The given problem involves finding points on the polar curve $r = 6 \cos(\theta)$ where the tangent is either horizontal or vertical. Here's the step-by-step solution:

Tangent Line Conditions:

1. Horizontal Tangent: The tangent line is horizontal when $\frac{dy}{d\theta} = 0$.
2. Vertical Tangent: The tangent line is vertical when $\frac{dx}{d\theta} = 0$.

The polar-to-cartesian transformations are: $x = r \cos(\theta), \quad y = r \sin(\theta)$ Substituting $r = 6 \cos(\theta)$, we get: $x = 6 \cos^2(\theta), \quad y = 6 \cos(\theta) \sin(\theta)$

Step 1: Compute Derivatives

For $x$ and $y$: $x = r \cos(\theta) = 6 \cos^2(\theta), \quad y = r \sin(\theta) = 6 \cos(\theta) \sin(\theta)$ $\frac{dx}{d\theta} = -12 \cos(\theta) \sin(\theta)$ $\frac{dy}{d\theta} = 6 [\cos^2(\theta) - \sin^2(\theta)]$

Step 2: Horizontal Tangent ($\frac{dy}{d\theta} = 0$)

$6 [\cos^2(\theta) - \sin^2(\theta)] = 0$ $\cos^2(\theta) = \sin^2(\theta)$ $\tan(\theta) = 1 \quad \Rightarrow \quad \theta = \frac{\pi}{4}$

Substitute $\theta = \frac{\pi}{4}$ into $r = 6 \cos(\theta)$: $r = 6 \cos\left(\frac{\pi}{4}\right) = 6 \cdot \frac{\sqrt{2}}{2} = 3\sqrt{2}$ The point is $\left(3\sqrt{2}, \frac{\pi}{4}\right)$.

Step 3: Vertical Tangent ($\frac{dx}{d\theta} = 0$)

$-12 \cos(\theta) \sin(\theta) = 0$ $\cos(\theta) = 0 \quad \text{or} \quad \sin(\theta) = 0$

• If $\cos(\theta) = 0$: $\theta = \frac{\pi}{2}$, $r = 6 \cos\left(\frac{\pi}{2}\right) = 0$. This gives no point.
• If $\sin(\theta) = 0$: $\theta = 0$, $r = 6 \cos(0) = 6$. The point is $(6, 0)$.

Final Answer:

• Horizontal Tangent: $\left(3\sqrt{2}, \frac{\pi}{4}\right)$
• Vertical Tangent: $(6, 0)$

Would you like further clarification or more examples?

1. What are the conditions for horizontal and vertical tangents in polar coordinates?
2. How do polar equations convert to Cartesian form for analysis?
3. Can horizontal or vertical tangents occur at the origin in this problem?
4. How do trigonometric identities simplify derivative-based problems in polar coordinates?
5. How does symmetry affect polar curves like $r = 6 \cos(\theta)$?

Tip: Use symmetry properties of polar equations to quickly predict tangent behavior.