Question

Find the points at which the following polar curves have a horizontal or a vertical tangent line. $$r=4 \cos \theta$$

Short answer

Short Answer: 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 $(4,0)$, $(0,\frac{\pi}{2})$, $(-4,\pi)$, $(0,\frac{3\pi}{2})$.

Step-by-step solution

Convert the polar equation to Cartesian form

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: $$x = 4\cos^2\theta$$ $$y = 4\cos\theta\sin\theta$$

Take the derivatives of the Cartesian parametric equations

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

Find when the tangent is horizontal or vertical

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: $$\cos\theta\sin\theta = 0$$

Solve for the points

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: $$\theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$$

Find the corresponding points on the polar curve

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 \((4,0)\), \((0,\frac{\pi}{2})\), \((-4,\pi)\), \((0,\frac{3\pi}{2})\).