin \theta
\]

### 2. Find the derivative \(\frac{dy}{dx}\)
To find the slope of the tangent line, we first find \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\):

\[
\frac{dx}{d\theta} = \frac{d}{d\theta} \left( e^\theta \cos \theta \right) = e^\theta \cos \theta - e^\theta \sin \theta
\]

\[
\frac{dy}{d\theta} = \frac{d}{d\theta} \left( e^\theta \sin \theta \right) = e^\theta \sin \theta + e^\theta \cos \theta
\]

Now, the slope of the tangent line is:

\[
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{e^\theta (\sin \theta + \cos \theta)}{e^\theta (\cos \theta - \sin \theta)} = \frac{\sin \theta + \cos \theta}{\cos \theta - \sin \theta}
\]

### 3. Evaluate the slope at \( \theta = 0 \)
Substitute \( \theta = 0 \) into the expression for \(\frac{dy}{dx}\):

\[
\left. \frac{dy}{dx} \right|_{\theta = 0} = \frac{\sin 0 + \cos 0}{\cos 0 - \sin 0} = \frac{0 + 1}{1 - 0} = 1
\]

The slope of the tangent line at \( \theta = 0 \) is 1.

### 4. Find the slope of the normal line
The slope of the normal line is the negative reciprocal of the tangent line's slope:

\[
\text{Slope of the normal line} = -\frac{1}{1} = -1
\]

### 5. Find the Cartesian coordinates at \( \theta = 0 \)
At \( \theta = 0 \), \( r = e^0 = 1 \). The Cartesian coordinates \((x_0, y_0)\) are:

\[
x_0 = 1 \cdot \cos(0) = 1
\]
\[
y_0 = 1 \cdot \sin(0) = 0
\]

So, the point of tangency is \( (1, 0) \).

### 6. Write the equation of the normal line
The equation of the normal line in point-slope form is:

\[
y - y_0 = m_{\text{normal}}(x - x_0)
\]

Substituting the values:

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

Simplifying:

\[
y = -x + 1
\]

Thus, the equation of the normal line at \( \theta = 0 \) is \( \boxed{y = -x + 1} \).

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Would you like to go over any specific part of this process, or do you have any other questions?

Here are some related questions:

1. How do you find the tangent line for any polar curve at a specific angle?
2. What is the significance of the normal line in geometry?
3. Can you find the normal line for a curve defined by a different polar equation, such as \( r = \sin \theta \)?
4. How do you convert between polar and Cartesian coordinates for more complex functions?
5. How would the normal line equation change if the point of tangency were different?

**Tip:** Always check whether your normal line is perpendicular to the tangent line by verifying their slopes multiply to \(-1\).

Learn how to calculate the equation of the normal line for the polar curve r = e^theta at theta = 0. Detailed step-by-step solution involving conversion to Cartesian coordinates, finding derivatives, evaluating slopes, and deriving the equation of the normal line. Suitable for undergraduate students studying polar coordinates and calculus.

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