Applications of First Order ODEs: Orthogonal Trajectories, Newton's Law of Cooling & Growth Models

Explanation

Orthogonal trajectories have profound physical significance across physics and engineering: lines of electric force are orthogonal to equipotential surfaces; heat flow lines are orthogonal to isothermal curves; fluid streamlines are orthogonal to velocity potential lines. The algebraic replacement of slope m with -1/m enforces perpendicularity at every differential tangency point.

Key points

• Cartesian Procedure: (1) Differentiate family equation F(x, y, c) = 0, (2) Eliminate constant c to get f(x, y, dy/dx) = 0, (3) Replace dy/dx with -dx/dy, (4) Integrate to find orthogonal trajectories.
• Polar Procedure: (1) Differentiate F(r, θ, c) = 0 with respect to θ, (2) Eliminate c, (3) Replace dr/dθ with -r² (dθ/dr), (4) Integrate.
• Newton's Law of Cooling: dT/dt = -k (T - T_m) where T is temperature, T_m is ambient medium temperature, and k > 0 is heat transfer coefficient. Solution: T(t) = T_m + (T₀ - T_m) e^{-kt}.
• Malthusian Growth & Decay: dN/dt = k N => N(t) = N₀ e^{kt}. Half-life t_{1/2} = (ln 2) / k.
• Physical Equipotential & Streamlines: In fluid dynamics and electrostatics, streamlines and equipotential curves form orthogonal trajectory systems.

Example

Code examples

// Derivation: Orthogonal Trajectories of Confocal Conics
// Problem: Find the orthogonal trajectories of the family of coaxial circles:
//    x^2 + y^2 + 2gx = 0  (circles passing through origin with centers on x-axis)
// Step 1: Differentiate with respect to x:
//    2x + 2y (dy/dx) + 2g = 0 => g = -(x + y y')
// Step 2: Eliminate parameter g by substituting into original equation:
//    x^2 + y^2 + 2(-x - y y') x = 0
//    x^2 + y^2 - 2x^2 - 2xy y' = 0 => y^2 - x^2 - 2xy y' = 0
//    dy/dx = (y^2 - x^2) / (2xy)
// Step 3: Replace dy/dx with -dx/dy for orthogonal family:
//    -dx/dy = (y^2 - x^2) / (2xy) => dx/dy = (x^2 - y^2) / (2xy)
//    2xy dx + (y^2 - x^2) dy = 0
// Step 4: Solve the homogeneous ODE (or recognize exact form):
//    Divide by y^2: (2x/y) dx + [1 - (x/y)^2] dy = 0
//    Notice: d [ (x^2 + y^2) / y ] = 0
//    Therefore: x^2 + y^2 = 2cy  (or x^2 + y^2 - 2cy = 0)
// Conclusion: The orthogonal trajectories are a family of circles passing through the origin with centers on the y-axis!

Exam questions

• [8 Marks] Find the orthogonal trajectories of the family of parabolas y² = 4ax. Sketch both families.
• [7 Marks] Find the orthogonal trajectories of the family of cardioids r = a (1 + cos θ).
• [7 Marks] A body at temperature 80°F is placed in an atmosphere of constant temperature 50°F. After 5 minutes, body temperature is 70°F. Find: (i) temperature after 15 minutes, (ii) time when temperature is 60°F.