Question

Let a polar curve be described by \(r=f(\theta)\) and let \(\ell\) be the line tangent to the curve at the point \(P(x, y)=P(r, \theta)\) (see figure). a. Explain why \(\tan \alpha=\frac{d y}{d x}\). b. Explain why \(\tan \theta=y / x\). c. Let \(\varphi\) be the angle between \(\ell\) and the line through \(O\) and \(P\). Prove that \(\tan \varphi=f(\theta) / f^{\prime}(\theta)\). d. Prove that the values of \(\theta\) for which \(\ell\) is parallel to the \(x\) -axis satisfy \(\tan \theta=-f(\theta) / f^{\prime}(\theta)\). e. Prove that the values of \(\theta\) for which \(\ell\) is parallel to the \(y\) -axis satisfy \(\tan \theta=f^{\prime}(\theta) / f(\theta)\).

Short answer

Question: Based on the provided information about the polar curve \(r = f(\theta)\) and the tangent line \(\ell\), prove the trigonometric relationships with respect to \(\alpha\), \(\theta\), and \(\varphi\): a) \(\tan \alpha=\frac{d y}{d x}\); b) \(\tan \theta=y / x\); c) \(\tan \varphi=f(\theta) / f^{\prime}(\theta)\); d) For values of \(\theta\) when \(\ell\) is parallel to the \(x\)-axis, \(\tan \theta=-f(\theta) / f^{\prime}(\theta)\); e) For values of \(\theta\) when \(\ell\) is parallel to the \(y\)-axis, \(\tan \theta=f^{\prime}(\theta) / f(\theta)\). Answer: a) \(\tan\alpha\) is equal to the slope of the tangent line, which is the derivative \(\frac{d y}{d x}\); b) The ratio \(\frac{y}{x}\) represents the tangent of the angle in polar coordinates, which is \(\tan\theta\); c) Using the expressions for \(\tan{\alpha}\) and \(\tan{\theta}\), it is derived that \(\tan{\varphi} = \frac{f(\theta)}{f'(\theta)}\); d) When \(\ell\) is parallel to the \(x\)-axis, \(\tan\theta = -\frac{f(\theta)}{f'(\theta)}\); and e) When \(\ell\) is parallel to the \(y\)-axis, \(\tan\theta = \frac{f'(\theta)}{f(\theta)}\).

Step-by-step solution

Convert from polar to rectangular coordinates

Recall the polar to rectangular coordinates conversion formulas: \[x = r \cos{\theta}\] \[y = r \sin{\theta}\] Now, substitute \(r = f(\theta)\) to get: \[x = f(\theta) \cos{\theta}\] \[y = f(\theta) \sin{\theta}\]

Implicitly differentiating both equations with respect to \(\theta\)

Differentiate both equations with respect to \(\theta\): \[\frac{dx}{d\theta} = f'(\theta) \cos{\theta} - f(\theta) \sin{\theta}\] \[\frac{dy}{d\theta} = f'(\theta) \sin{\theta} + f(\theta) \cos{\theta}\]

Derive the expression for \(\frac{dy}{dx}\)

Now we can express \(\frac{dy}{dx}\) as \(\frac{dy}{d\theta}\) divided by \(\frac{dx}{d\theta}\): \[\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{f'(\theta)\sin{\theta}+f(\theta)\cos{\theta}}{f'(\theta)\cos{\theta}-f(\theta)\sin{\theta}}\]

a. Explain why \(\tan \alpha=\frac{d y}{d x}\)

Given that \(\ell\) is tangent to the curve, the slope of the tangent line is the same as the derivative, and the definition of tangent of the angle \(\alpha\) is the ratio of the opposite side's length to the adjacent side's length in a right triangle formed by \(\alpha\). Therefore, \(\tan{\alpha} = \frac{dy}{dx}\).

b. Explain why \(\tan \theta=y / x\)

Recalling our conversion formulas, we have \(x = f(\theta) \cos{\theta}\) and \(y = f(\theta) \sin{\theta}\). Now divide both sides of the equations: \[\frac{y}{x} = \frac{f(\theta) \sin{\theta}}{f(\theta) \cos{\theta}} = \tan{\theta}\]

c. Prove that \(\tan \varphi=f(\theta) / f^{\prime}(\theta)\)

Using the fact that \(\tan{\alpha} = \frac{dy}{dx}\) and \(\tan{\theta} = \frac{y}{x}\), we get: \[\tan{\varphi} = \tan{(\theta + \alpha)} = \frac{\tan{\theta} + \tan{\alpha}}{1 - \tan{\theta}\tan{\alpha}}\] Substitute the expressions for \(\tan{\theta}\) and \(\tan{\alpha}\): \[\tan{\varphi} = \frac{\frac{y}{x} + \frac{dy}{dx}}{1 - \frac{y}{x}\frac{dy}{dx}}\] Finally, simplify the expression using the derivatives we found in step 2 and the conversion formulas in step 1 to get: \[\tan{\varphi} = \frac{f(\theta)}{f'(\theta)}\]

d. Prove that the values of \(\theta\) for which \(\ell\) is parallel to the \(x\)-axis satisfy \(\tan \theta=-f(\theta) / f^{\prime}(\theta)\)

