Differential Equations - Problems on one parameter family of curves

In this topic, we will solve problems related to the one parameter family of curves.

  • Definition of a one parameter family of curves
  • Derivation of the general solution of a one parameter family of curves
  • Finding the particular solution of a one parameter family of curves
  • Solving problems using the concept of a one parameter family of curves

Definition of a one parameter family of curves

A one parameter family of curves is a set of curves that can be represented by a single equation involving a constant parameter. Examples:

  • All straight lines passing through the origin (0,0) can be represented by the equation y = mx, where m is the slope of the line.
  • All circles with center (h,k) can be represented by the equation (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

Derivation of the general solution of a one parameter family of curves

To derive the general solution of a one parameter family of curves, we follow these steps:

  1. Assume the constant parameter as a variable in the equation.
  1. Solve the equation to find the general solution involving the variable parameter. Example: Consider the equation of a one parameter family of curves as y = mx^2. Assuming m as a variable, we can solve this equation to find the general solution y = cx^2, where c is the constant parameter.

Finding the particular solution of a one parameter family of curves

To find the particular solution of a one parameter family of curves, we need to know the value of the parameter. Example: For the family of lines y = mx, if we are given the value of m as 2, we can find the particular solution as y = 2x.

Solving problems using the concept of a one parameter family of curves

We can solve various mathematical problems using the concept of a one parameter family of curves. Let’s look at some examples: Example 1: Find the equation of the tangent to the curve y = 3x^2 passing through the point (1,6). Solution:

  1. Differentiate the equation y = 3x^2 to find the slope of the tangent, which is 6x.
  1. Substitute the coordinates of the given point into the equation y = mx + c to find the value of c.
  1. Substitute the slope and the value of c into the equation y = mx + c to find the equation of the tangent. Example 2: Solve the differential equation (x + y) dx - y dy = 0. Solution:
  1. Rewrite the equation as (x + y) dx = y dy.
  1. Integrate both sides to find the general solution.
  1. Substitute initial conditions to find the particular solution.

Solving problems using the concept of a one parameter family of curves

  • Example 3: Find the equation of a circle passing through the points (1,2) and (4,5).
    • Assume the equation of the circle as (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center and r is the radius.
    • Substitute the coordinates of the given points into the equation to form two equations.
    • Solve the system of equations to find the values of h, k, and r.
    • Substitute the values into the equation to find the equation of the circle.
  • Example 4: Find the equation of a parabola with vertex (2,-3) and passing through the point (4,1).
    • Assume the equation of the parabola as y = a(x-h)^2 + k, where (h,k) is the vertex and a is a constant.
    • Substitute the coordinates of the given point into the equation to form an equation.
    • Substitute the values of the vertex and the given point into the equation to form another equation.
    • Solve the system of equations to find the value of a.
    • Substitute the value of a into the equation to find the equation of the parabola.
  • Example 5: Find the equation of a hyperbola with vertices (0,±2) and passing through the point (3,4).
    • Assume the equation of the hyperbola as (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) is the center and a and b are constants.
    • Substitute the coordinates of the given points into the equation to form two equations.
    • Solve the system of equations to find the values of h, k, a, and b.
    • Substitute the values into the equation to find the equation of the hyperbola.

Solving problems using the concept of a one parameter family of curves (contd.)

  • Example 6: Find the equation of an ellipse with foci (-2,0) and (2,0) passing through the point (1,3).
    • Assume the equation of the ellipse as (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) is the center and a and b are constants.
    • Substitute the coordinates of the given points into the equation to form three equations.
    • Solve the system of equations to find the values of h, k, a, and b.
    • Substitute the values into the equation to find the equation of the ellipse.
  • Example 7: Find the equation of a line tangent to the curve y = x^3 + 4x^2 + 3x at the point (-1,2).
    • Differentiate the equation y = x^3 + 4x^2 + 3x to find the slope of the tangent at any point.
    • Substitute the coordinates of the given point into the derivative to find the slope of the tangent at the given point.
    • Use the point-slope form of a line to find the equation of the tangent.
  • Example 8: Find the condition on parameter m for which the line y = mx is tangent to the curve x^2 + y^2 + 4x - 6y - 12 = 0.
    • Substitute y = mx into the equation of the curve to form a quadratic equation in terms of x.
    • Find the discriminant of the quadratic equation for the line to be tangent.
    • Set the discriminant equal to zero and solve for m to find the condition.

Solving problems using the concept of a one parameter family of curves (contd.)

  • Example 9: Find the equation of a line perpendicular to the curve y = 2x^2 - 3x + 5 at the point (1,4).
    • Differentiate the equation y = 2x^2 - 3x + 5 to find the slope of the tangent at any point.
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the tangent.
    • Use the point-slope form of a line to find the equation of the perpendicular line.
  • Example 10: Find the equation of a tangent to the curve y = e^x at the point (0,1).
    • Differentiate the equation y = e^x to find the slope of the tangent at any point.
    • Substitute the coordinates of the given point into the derivative to find the slope of the tangent at the given point.
    • Use the point-slope form of a line to find the equation of the tangent.
  • Example 11: Find the equation of a normal to the curve y = ln(x) at the point (1,0).
    • Differentiate the equation y = ln(x) to find the slope of the tangent at any point.
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the tangent.
    • Use the point-slope form of a line to find the equation of the normal.

Solving problems using the concept of a one parameter family of curves (contd.)

