91Ó°ÊÓ

To convert the given polar equation to Cartesian form, we use the two polar to Cartesian conversion formulas: $$x = r \cos \theta$$ $$y = r \sin \theta$$ Given the polar equation: $$r= 4 \cos \theta$$ We can substitute this into the Cartesian conversion formulas: $$x = (4\cos\theta) \cos\theta$$ $$y = (4\cos\theta) \sin\theta$$ Which results in: $$x = 4\cos^2\theta$$ $$y = 4\cos\theta\sin\theta$$

Next, we need to find the derivatives of the Cartesian parametric equations with respect to the parameter, \(\theta\). Using the Chain Rule, we have: $$\frac{dx}{d\theta} = -8\cos\theta\sin\theta$$ $$\frac{dy}{d\theta} = 4\sin^2\theta - 4\cos^2\theta$$

For a horizontal tangent, we need \(\frac{dy}{d\theta} = 0\). This is true when: $$4\sin^2\theta - 4\cos^2\theta = 0$$ Solving for \(\sin^2\theta\), we get: $$\sin^2\theta = \cos^2\theta$$ Since \(\sin^2\theta + \cos^2\theta = 1\), it follows that \(\sin^2\theta = \cos^2\theta = \frac{1}{2}\). For a vertical tangent, we need \(\frac{dx}{d\theta} = 0\). This is true when: $$-8\cos\theta\sin\theta = 0$$ Dividing both sides by \(-8\), we have: $$\cos\theta\sin\theta = 0$$

From the horizontal tangent condition: $$\cos^2\theta = \frac{1}{2}$$ Taking the square root, we have: $$\cos\theta = \pm\frac{1}{\sqrt{2}}$$ The corresponding \(\theta\) values are: $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$$ From the vertical tangent condition: $$\cos\theta\sin\theta = 0$$ This occurs when either \(\sin\theta\) is \(0\) or \(\cos\theta\) is \(0\). Hence, the values for \(\theta\) are: $$\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$$

Now we'll plug these values of \(\theta\) back into the polar curve equation \(r= 4\cos\theta\) to find the corresponding points \((r, \theta)\): For horizontal tangents: $$\theta = \frac{\pi}{4} \Rightarrow r=4\cos\left(\frac{\pi}{4}\right) = 2\sqrt{2} \Rightarrow (2\sqrt{2},\frac{\pi}{4})$$ $$\theta = \frac{3\pi}{4} \Rightarrow r=4\cos\left(\frac{3\pi}{4}\right) = -2\sqrt{2} \Rightarrow (-2\sqrt{2},\frac{3\pi}{4})$$ $$\theta = \frac{5\pi}{4} \Rightarrow r=4\cos\left(\frac{5\pi}{4}\right) = -2\sqrt{2} \Rightarrow (-2\sqrt{2},\frac{5\pi}{4})$$ $$\theta = \frac{7\pi}{4} \Rightarrow r=4\cos\left(\frac{7\pi}{4}\right) = 2\sqrt{2} \Rightarrow (2\sqrt{2},\frac{7\pi}{4})$$ For vertical tangents: $$\theta = 0 \Rightarrow r=4\cos\left(0\right) = 4 \Rightarrow (4,0)$$ $$\theta = \frac{\pi}{2} \Rightarrow r=4\cos\left(\frac{\pi}{2}\right) = 0 \Rightarrow (0,\frac{\pi}{2})$$ $$\theta = \pi \Rightarrow r=4\cos\left(\pi\right) = -4 \Rightarrow (-4,\pi)$$ $$\theta = \frac{3\pi}{2} \Rightarrow r=4\cos\left(\frac{3\pi}{2}\right) = 0 \Rightarrow (0,\frac{3\pi}{2})$$ Thus, the points on the polar curve where the tangent is horizontal are \((2\sqrt{2},\frac{\pi}{4})\), \((-2\sqrt{2},\frac{3\pi}{4})\), \((-2\sqrt{2},\frac{5\pi}{4})\), \((2\sqrt{2},\frac{7\pi}{4})\), and the points where the tangent is vertical are \((4,0)\), \((0,\frac{\pi}{2})\), \((-4,\pi)\), \((0,\frac{3\pi}{2})\).