Question

Find the slope of the line tangent to the following polar curves at the given points. At the points where the curve intersects the origin (when this occurs), find the equation of the tangent line in polar coordinates. $$r=4+\sin \theta ;(4,0) \text { and }\left(3, \frac{3 \pi}{2}\right)$$

Short answer

Answer: At (4,0), the tangent slope is 0. At (3, 3π/2), the tangent slope is -1.

Step-by-step solution

Calculate \(\frac{dr}{d\theta}\)

Differentiate \(r(θ)\) with respect to \(θ\) to find the rate of change of \(r\). $$\frac{dr}{d\theta}=\frac{d(4+\sin\theta)}{d\theta}=\cos\theta$$

Polar representation of tangent slope

To find the tangent slope for a curve in polar coordinates we can use the following formula: $$\frac{dy}{dx} = \frac{r(\theta)\frac{dr}{d\theta}\cos(\theta)+r(\theta)\sin(\theta)}{-r(\theta)\frac{dr}{d\theta}\sin(\theta)+r(\theta)\cos(\theta)}$$ Now, plug in the values of r and \(\frac{dr}{d\theta}\) and simplify the expression: $$\frac{dy}{dx} = \frac{(4+\sin\theta)\cos(\theta)^2+(4+\sin\theta)\sin(\theta)\cos(\theta)}{-(4+\sin\theta)\cos(\theta)\sin(\theta)+(4+\sin\theta)\cos(\theta)^2}$$

Evaluate the tangent slope at the given points

Now, we will find the tangent slope at the two given points, by plugging in their corresponding values of \(θ\) into \(\frac{dy}{dx}\). For point \((4,0)\): $$\theta = 0, \quad r = 4$$ $$\frac{dy}{dx} = \frac{(4+\sin(0))\cos(0)^2+(4+\sin(0))\sin(0)\cos(0)}{-(4+\sin(0))\cos(0)\sin(0)+(4+\sin(0))\cos(0)^2}$$ $$\frac{dy}{dx} = 0$$ For point \(\left(3, \frac{3 \pi}{2}\right)\): $$\theta = \frac{3\pi}{2}, \quad r = 3$$ $$\frac{dy}{dx} = \frac{(3+\sin(\frac{3\pi}{2}))\cos(\frac{3\pi}{2})^2+(3+\sin(\frac{3\pi}{2}))\sin(\frac{3\pi}{2})\cos(\frac{3\pi}{2})}{-(3+\sin(\frac{3\pi}{2}))\cos(\frac{3\pi}{2})\sin(\frac{3\pi}{2})+(3+\sin(\frac{3\pi}{2}))\cos(\frac{3\pi}{2})^2}$$ $$\frac{dy}{dx} = -1$$ So, the tangent slopes at the given points are: $$\text{At }(4,0): \frac{dy}{dx} = 0$$ $$\text{At }\left(3, \frac{3 \pi}{2}\right): \frac{dy}{dx} = -1$$

Intersection with the origin and tangent line equation

To find where the curve intersects the origin, set \(r = 0\) and solve for \(θ\): $$0 = 4+\sin \theta$$ $$\sin \theta = -4$$ Since \(\sin \theta\) is always between -1 and 1, this equation has no solution, so there are no points of intersection with the origin. Therefore, there is no need to calculate the equation of the tangent line in polar coordinates. The final answers are: $$\text{At }(4,0): \frac{dy}{dx} = 0$$ $$\text{At }\left(3, \frac{3 \pi}{2}\right): \frac{dy}{dx} = -1$$

Key concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point. It doesn't cross the curve or go through it at that point. Instead, it "just touches" the curve, letting us see the direction the curve is heading. In simpler terms, it is like placing a straight-edge ruler against the curve without cutting through it.

At any given point on the polar curve, the tangent line represents the direction in which the curve is heading. To visualize this, imagine zooming in on the point of tangency so that the curve looks nearly like a straight line. That's the tangent line. This concept is crucial because it helps us understand how the curve behaves locally at that point.

In mathematical terms, to find a tangent line to a curve in polar coordinates, we often use derivatives. By using the slope of the tangent line calculated via derivatives, we can determine exactly how the curve approaches the tangent point.

Slope of Tangent Line

The slope of a tangent line measures how steep the tangent is at a particular point on a curve. It tells us how fast and in which direction the curve is changing at that point. In rectangular coordinates, you might know this as \(\frac{dy}{dx}\). But when it comes to polar coordinates, things get a bit trickier.

In polar coordinates, each point is defined by \(r\) and \(\theta\). To find the slope of the tangent line to a polar curve, we need to shift our perspective slightly and use a specific formula:
- The formula we use is: \ \[ \frac{dy}{dx} = \frac{r(\theta)\frac{dr}{d\theta}\cos(\theta)+r(\theta)\sin(\theta)}{-r(\theta)\frac{dr}{d\theta}\sin(\theta)+r(\theta)\cos(\theta)} \]- Here, \(\frac{dr}{d\theta}\) is the derivative of \(r\) with respect to \(\theta\), giving us the rate of change of \(r\) as \(\theta\) changes.

By using this formula, we can plug in the specific \(\theta\) values to find the slope of the tangent line for our polar curve. For example, as we solved in the original exercise, at the point \(4, 0\), the slope came out to be zero, while at \(3, \frac{3\pi}{2}\), the slope was -1.

Polar Curve Intersections

The concept of intersections in polar curves can sometimes be a bit counter-intuitive, especially when it comes to locating where a curve intersects with the origin. In polar coordinates, the curve intersects the origin when the radial distance \(r\) is zero. This essentially means that the curve "starts" at the origin.

To find intersections, we set \(r = 0\) and solve the equation. If the resulting values for \(\theta\) are valid, those angles represent points where the curve passes through the origin.

Let’s break this down a bit:

• When you solve \(r(\theta) = 0\), you're looking for \(\theta\) values where the radial distance from the origin is zero.
  
• However, there might be no solution at all if, for example, the range of \(\sin\theta\) or \(\cos\theta\) doesn’t reach the calculated value needed for \(r = 0\).
  

In the exercise we solved, the equation \sin \theta = -4\ didn't have a solution because the sine function only takes values from -1 to 1. Thus, it was determined that the polar curve doesn't intersect the origin at any point. Understanding this helps prevent unnecessary confusion when visualizing or analyzing polar curves.