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

Answer: The equation of the tangent line in polar coordinates is $$r_t(\cos \theta_t - 2\sin \theta_t) = 4\cos(\frac{5\pi}{6}) - 8\sin(\frac{5\pi}{6})$$.

Step-by-step solution

Convert polar equation to Cartesian coordinates

Using the given polar equation (r = 8 sinθ), we can write the expressions for x and y: $$x = (8\sin\theta)\cos\theta$$ $$y = (8\sin\theta)\sin\theta$$ Then find the derivative dy/dx.

Find dy/dx

We will first find dy/dθ and dx/dθ, and then find dy/dx using the chain rule: $$\frac{dy}{d\theta} = 8\cos^2\theta - 8\sin^2\theta$$ $$\frac{dx}{d\theta} = -8\sin\theta\cos\theta + 8\cos\theta\sin\theta$$ Now, we'll find dy/dx: $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$ $$\frac{dy}{dx} = \frac{8\cos^2\theta - 8\sin^2\theta}{-8\sin\theta\cos\theta + 8\cos\theta\sin\theta}$$ Simplifying, we get: $$\frac{dy}{dx} = \frac{\cos^2\theta - \sin^2\theta}{-\sin\theta\cos\theta + \cos\theta\sin\theta}$$

Evaluate the slope at the given point

The given point is (4, 5π/6). We need to evaluate the slope at this point: $$\frac{dy}{dx}\left(\frac{5\pi}{6}\right) = \frac{\cos^2\left(\frac{5\pi}{6}\right) - \sin^2\left(\frac{5\pi}{6}\right)}{-\sin\left(\frac{5\pi}{6}\right)\cos\left(\frac{5\pi}{6}\right) + \cos\left(\frac{5\pi}{6}\right)\sin\left(\frac{5\pi}{6}\right)}$$ Evaluating, we get: $$\frac{dy}{dx}\left(\frac{5\pi}{6}\right) = \frac{1}{2}$$ So the slope of the tangent line at the given point is 1/2.

Find the equation of the tangent line in polar coordinates

Since the slope is 1/2, we will now find the tangent line equation in polar coordinates. Let \((x_t, y_t)\) be a point on the tangent line, and \((x_1, y_1)\) be the Cartesian coordinates of the given polar point. $$x_t = r_t\cos \theta_t$$ $$y_t = r_t\sin \theta_t$$ $$x_1 = r_1\cos \theta_1$$ $$y_1 = r_1\sin \theta_1$$ Slope of the tangent line (m) at \((x_1, y_1)\) is: $$m = \frac{y_t - y_1}{x_t - x_1}$$ We have the slope, m, and we know that the tangent line passes through the point (4, 5π/6). We will use these facts to find the equation of the tangent line in polar coordinates. Plugging in the values, we get: $$\frac{1}{2} = \frac{r_t\sin \theta_t - 4\sin(\frac{5\pi}{6})}{r_t\cos \theta_t - 4\cos(\frac{5\pi}{6})}$$ Rearranging, we have the equation of the tangent line: $$r_t(\cos \theta_t - 2\sin \theta_t) = 4\cos(\frac{5\pi}{6}) - 8\sin(\frac{5\pi}{6})$$

Key concepts

Polar Coordinates

Polar coordinates represent a way of describing the location of a point in a plane through the values of two variables: the radial distance (r) from the origin and the angle (\(\theta\)) from the positive x-axis. Unlike Cartesian coordinates, which use x and y axes, polar coordinates are particularly useful when dealing with problems possessing circular or rotational symmetry.
Imagine knowing how far you are from the center of a circle and the angle you are facing. This is how polar coordinates guide us. For example, in the exercise, the point \((4, \frac{5\pi}{6})\) indicates that the point is 4 units from the origin, and at an angle of \(\frac{5\pi}{6}\) radians from the positive x-axis.
Understanding these coordinates is crucial when dealing with curves or shapes that naturally align in radial patterns like spirals and circles, making many calculations more intuitive.

Calculus Derivatives

Derivatives in calculus are used to find the rate at which one quantity changes with respect to another. When considering polar coordinates, understanding derivatives helps us to determine how the function changes as \(\theta\) changes. This is crucial when finding the slope of curves or the tangent lines on these curves.
In the step-by-step solution provided, we derive both \(dy/d\theta\) and \(dx/d\theta\) first, which involves differentiating the expressions for x and y with respect to \(\theta\). This helps in finding \(dy/dx\) using the chain rule:

• \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\)

The ability to find \(dy/dx\) is essential because it describes the slope of the line that is tangent to the curve.
Differentiation in polar coordinates can sometimes be more complex due to the involvement of both trigonometric functions and products of these functions, but the theory remains the same and is pivotal in analyzing the behavior of curves.

Polar to Cartesian Conversion

Converting polar coordinates to Cartesian coordinates involves using trigonometric relationships to express a point's coordinates in terms of x and y. The conversion formulas used are:

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

These relationships allow us to switch between systems to conveniently apply calculus concepts like derivatives, as seen in the provided solution.
In the exercise, the curve \(r = 8 \sin \theta\) is converted to its Cartesian counterpart to easily compute derivatives. By substituting anything concerning \(\theta\) and solving, we can derive proper representations for further calculations.
By mastering this conversion process, you can analyze problems that are naturally configured in polar coordinates using Cartesian coordinate tools, which can be handier for finding derivatives and integrating calculus principles into the analysis.