To find the equation of the line tangent to the polar curve $r = 2\cos\theta$ at $\theta = \frac{7\pi}{4}$, we need to follow these steps:

1. Convert to Cartesian Coordinates

To write the equation of the tangent line in Cartesian form, we'll first need to express the polar curve $r = 2\cos\theta$ in Cartesian coordinates. The standard conversions are:

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

For the point at $\theta = \frac{7\pi}{4}$, we calculate $r$ and convert it to Cartesian form.

• $r = 2\cos\left(\frac{7\pi}{4}\right)$
• $\cos\left(\frac{7\pi}{4}\right) = \frac{1}{\sqrt{2}}$, so: $r = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2}$

Thus, the polar point $(\sqrt{2}, \frac{7\pi}{4})$ corresponds to the Cartesian point:

$x = \sqrt{2}\cos\left(\frac{7\pi}{4}\right) = \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 1$ $y = \sqrt{2}\sin\left(\frac{7\pi}{4}\right) = \sqrt{2} \cdot \left(-\frac{1}{\sqrt{2}}\right) = -1$

So, the Cartesian coordinates of the point are $(1, -1)$.

2. Find the Slope of the Tangent Line

To find the slope of the tangent line at $\theta = \frac{7\pi}{4}$, we can use the following formula for the slope of the tangent line to a polar curve:

$\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin\theta + r\cos\theta}{\frac{dr}{d\theta}\cos\theta - r\sin\theta}$

First, compute $\frac{dr}{d\theta}$ for $r = 2\cos\theta$:

$\frac{dr}{d\theta} = -2\sin\theta$

Now, substitute $\theta = \frac{7\pi}{4}$ into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{(-2\sin\left(\frac{7\pi}{4}\right))\sin\left(\frac{7\pi}{4}\right) + 2\cos\left(\frac{7\pi}{4}\right)\cos\left(\frac{7\pi}{4}\right)}{(-2\sin\left(\frac{7\pi}{4}\right))\cos\left(\frac{7\pi}{4}\right) - 2\cos\left(\frac{7\pi}{4}\right)\sin\left(\frac{7\pi}{4}\right)}$

Substituting $\sin\left(\frac{7\pi}{4}\right) = -\frac{1}{\sqrt{2}}$ and $\cos\left(\frac{7\pi}{4}\right) = \frac{1}{\sqrt{2}}$:

$\frac{dy}{dx} = \frac{(-2 \cdot -\frac{1}{\sqrt{2}}) \cdot -\frac{1}{\sqrt{2}} + 2 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}{(-2 \cdot -\frac{1}{\sqrt{2}}) \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{1}{\sqrt{2}} \cdot -\frac{1}{\sqrt{2}}}$

Simplify the expression:

$\frac{dy}{dx} = \frac{2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2}}{2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2}} = \frac{1 + 1}{1 + 1} = \frac{2}{2} = 1$

So, the slope of the tangent line is 1.

3. Equation of the Tangent Line

The tangent line passes through the point $(1, -1)$ and has a slope of 1. The equation of the line is given by:

$y - y_1 = m(x - x_1)$

Substitute $m = 1$, $x_1 = 1$, and $y_1 = -1$:

$y + 1 = 1(x - 1)$ $y = x - 2$

Thus, the equation of the tangent line is:

$y = x - 2$

Summary:

The equation of the tangent line to the curve $r = 2\cos\theta$ at $\theta = \frac{7\pi}{4}$ is:

$y = x - 2$

Would you like more details on any of these steps or concepts?

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Here are 5 related questions:

1. How do you find the tangent line to a parametric curve?
2. How do you convert a polar equation to Cartesian coordinates?
3. What is the relationship between polar and Cartesian coordinate systems?
4. How is the derivative in polar coordinates calculated?
5. How do you find the slope of a curve given in polar coordinates?

Tip: When working with polar coordinates, always convert to Cartesian if you need slopes or tangent lines for easier interpretation.