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. $$r=8 \sin \theta ;\left(4, \frac{5 \pi}{6}\right)$$

Short answer

#tag_title#Step 5: Convert the equation back to polar coordinates and find the intersection with the origin#tag_content# Now we have the equation of the tangent line in Cartesian coordinates: $$y - \frac{8}{3} = -\frac{1}{3}(x - 4)$$ We need to convert the equation back to polar coordinates. First, rewrite the above equation as: $$y = -\frac{1}{3}x + \frac{20}{3}$$ Recall that $$y = r\sin \theta$$ and $$x = r\cos \theta$$. Substitute these expressions into the equation: $$r\sin\theta = -\frac{1}{3}(r\cos \theta) + \frac{20}{3}$$ To find the intersection with the origin, we let $$\theta = 0$$. Plug in this value into the formula: $$r\sin(0) = -\frac{1}{3}(r\cos(0)) + \frac{20}{3}$$ Solving for r, we get: $$r = \frac{20}{3}$$ Thus, the intersection of the tangent line with the origin in polar coordinates is $$\left(\frac{20}{3}, 0\right)$$.

Step-by-step solution

Convert polar curve to Cartesian coordinates

Given the polar curve $$r = 8\sin \theta$$, we can convert it to Cartesian coordinates using the following relationships: $$x = r\cos \theta$$ and $$y = r\sin \theta$$ Substitute the polar curve into the relationships to get: $$x = 8\sin \theta \cos \theta$$ and $$y = 8\sin^2 \theta$$

Compute the derivative with respect to x

To compute the derivative \(\frac{dy}{dx}\), we will use the chain rule. First, compute the derivatives of y and x with respect to \(\theta\): $$\frac{dy}{d\theta} = 16\sin \theta \cos \theta$$ and $$\frac{dx}{d\theta} = 8\cos^2 \theta - 8\sin^2 \theta$$ Next, apply the chain rule: $$\frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx} = \frac{16\sin \theta \cos \theta}{8\cos^2 \theta - 8\sin^2 \theta}$$

Find the slope of the tangent line at point (4, 5π/6)

Now we need to find the slope of the tangent line at the given point $$\left(4, \frac{5 \pi}{6}\right)$$. The second coordinate in the point represents the angle \(\theta\), which is $$\theta = \frac{5 \pi}{6}$$. Plug the value of \(\theta\) into the slope equation to get: $$m = \frac{16\sin \frac{5 \pi}{6} \cos \frac{5 \pi}{6}}{8\cos^2 \frac{5 \pi}{6} - 8\sin^2 \frac{5 \pi}{6}}$$ Simplify to obtain the slope: $$m = -\frac{1}{3}$$

Write the Cartesian equation of the tangent line

Using the Cartesian point-slope formula, $$(y-y_1)=m(x-x_1)$$, we can compute the equation of the tangent line. We have the slope $$m = -\frac{1}{3}$$ and the given point in Cartesian coordinates is $$(x_1, y_1) = (4,16\sin^2(\frac{5\pi}{6})) = \left(4,\frac{8}{3}\right)$$. Substitute the slope and point into the formula: $$y - \frac{8}{3} = -\frac{1}{3}(x - 4)$$ To find the intersection with the origin, the equation will be in the form $$y = mx + r_\theta$$, where $$r_\theta$$ represents the distance from the origin at angle $$\theta$$. In polar coordinates, the equation for the tangent line will be: $$r = -\frac{1}{3}\theta + r_\theta$$

Key concepts

Tangent Line

A tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. This line has the same slope as the curve at the contact point, allowing it to represent the immediate direction of the curve.

When talking about tangent lines in the context of polar coordinates, the concept remains, but the process for finding them differs slightly from Cartesian coordinates. In polar coordinates, points are defined by a radius and an angle instead of x and y. Therefore, the tangent line in polar coordinates introduces complications due to this radial aspect.

Finding the equation of a tangent in polar form sometimes involves a multi-step process since you may need to convert between polar and Cartesian forms to properly understand and compute the results.

Slope of Tangent

The slope of a tangent line is a crucial concept in understanding how curves change. In Cartesian coordinates, the slope is the rate of change of y with respect to x, simplified as \(\frac{dy}{dx}\). The value of this slope tells you how steep or flat a line is. A larger absolute slope indicates a steeper line.

In polar coordinates, finding this slope takes an additional step as the coordinates are not immediately aligned to an x,y grid. The process typically involves first converting the polar coordinates into Cartesian coordinates. Once this conversion is made, you can find the derivative and thus the slope of the tangent line.

In our exercise, after conversion and calculation, we find the slope to be \(m = -\frac{1}{3}\). This negative value tells us the tangent line decreases as it moves from left to right.

Cartesian Coordinates

Cartesian coordinates describe points on a plane using two values: \(x\) and \(y\), representing horizontal and vertical positions. Converting polar coordinates to Cartesian coordinates is often necessary to find the tangent slope or the equation of the tangent line.

In this exercise, the polar equation \(r = 8\sin \theta\) transforms into Cartesian form using the formulas \(x = r\cos \theta\) and \(y = r\sin \theta\). When these transformations are applied, it allows us to more easily calculate derivatives and thus ascertain the behavior of the curve in a more familiar Cartesian framework.

For the given example, the point \(\left(4, \frac{5 \pi}{6}\right)\) transformed in this manner guides the subsequent steps to discover the characteristics of the tangent line.

Derivative in Polar Form

Derivatives, which measure rates of change, can establish how quickly a curve twists or turns. In polar coordinates, the derivative allows us to calculate the slope of the tangent without immediately translating the curve into Cartesian form, although conversion is often more straightforward for application purposes.

The derivative for polar functions can be complex because they depend on both \(r\) and \(\theta\) — the radius and angle. Calculating \(\frac{dy}{dx}\) involves the chain rule, where both the numerator and the denominator must be separately derived with respect to \(\theta\). Bringing them together as \(\frac{dy}{d\theta} \cdot \frac{d\theta}{dx}\) efficiently arrives at the change from radial to linear criteria.

In this exercise, we compute these derivatives, apply them as fractions, and simplify to get the slope. This process highlights how derivatives play a role in transitioning between or interpreting the behavior of curves in different coordinate systems.