Question

Find: (a) \(\frac{d y}{d x}\) (b) the equation of the tangent and normal lines to the curve at the indicated \(\theta\) -value. \(r=\cos (3 \theta) ; \quad \theta=\pi / 6\)

Short answer

(a) \(-\sqrt{3}\); (b) Tangent: \(y=-\sqrt{3}x\), Normal: \(y=\frac{1}{\sqrt{3}}x\).

Step-by-step solution

Find the Polar Derivative Formula

The polar form of a function is given by the relation \( r = f(\theta) \). To find \( \frac{dy}{dx} \), we use the polar derivative formulas:\( x = r \cos(\theta) \) and \( y = r \sin(\theta) \). The derivative \( \frac{dy}{dx} \) is given by \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).Now, substitute and differentiate: - \( \frac{dx}{d\theta} = \cos(3\theta) \cos(\theta) - 3\sin(3\theta) \sin(\theta) \)- \( \frac{dy}{d\theta} = \cos(3\theta) \sin(\theta) + 3\sin(3\theta) \cos(\theta) \)

Evaluate Derivatives at \(\theta = \pi/6\)

First, substitute \( \theta = \frac{\pi}{6} \) into the expressions for \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).- \( \cos (\pi/2) = 0 \), \( \sin (\pi/2) = 1 \). Therefore: - \( \frac{dx}{d\theta} = 0 \cdot \cos(\pi/6) - 3 \cdot 1 \cdot 1/2 = -\frac{3}{2} \) - \( \frac{dy}{d\theta} = 0 \cdot \sin(\pi/6) + 3 \cdot 1 \cdot \cos(\pi/6) = \frac{3\sqrt{3}}{2} \).Hence, \( \frac{dy}{dx} = \frac{\frac{3\sqrt{3}}{2}}{-\frac{3}{2}} = -\sqrt{3} \).

Find the Cartesian Coordinates at \(\theta = \pi/6\)

Substitute \( \theta = \frac{\pi}{6} \) into \( r \) to get the point of tangency:- \( r = \cos(3 \cdot \frac{\pi}{6}) = \cos(\frac{\pi}{2}) = 0 \).Thus, the point in the Cartesian plane is \((0,0)\), since both \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \) are zero.

Calculate Equation of Tangent Line

The general equation of the tangent line is given by \( y - y_1 = m(x - x_1) \), where \( m = \frac{dy}{dx} \) and \((x_1, y_1)\) is the point of tangency. Here \( m = -\sqrt{3} \) and \((x_1, y_1) = (0,0)\).Thus, the equation of the tangent line is \( y = -\sqrt{3}x \).

Calculate Equation of Normal Line

The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, it is \( m = \frac{1}{\sqrt{3}} \). Using the point \((0, 0)\) and the slope \( m = \frac{1}{\sqrt{3}} \), the equation of the normal line is \( y = \frac{1}{\sqrt{3}}x \).

Key concepts

Polar Derivative

When dealing with curves represented in polar coordinates, understanding the derivative helps us analyze the behavior of the curve. Polar coordinates use the angle \(\theta\) and radius \(r\) of the curve, expressed as \(r=f(\theta)\). To convert polar derivatives into ones familiar in Cartesian coordinates, we need a couple of formulas. The challenge lies in finding \(\frac{dy}{dx}\), which represents how the curve changes in the Cartesian plane (the typical \(x-y\) graph).
By using the conversion formulas: \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\), we derive \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\). This method utilizes the tools of calculus to break the changes in \(y\) and \(x\) down to changes in \(\theta\) - offering the familiar slope of a line in Cartesian terms.

• Differentiate \(x\) and \(y\) concerning \(\theta\)
• Plug in the value of \(\theta\)
• Calculate the resulting ratio \(\frac{dy}{dx}\)

These steps help convert the interesting dance of a polar curve into something you can graph and understand on the more familiar Cartesian plane.

Tangent Line

After discovering the slope of the curve using the polar derivative, the challenge is to find the tangent line. This is the straight line that just "touches" the curve at the point of interest. The tangent line gives a great approximation of the curve near that point.
The general formula for any tangent line is \(y - y_1 = m(x - x_1)\), where \(m\) is the calculated slope \(\frac{dy}{dx}\). In this case, we found that \(m = -\sqrt{3}\). The point of tangency,\( (x_1, y_1)\), can be found by evaluating \(r\) at \(\theta = \pi/6\), which turned out to be \((0,0)\).
To wrap it up:

• Plug the slope into the formula
• Use the Cartesian coordinates at the point

The tangent line equation becomes \(y = -\sqrt{3}x\). This linear relationship helps us see how the curve behaves locally - like zooming in with a microscope on our curve.

Normal Line

The normal line is always perpendicular to the tangent line at the point of tangency. If you think of standing on a curve, the tangent is the direction you'd glide along; the normal line is where you add a perpendicular stepping stone.
Mathematically, for a line that's perpendicular, its slope is the negative reciprocal of the tangent's slope. Thus, if our tangent slope is \(-\sqrt{3}\), the normal slope becomes \(\frac{1}{\sqrt{3}}\).
Using the point-slope form, with our slope and the same touch-point \((0,0)\), we form the normal line's equation:

• Find the negative reciprocal of the tangent slope
• Use the same point of tangency

This leads us to the equation \(y = \frac{1}{\sqrt{3}}x\), showing a line perpendicular to our original tangent, offering a complete perspective at that specific point.

Cartesian Coordinates

To visualize curves expressed in polar form using the traditional \(x-y\) graphing coordinates, it's important to perform a conversion. This helps transform our understanding and apply familiar concepts like slopes and lines to curves originally seen from polar perspectives.
You might start with polar coordinates known as \((r, \theta)\) where \(r\) is the distance from the origin and \(\theta\) is the angle from the positive x-axis. Through the formulas:

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

We translate our polar representation into Cartesian \((x, y)\) coordinates, typically for ease of graphing and analysis.
In the exercise provided, converting polar results allowed finding direct points like \((0,0)\), supporting the application of tangent and normal lines. This translation bridges the gap between different coordinate systems, granting more tools to tackle curve-related problems effectively.