Section 2.5-2.6 lecture notes, Lecture notes of Engineering Mathematics

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Chapter 2, Part 2

2.5. Applications (Text, Section 2.5)

Orthogonal trajectories

Exponential Growth/Decay

Newton’s Law of Cooling/Heating

Limited Growth (Logistic Equation)

Miscellaneous Models 1

2.5.1. Orthogonal Trajectories

Example: Family of circles, center at (1, 2):

(x − 1)^2 + (y − 2)^2 = C

• 1 1 2 3 x

1

2

3

4

y

Family of lines through (1, 2):

y − 2 = K(x − 1)

• 1 1 2 3 x

1

2

3

4

y

DE for this family:

y − 2 = K(x − 1)

y′^ = y x^ −−^21

The lines and the circles:

• 1 1 2 3 x

1

2

3

4

y

Given a one-parameter family of curves

F (x, y, C) = 0.

A curve that intersects each mem- ber of the family at right angles (or- thogonally) is called an orthogonal trajectory of the family.

A procedure for finding a family of orthogonal trajectories

G(x, y, K) = 0

for a given family of curves

F (x, y, C) = 0

Step 1. Determine the differential equation for the given family (recall Chapter 1 problems)

F (x, y, C) = 0.

Step 2. Replace y′^ in that equa- tion by − 1 /y′; the resulting equa- tion is the differential equation for the family of orthogonal trajecto- ries.

Step 3. Find the general solu- tion of the new differential equation. This is the family of orthogonal tra- jectories.

• y^3 = Cx^2 +

Orthogonal trajectories:

3 x^2 + 2y^2 +^8 y = C

• 2 - 1 1 2 x
  • 2
  • 1

1

y

2. Find the orthogonal trajecto- ries of the family of parabolas with vertical axis and vertex at the point (− 1 , 3).
   • 1 x

3

y

Differential equation for the family:

2 (x^ + 1)

(^2) + (y − 3) (^2) = C (ellipses)

- ^ - 3 - 1 1 3 ^ x

1

3

y

Both families:

-
- 4 - 2 2 4 x

2

4

y