Question

The orthogonal trajectory of the cardioid \( r = a(1 - \cos\theta) \), where 'a' is an arbitrary constant is:

• A. \( r = b(1 + \cos\theta) \), where b is an arbitrary constant
• B. \( r = b(1 - \cos\theta) \), where b is an arbitrary constant
• C. \( r = b(1 + \sin\theta) \), where b is an arbitrary constant
• D. \( r = b(1 - \sin\theta) \), where b is an arbitrary constant

Hint:

The process for finding orthogonal trajectories in polar coordinates involves replacing \( \frac{dr}{d\theta} \) with \( -r^2/\frac{dr}{d\theta} \). Memorizing this transformation is key. Also, be ready to use trigonometric half-angle identities to simplify and solve the resulting differential equation.

Answer 1

The Correct Option is A

Step 1: Introduction
Orthogonal trajectories intersect a family of curves at right angles. For a family of curves in polar coordinates \( F(r, \theta, a) = 0 \), the differential equation is found. Then, \( \frac{dr}{d\theta} \) is replaced with \( -r^2 \frac{d\theta}{dr} \) to get the orthogonal family's differential equation.

Step 2: Differential Equation of the Given Family
Given family: \( r = a(1 - \cos\theta) \). Differentiate with respect to \( \theta \): \[ \frac{dr}{d\theta} = a(\sin\theta) \quad (1) \] Eliminate \(a\) using \( a = \frac{r}{1 - \cos\theta} \). Substitute into (1): \[ \frac{dr}{d\theta} = \frac{r \sin\theta}{1 - \cos\theta} \] This is the differential equation for the cardioid family.

Step 3: Differential Equation of the Orthogonal Trajectory
Replace \( \frac{dr}{d\theta} \) with \( -r^2 \frac{d\theta}{dr} \): \[ -r^2 \frac{d\theta}{dr} = \frac{r \sin\theta}{1 - \cos\theta} \] \[ -r \frac{d\theta}{dr} = \frac{\sin\theta}{1 - \cos\theta} \] Separate variables: \[ \frac{1 - \cos\theta}{\sin\theta} d\theta = -\frac{dr}{r} \] Using half-angle identities: \(1-\cos\theta = 2\sin^2(\theta/2)\) and \(\sin\theta = 2\sin(\theta/2)\cos(\theta/2)\). \[ \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} d\theta = -\frac{dr}{r} \] \[ \tan(\theta/2) d\theta = -\frac{dr}{r} \]
Step 4: Solve the New Differential Equation
Integrate both sides: \[ \int \tan(\theta/2) d\theta = - \int \frac{1}{r} dr \] \[ -2\ln|\cos(\theta/2)| = -\ln|r| + C_1 \] \[ 2\ln|\cos(\theta/2)| = \ln|r| - C_1 \] \[ \ln(\cos^2(\theta/2)) = \ln|r| + \ln(e^{-C_1}) \] \[ \cos^2(\theta/2) = r \cdot C_2 \quad (\text{where } C_2 = e^{-C_1}) \] Using \( \cos^2(\theta/2) = \frac{1+\cos\theta}{2} \): \[ \frac{1+\cos\theta}{2} = r C_2 \] \[ r = \frac{1}{2C_2}(1+\cos\theta) \] Let \( b = \frac{1}{2C_2} \). The orthogonal trajectories are: \[ r = b(1+\cos\theta) \]
Final Answer: The orthogonal trajectory is \( r = b(1 + \cos\theta) \).