Question

Find an equation of the line tangent to the following curves at the given point. $$r=\frac{1}{1+\sin \theta} ;\left(\frac{2}{3}, \frac{\pi}{6}\right)$$

Short answer

The equation of the line tangent to the curve at the given point in Cartesian coordinates is: $$y-\frac{1}{3}=\frac{11\sqrt{3}}{4}\left(x-\frac{2\sqrt{3}}{6}\right)$$

Step-by-step solution

1. Convert polar coordinates to Cartesian coordinates

Recall that the conversion formulas for polar coordinates to Cartesian coordinates are \(x = r\cos\theta\) and \(y = r\sin\theta\). So, in this case, we have: $$r = \frac{1}{1+\sin\theta}$$ $$x = \frac{\cos\theta}{1+\sin\theta}$$ $$y = \frac{\sin\theta}{1+\sin\theta}$$

2. Differentiate both x and y with respect to θ

Next, we need to compute \(\frac{dy}{dx}\) to find the slope of the tangent line. To do this, we first find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\): $$\frac{dx}{d\theta}=-\frac{\sin\theta(\sin\theta+1)-\cos\theta\cos\theta}{(\sin\theta+1)^2}$$ $$\frac{dy}{d\theta}=\frac{\cos\theta(\sin\theta+1)-\sin\theta\cos\theta}{(\sin\theta+1)^2}$$

3. Compute the derivatives at the given point

Now, we plug in the given point \(\left(\frac{2}{3}, \frac{\pi}{6}\right)\) into our derivatives: $$\frac{dx}{d\theta}\bigg|_{\theta=\frac{\pi}{6}}=-\frac{\frac{\sqrt{3}}{4} +\frac{1}{4}-\frac{3}{4}}{\left(\frac{1}{2}+1\right)^2}=-\frac{-\frac{1}{2}}{\frac{9}{4}}=\frac{2}{9}$$ $$\frac{dy}{d\theta}\bigg|_{\theta=\frac{\pi}{6}}=\frac{\frac{1}{2}+\frac{1}{2\sqrt{3}}-\frac{\sqrt{3}}{6}}{\left(\frac{1}{2}+1\right)^2}=\frac{11}{18\sqrt{3}}$$

4. Compute the slope of the tangent line

Now, we can compute the slope of the tangent line using \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\frac{11}{18\sqrt{3}}}{\frac{2}{9}} = \frac{11\sqrt{3}}{4}$$

5. Write the equation of the tangent line in point-slope form

Now that we have the slope of the tangent line, we can use the point-slope form of a line to find its equation. The given point is \(\left(\frac{2}{3}, \frac{\pi}{6}\right)\), but we need to convert this point back to Cartesian coordinates. Using \(x = r\cos\theta\) and \(y = r\sin\theta\), we get: $$x = \frac{2}{3}\cos\left(\frac{\pi}{6}\right) = \frac{2\sqrt{3}}{6}$$ $$y = \frac{2}{3}\sin\left(\frac{\pi}{6}\right) = \frac{1}{3}$$ So the equation of the tangent line is: $$y-\frac{1}{3}=\frac{11\sqrt{3}}{4}\left(x-\frac{2\sqrt{3}}{6}\right)$$ This is the equation of the line tangent to the curve at the given point in Cartesian coordinates.

Key concepts

Polar Coordinates

Polar coordinates are a way of representing points in a plane using a distance and angle, different from the usual Cartesian coordinates (x, y). In polar coordinates, a point is represented as \((r, \theta)\), where:

• \(r\) is the distance from the origin to the point, called the radial coordinate.
• \(\theta\) is the angle measured from the positive x-axis to the line segment connecting the origin to the point, called the angular coordinate.

This system is especially useful in situations with circular or rotational symmetry. In our exercise, we start with a polar equation \(r=\frac{1}{1+\sin \theta}\) to describe a curve and a specific point given in polar form. By understanding polar coordinates, we can interpret and work with curves effectively in various mathematical problems.

Cartesian Coordinates

Cartesian coordinates are the most common way to represent points in a plane. They are composed of an x-coordinate and a y-coordinate, which specify the point's horizontal and vertical position, respectively. To convert from polar coordinates \((r, \theta)\) to Cartesian coordinates, we use the formulas:

• \(x = r\cos\theta\)
• \(y = r\sin\theta\)

For our problem, we applied these formulas to transform the given polar point \(\left(\frac{2}{3}, \frac{\pi}{6}\right)\) into Cartesian form. The conversion is crucial for many types of mathematical analysis, including finding tangent lines, as it allows us to utilize Cartesian coordinate methods.

Slope of the Tangent

The slope of the tangent line to a curve at a given point provides information about the direction in which the curve is heading at that point. It is defined as \(\frac{dy}{dx}\), the rate of change in y with respect to x. In our example, we first needed to differentiate both x and y with respect to \(\theta\), yielding \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\). Then, the slope of the tangent line was found by dividing these derivatives: \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \].By calculating\( \frac{dy}{dx} \) at the specific point, in this case, \(\theta = \frac{\pi}{6}\), we found the tangent line's slope to be \( \frac{11\sqrt{3}}{4} \), which helps define the line precisely in the context of the Cartesian coordinate system.

Point-Slope Form

The point-slope form is a way to express the equation of a straight line. It is particularly useful when you know a single point on the line and its slope. The standard formula is \( y - y_1 = m(x - x_1) \), where:

• \(m\) is the slope of the line.
• \((x_1, y_1)\) is a particular point on the line.

Once we establish the slope of the tangent line and convert the given point from polar to Cartesian coordinates, we use the point-slope form to write the equation of the tangent line. In our problem, the specific slope \(\frac{11\sqrt{3}}{4}\) and the Cartesian point derived from \(\left(\frac{2}{3}, \frac{\pi}{6}\right)\) lead us to the final equation \(y-\frac{1}{3}=\frac{11\sqrt{3}}{4}\left(x-\frac{2\sqrt{3}}{6}\right)\), completing our task to find the tangent line to the original polar curve.