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. \(r=\theta ; \quad \theta=\pi / 2\)

Short answer

(a) \(\frac{dy}{dx} = -\frac{2}{\pi}\); (b) Tangent: \(y = -\frac{2}{\pi}x + \frac{\pi}{2}\); Normal: \(y = \frac{\pi}{2}x + \frac{\pi}{2}\).

Step-by-step solution

Express polar equation in Cartesian form

Given the polar equation is \(r = \theta\). To find the Cartesian equation:- \(x = r \cos \theta = \theta \cos \theta\)- \(y = r \sin \theta = \theta \sin \theta\)Now we have the relations \(x = \theta \cos \theta\) and \(y = \theta \sin \theta\).

Differentiate to find \(\frac{dy}{dx}\)

We use the chain rule and the derivatives of parametric equations:- \(\frac{dx}{d\theta} = \cos\theta - \theta\sin\theta\)- \(\frac{dy}{d\theta} = \sin\theta + \theta\cos\theta\)Thus, \[\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{\sin\theta + \theta\cos\theta}{\cos\theta - \theta\sin\theta}\]

Evaluate \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{2}\)

Substitute \(\theta = \frac{\pi}{2}\):- \(\sin\left(\frac{\pi}{2}\right) = 1\) and \(\cos\left(\frac{\pi}{2}\right) = 0\)- \(\frac{dx}{d\theta} = 0 - \frac{\pi}{2} \times 1 = -\frac{\pi}{2}\)- \(\frac{dy}{d\theta} = 1 + \frac{\pi}{2} \times 0 = 1\)Thus, \[\frac{dy}{dx} = \frac{1}{-\frac{\pi}{2}} = -\frac{2}{\pi}\]

Find the Cartesian coordinates at \(\theta = \frac{\pi}{2}\)

Calculate \(x\) and \(y\) when \(\theta = \frac{\pi}{2}\):- \(x = \frac{\pi}{2} \times 0 = 0\)- \(y = \frac{\pi}{2} \times 1 = \frac{\pi}{2}\)The point on the curve is \((0, \frac{\pi}{2})\).

Find the equation of the tangent line

The general equation for a line is \(y - y_1 = m(x - x_1)\) where \(m\) is the slope:- \(y - \frac{\pi}{2} = -\frac{2}{\pi} (x - 0)\)Thus, the equation of the tangent is \[y = -\frac{2}{\pi}x + \frac{\pi}{2}\]

Find the equation of the normal line

The normal line is perpendicular to the tangent, so its slope is the negative reciprocal:- The slope is \(\frac{\pi}{2}\) since the negative reciprocal of \(-\frac{2}{\pi}\) is \(\frac{\pi}{2}\)- Use the point \((0, \frac{\pi}{2})\):The equation becomes \[y - \frac{\pi}{2} = \frac{\pi}{2}(x - 0)\]Or \[y = \frac{\pi}{2}x + \frac{\pi}{2}\]

Key concepts

Polar to Cartesian Conversion

When dealing with curves, we often represent points using different coordinate systems. Two common systems are polar and Cartesian coordinates. Polar coordinates use a radius and an angle \((r, \theta)\) to specify a point. Meanwhile, Cartesian coordinates use \((x, y)\) pairs. These two systems can be related to one another with simple trigonometric functions:
\[ x = r \cos \theta \]
\[ y = r \sin \theta \]
In this conversion, the role of \(r\) is crucial, as it represents the distance from the origin to the point. Angle \(\theta\) indicates the direction from the positive x-axis.
By understanding how to switch between these forms, you can translate different perspectives into a unified view that is often easier to manipulate in calculus.

Chain Rule

The chain rule is a powerful differentiation technique. It allows the computation of a derivative of a function that is composed of other functions. When a function is expressed parametrically, this rule becomes essential. Let's explore this through the given task:
The derivatives \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\) represent how \(x\) and \(y\) change with \(\theta\):
\[ \frac{dx}{d\theta} = \cos\theta - \theta\sin\theta \]
\[ \frac{dy}{d\theta} = \sin\theta + \theta\cos\theta \]
Then, the chain rule helps find \(\frac{dy}{dx}\) by dividing these derivatives:
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \]
By breaking down complex equations into simpler ones, the chain rule provides a robust method to tackle differentiation in multivariable calculus.

Tangent and Normal Lines

Tangent lines touch a curve at a single point without crossing it. They represent the immediate slope or direction of the curve at that point. In the given exercise, we find the tangent by first calculating \(\frac{dy}{dx}\) and then using it in the point-slope equation:
\[ y - y_1 = m(x - x_1) \]
where \(m\) is the slope. With our solution:
\[ y = -\frac{2}{\pi}x + \frac{\pi}{2} \]
The normal line is perpendicular to the tangent, hence it has a slope that is the negative reciprocal of the tangent slope. For our solution, this results in:
\[ y = \frac{\pi}{2}x + \frac{\pi}{2} \]
Understanding how to derive these lines is crucial for graphs or situations where describing the immediate slope or perpendicular direction is necessary.

#Angle Theta

Angles in polar coordinates, denoted as \(\theta\), are central in describing the position of a point in space relative to the origin. In the problem, \(\theta = \frac{\pi}{2}\) represents a 90-degree angle from the positive x-axis. This particular angle simplifies many calculations because:

• The cosine of \(\frac{\pi}{2}\) is \(0\).
• The sine of \(\frac{\pi}{2}\) is \(1\).

Knowing these trigonometric values helps simplify expressions and calculus operations involving polar coordinates. As a foundational aspect of trigonometry, mastering the unit circle and these angle relationships will facilitate work in both pure and applied mathematics.