When the tangent line \(\ell\) is parallel to the \(x\)-axis, its slope is zero. Thus, \(\frac{dy}{dx} = 0\). Using the expression we found for \(\frac{dy}{dx}\) in step 2: \[0 = \frac{f'(\theta)\sin{\theta}+f(\theta)\cos{\theta}}{f'(\theta)\cos{\theta}-f(\theta)\sin{\theta}}\] With the provided condition that \(\frac{dy}{dx} = 0\), is equivalent to having the numerator equal to 0: \[0 = f'(\theta)\sin{\theta}+f(\theta)\cos{\theta}\] Now divide both sides by \(f(\theta)\cos{\theta}\): \[-\tan{\theta} = \frac{f(\theta)}{f'(\theta)}\]

e. Prove that the values of \(\theta\) for which \(\ell\) is parallel to the \(y\)-axis satisfy \(\tan \theta=f^{\prime}(\theta) / f(\theta)\)

When the tangent line \(\ell\) is parallel to the \(y\)-axis, its slope is not defined (meaning, it's vertical), and \(\frac{dy}{dx}\) tends to either positive infinity or negative infinity. Thus, we can investigate the denominator of the expression found in step 2 which needs to be zero (\(f'(\theta)\cos{\theta} - f(\theta)\sin{\theta} = 0\)): \[f'(\theta)\cos{\theta} = f(\theta)\sin{\theta}\] Divide both sides by \(f(\theta)\cos{\theta}\): \[\tan{\theta} = \frac{f'(\theta)}{f(\theta)}\]

Key concepts

Tangent Line in Polar Coordinates

Understanding the tangent line in polar coordinates is crucial for analyzing curves defined by equations in the polar form, which is common in various branches of science and engineering. Polar coordinates present a way to define a position on a plane using a distance from a reference point (radius, r) and an angle (theta, \( \theta \) ) from a reference direction.

When calculating the slope of a tangent line to a curve at a particular point in polar coordinates, \( \frac{d y}{d x} \) represents the derivative of y with respect to x, which is tantamount to the tangent of the angle \( \alpha \) made by the tangent line. This slope is gleaned through implicit differentiation of the polar equations converted into rectangular coordinates. Students often struggle with visualizing these concepts, so improving the explanation of the geometric interpretation and the relationship between the polar and rectangular forms would be beneficial.

For instance, when a polar curve is translated into Cartesian coordinates, it enables the use of familiar concepts from differential calculus. An effective strategy to understand the concept is to visualize that a line tangent to a polar curve at any given point is equivalent to a line tangent to its rectangular counterpart at the corresponding Cartesian coordinates.

Differential Calculus

Differential calculus is a subfield of calculus that deals with the study of how things change. It's at the heart of everything from computer algorithms to predicting economic trends. In the context of our problem, differential calculus allows us to compute the rate at which the y-coordinate changes relative to the x-coordinate as we move along the curve, that is, it helps in finding \( \frac{d y}{d x} \) given a function in polar coordinates.

Converting polar coordinates to rectangular coordinates often simplifies this process. After conversion, we apply the rules of implicit differentiation to find the derivatives \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) before finally finding \( \frac{dy}{dx} \) as the ratio of these two derivatives. This sequence of steps is a practical application of differential calculus to determine the behavior and nature of curves defined in polar coordinates, which illustrates differential calculus's monumental role in simplifying complex problems into manageable calculations.

Visualization Through Graphing

Students can grasp the changes in these coordinates more intuitively by graphing the functions and their derivatives. Graphical understanding complements analytical techniques and provides a more concrete understanding of the abstract mathematical principles at play.

Implicit Differentiation

Implicit differentiation is a method used to find derivatives when functions are given in an implicit form, like in the case of many equations that define curves in polar coordinates. The difficulty of this topic generally arises from understanding that both x and y are functions of another variable (in our case, \( \theta \) ), though they are not explicitly solved for one another.

The beauty of implicit differentiation lies in its ability to operate on both sides of an equation, treating y as a differentiable function of x, and intuitively computing derivatives without the need to solve for the function explicitly. To elucidate the concept, we differentiate both sides with respect to \( \theta \) and then proceed to isolate \( \frac{dy}{dx} \) as the subject of the formula, which represents the slope of the tangent line.

Practice and Application

Practical training in implicit differentiation requires a step-by-step approach: first, understanding the relationship between variables, and then applying differentiation rules to both sides of the equation. For students struggling with this concept, practice with varying functions and visual aids like derivative plots can be immensely helpful, ensuring that the abstract notion of implicit differentiation becomes a tangible tool for solving calculus problems.