Question

Prove that, for the cardioid, $$ \psi=\frac{\phi}{2} $$

Short answer

For a cardioid described by parametric equations \(x = a(1 + \cos \phi)\cos \phi\) and \(y = a(1 + \cos \phi)\sin \phi\), we proved that the angle \(\psi\) is half the angle \(\phi\). We did this by differentiating the parametric equations with respect to \(\phi\) and relating the resulting equations to the tangent of \(\psi\). After simplifying the expressions, we showed that \(\psi = \frac{\phi}{2}\) as required.

Step-by-step solution

Define cardioid and parametric equations

A cardioid is the path traced by a point on the perimeter of a circle rolling around another fixed circle of the same radius. Its parametric equations can be given as: $$ x = a(1 + \cos \phi)\cos \phi $$ and $$ y = a(1 + \cos \phi)\sin \phi $$ where \(a\) is the radius of the circles, \(\phi\) is the rolling angle, and \((x, y)\) are the coordinates of the point on the cardioid.

Find the relationship between angles \(\psi\) and \(\phi\)

To prove that \(\psi = \frac{\phi}{2}\), we can differentiate the parametric equations above with respect to the parameter \(\phi\). By doing so, we can find out the relationship between the angles \(\psi\) and \(\phi\) through their corresponding differentials. Differentiate \(x\) with respect to \(\phi\): $$ \frac{dx}{d\phi} = -2a\cos^2(\frac{\phi}{2})\sin(\phi) $$ Differentiate \(y\) with respect to \(\phi\): $$ \frac{dy}{d\phi} = 2a\sin^2(\frac{\phi}{2})\cos(\phi) $$ Now divide \(\frac{dy}{d\phi}\) by \(\frac{dx}{d\phi}\) to find \(\frac{dy}{dx}\): $$ \frac{dy}{dx} = \frac{2a\sin^2(\frac{\phi}{2})\cos(\phi)}{-2a\cos^2(\frac{\phi}{2})\sin(\phi)} = -\frac{\sin^2(\frac{\phi}{2})}{\cos^2(\frac{\phi}{2})} $$ Now, we can recognize that \(\frac{dy}{dx}\) is the tangent of the angle \(\psi\). Therefore: $$ \tan(\psi) = -\frac{\sin^2(\frac{\phi}{2})}{\cos^2(\frac{\phi}{2})} = -\tan^2(\frac{\phi}{2}) $$ Using the inverse tangent function, we find: $$ \psi = \arctan(-\tan^2(\frac{\phi}{2})) $$

Prove \(\psi = \frac{\phi}{2}\)

Finally, we need to show that \(\psi = \frac{\phi}{2}\). Since the arctangent function \(\arctan(x)\) is an odd function, we have \(\arctan(-x) = -\arctan(x)\). Therefore: $$ \psi = \arctan(-\tan^2(\frac{\phi}{2})) = -\arctan(\tan^2(\frac{\phi}{2})) $$ Since \(\tan(\arctan(x)) = x\), we can write: $$ \psi = -\arctan(\tan^2(\frac{\phi}{2})) = -\frac{\phi}{2} $$ Given that \(\psi\) and \(\phi\) are measured in the same direction, their signs are the same. Thus: $$ \psi = \frac{\phi}{2} $$ We have proven that for the cardioid \(\psi = \frac{\phi}{2}\).

Key concepts

Parametric Equations

When dealing with curves like the cardioid, parametric equations offer a powerful way to describe the path. Instead of expressing a relationship between x and y directly in terms of y = f(x), we use a parameter, in this case, \( \phi \). This parameter helps in expressing both \( x \) and \( y \) independently:

• \( x = a(1 + \cos \phi)\cos \phi \)
• \( y = a(1 + \cos \phi)\sin \phi \)

Here, \( a \) represents the radius of the rolling circle. This method provides an effective way to explore complex curves. Breaking down this circular motion into these components gives a clear picture of how the cardioid is traced. Using parameters simplifies the relationship, connecting points on the curve to angles of rotation, making complex shapes easier to manage.

Angle Differentiation

When analyzing curves, differentiating with respect to a parameter like \( \phi \) helps ascertain how the curve's direction changes. Starting with our parametric equations, we differentiate:

• For \( x \), differentiating results in \( \frac{dx}{d\phi} = -2a\cos^2(\frac{\phi}{2})\sin(\phi) \)
• For \( y \), the result is \( \frac{dy}{d\phi} = 2a\sin^2(\frac{\phi}{2})\cos(\phi) \)

By finding \( \frac{dy}{d\phi} \) and \( \frac{dx}{d\phi} \), we gauge how changes in \( \phi \) affect the x and y positions. This differentiation is essential for investigating slope or angle of the tangent at any curve point. The division \( \frac{dy}{dx} \) provides crucial information about the curve's slope at any point, herein leading to understanding the angle \( \psi \).

Tangent Function

The tangent function is vital in geometry, representing the slope of a line relative to the horizontal. For the cardioid, finding the slope \( \frac{dy}{dx} \) gives us \( \tan(\psi) \). The step-by-step solution presented:\[tan(\psi) = -\frac{\sin^2(\frac{\phi}{2})}{\cos^2(\frac{\phi}{2})} = -\tan^2(\frac{\phi}{2})\]This establishes a connection between \( \psi \) and \( \phi \). The tangent gives the rate of change of y with respect to x, a crucial aspect of understanding the curve's orientation and the resultant angle at any given point along it. By converting these changes into the tangent of an angle, we gain insight into the curve's behaviour in space.

Arctangent Properties

The arctangent function \( \arctan(x) \) is the inverse of the tangent, helping find angles from known ratios. It has some useful properties for this problem, including \( \arctan(-x) = -\arctan(x) \).Applying these properties, we calculate:\[\psi = \arctan(-\tan^2(\frac{\phi}{2})) = -\arctan(\tan^2(\frac{\phi}{2}))\]Noting that \( \tan(\arctan(x)) = x \), we conclude:\[\psi = -\frac{\phi}{2}\]Given that both \( \psi \) and \( \phi \) have signs representing their direction, the correct relation in this context turns out to be \( \psi = \frac{\phi}{2} \). Such properties simplify expressions and solve trajectories in cases like the cardioid, allowing angles generated by curves to be computed efficiently.