A curve is defined in polar coordinates by the equation $r = 2 + \cos(2\theta)$. Find the coordinates of the point on the curve where the tangent line is perpendicular to the radius vector. Express your answer in rectangular coordinates.

We are tasked with finding the point on the curve defined by the polar equation $ r = 2 + \cos(2\theta) $ where the tangent line is perpendicular to the radius vector. The final answer should be expressed in rectangular coordinates.

### Step 1: Condition for the Tangent Line to be Perpendicular to the Radius Vector
In polar coordinates, the tangent line to a curve is perpendicular to the radius vector if the derivative of $ r $ with respect to $ \theta $ satisfies the condition:

\[
\frac{dr}{d\theta} = \frac{r}{\tan\theta}
\]

This condition arises because when the tangent is perpendicular to the radius vector, the slope of the tangent must be the negative reciprocal of the slope of the radius vector.

### Step 2: Compute the Derivative of $ r $
The given curve is $ r = 2 + \cos(2\theta) $. Let's differentiate this with respect to $ \theta $:

\[
\frac{dr}{d\theta} = \frac{d}{d\theta} \left( 2 + \cos(2\theta) \right)
= -2\sin(2\theta)
\]

### Step 3: Solve for $ \theta $
Using the condition $ \frac{dr}{d\theta} = \frac{r}{\tan\theta} $, substitute $ \frac{dr}{d\theta} = -2\sin(2\theta) $ and $ r = 2 + \cos(2\theta) $ into the equation:

\[
-2\sin(2\theta) = \frac{2 + \cos(2\theta)}{\tan\theta}
\]

Rearrange the equation:

\[
-2\sin(2\theta)\tan\theta = 2 + \cos(2\theta)
\]

Use the identity $ \sin(2\theta) = 2\sin\theta\cos\theta $:

\[
-2(2\sin\theta\cos\theta)\tan\theta = 2 + \cos(2\theta)
\]

This simplifies to:

\[
-4\sin\theta\cos\theta = 2 + \cos(2\theta)
\]

Using the identity $ \cos(2\theta) = 2\cos^2\theta - 1 $, substitute that into the equation:

\[
-4\sin\theta\cos\theta = 2 + 2\cos^2\theta - 1
\]

Simplify further:

\[
-4\sin\theta\cos\theta = 1 + 2\cos^2\theta
\]

This is the equation we need to solve for $ \theta $.

### Step 4: Solve for $ \theta $
Next, we'll look for specific values of $ \theta $ that satisfy this equation.

At $ \theta = \frac{\pi}{2} $, we have:

\[
r = 2 + \cos\left( 2 \times \frac{\pi}{2} \right) = 2 + \cos(\pi) = 2 - 1 = 1
\]

Thus, the corresponding point in polar coordinates is $ (r, \theta) = (1, \frac{\pi}{2}) $.

### Step 5: Convert to Rectangular Coordinates
To convert from polar to rectangular coordinates, use the relations:

\[
x = r \cos\theta \quad \text{and} \quad y = r \sin\theta
\]

For $ r = 1 $ and $ \theta = \frac{\pi}{2} $:

\[
x = 1 \cdot \cos\left(\frac{\pi}{2}\right) = 1 \cdot 0 = 0
\]
\[
y = 1 \cdot \sin\left(\frac{\pi}{2}\right) = 1 \cdot 1 = 1
\]

Therefore, the point in rectangular coordinates is $ (x, y) = (0, 1) $.

### Final Answer:
The coordinates of the point where the tangent line is perpendicular to the radius vector are $ (0, 1) $.

Would you like a deeper explanation of any of these steps?

Here are 5 related questions to explore this further:
1. What is the geometric interpretation of a tangent line being perpendicular to a radius vector in polar coordinates?
2. How do you find the tangent line to a curve in polar coordinates in general?
3. What is the significance of the condition $ \frac{dr}{d\theta} = \frac{r}{\tan\theta} $?
4. Can you use the rectangular coordinates method directly to solve similar polar coordinate problems?
5. How would the solution change if the polar equation had a different form, such as $ r = 1 + \cos(\theta) $?

**Tip:** In polar coordinate problems, always keep in mind the relationship between polar and rectangular coordinates: $ x = r \cos \theta $ and $ y = r \sin \theta $.

A curve is defined in polar coordinates by the equation r = 2 + cos(2θ). Find the coordinates of the point on the curve where the tangent line is perpendicular to the radius vector. Express your answer in rectangular coordinates.

Explore how to find the point on the curve defined by r = 2 + cos(2θ) where the tangent is perpendicular to the radius vector. This problem covers key concepts in polar coordinates, including tangents, derivatives, and conversions to rectangular coordinates. Suitable for Grades 11-12.

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