  • Example 12: Find the equation of a line parallel to the curve y = 4x^2 + 3x + 2 and passing through the point (1,5).
    • Find the slope of the curve y = 4x^2 + 3x + 2 by differentiating.
    • Find the equation of the line parallel to the curve by using the same slope.
    • Use the point-slope form of a line to find the equation of the line passing through the given point.
  • Example 13: Find the equation of a line perpendicular to the curve y = √x at the point (4,2).
    • Find the slope of the curve y = √x by differentiating.
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the tangent.
    • Use the point-slope form of a line to find the equation of the perpendicular line.
  • Example 14: Find the equation of a line parallel to the line 5x - 3y + 2 = 0 and passing through the point (2,1).
    • Find the slope of the given line by rearranging the equation in slope-intercept form y = (5/3)x + (2/3).
    • Find the equation of the line parallel to the given line by using the same slope.
    • Use the point-slope form of a line to find the equation of the parallel line passing through the given point.

Solving problems using the concept of a one parameter family of curves (contd.)

  • Example 15: Find the equation of a line perpendicular to the line 2x + 3y - 4 = 0 and passing through the point (-1,2).
    • Find the slope of the given line by rearranging the equation in slope-intercept form y = (-2/3)x + (4/3).
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the given line.
    • Use the point-slope form of a line to find the equation of the perpendicular line passing through the given point.
  • Example 16: Find the equation of a line parallel to the line 3x - 2y + 5 = 0 and passing through the point (-2,-3).
    • Find the slope of the given line by rearranging the equation in slope-intercept form y = (3/2)x - (5/2).
    • Find the equation of the line parallel to the given line by using the same slope.
    • Use the point-slope form of a line to find the equation of the parallel line passing through the given point.
  • Example 17: Find the equation of a line perpendicular to the line 4x + 5y - 6 = 0 and passing through the point (3,-2).
    • Find the slope of the given line by rearranging the equation in slope-intercept form y = (-4/5)x + (6/5).
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the given line.
    • Use the point-slope form of a line to find the equation of the perpendicular line passing through the given point.

Solving problems using the concept of a one parameter family of curves (contd.)

  • Example 18: Find the equation of a line parallel to the line 2x + y - 3 = 0 and passing through the point (4,1).
    • Find the slope of the given line by rearranging the equation in slope-intercept form y = -2x + 3.
    • Find the equation of the line parallel to the given line by using the same slope.
    • Use the point-slope form of a line to find the equation of the parallel line passing through the given point.
  • Example 19: Find the equation of a line perpendicular to the line x - 4y + 2 = 0 and passing through the point (-3,5).
    • Find the slope of the given line by rearranging the equation in slope-intercept form y = (1/4)x + (1/2).
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the given line.
    • Use the point-slope form of a line to find the equation of the perpendicular line passing through the given point.
  • Example 20: Find the equation of a line perpendicular to the line 3x + 2y - 5 = 0 and passing through the point (2,-1).
    • Find the slope of the given line by rearranging the equation in slope-intercept form y = (-3/2)x + (5/2).
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the given line.
    • Use the point-slope form of a line to find the equation of the perpendicular line passing through the given point.
  1. Problems on one parameter family of curves
  • Example 1: Find the equation of the tangent to the curve y = 2x^2 + 3x - 1 at the point (-1, -2).
    • Differentiate the equation y = 2x^2 + 3x - 1 to find the slope of the tangent, which is 4x + 3.
    • Substitute the coordinates of the given point into the equation y = mx + c to find the value of c.
    • Substitute the slope and the value of c into the equation y = mx + c to find the equation of the tangent.
  • Example 2: Find the equation of the tangent to the curve y = sin(x) at the point (π/2, 1).
    • Differentiate the equation y = sin(x) to find the slope of the tangent, which is cos(x).
    • Substitute the coordinates of the given point into the equation y = mx + c to find the value of c.
    • Substitute the slope and the value of c into the equation y = mx + c to find the equation of the tangent.
  • Example 3: Find the equation of the perpendicular line to the curve y = x^3 + 2x^2 - x + 3 at the point (1, 5).
    • Differentiate the equation y = x^3 + 2x^2 - x + 3 to find the slope of the tangent at any point.
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the tangent.
    • Use the point-slope form of a line to find the equation of the perpendicular line.
  • Example 4: Find the equation of the line parallel to the curve y = e^x at the point (0, 1).
    • Differentiate the equation y = e^x to find the slope of the tangent at any point.
    • Find the equation of the line parallel to the tangent by using the same slope.
    • Use the point-slope form of a line to find the equation of the parallel line passing through the given point.
  • Example 5: Find the equation of the tangent to the curve y = ln(x) at the point (1, 0).
    • Differentiate the equation y = ln(x) to find the slope of the tangent at any point.
    • Substitute the coordinates of the given point into the equation y = mx + c to find the value of c.
    • Substitute the slope and the value of c into the equation y = mx + c to find the equation of the tangent.
  1. Problems on one parameter family of curves (contd.)
  • Example 6: Find the equation of the tangent to the curve y = √x at the point (4, 2).
    • Differentiate the equation y = √x to find the slope of the tangent at any point.
    • Substitute the coordinates of the given point into the equation y = mx + c to find the value of c.
    • Substitute the slope and the value of c into the equation y = mx + c to find the equation of the tangent.
  • Example 7: Find the equation of the tangent to the curve y = 1/x at the point (2, 1/2).
    • Differentiate the equation y = 1/x to find the slope of the tangent at any point.
    • Substitute the coordinates of the given point into the equation y = mx + c to find the value of c.
    • Substitute the slope and the value of c into the equation y = mx + c to find the equation of the tangent.
  • Example 8: Find the equation of the tangent to the curve y = 1/x^2 at the point (3, 1/9).
    • Differentiate the equation y = 1/x^2 to find the slope of the tangent at any point.
    • Substitute the coordinates of the given point into the equation y = mx + c to find the value of c.
    • Substitute the slope and the value of c into the equation y = mx + c to find the equation of the tangent.
  • Example 9: Find the equation of the normal to the curve y = sin(x) at the point (π/2, 1).
    • Differentiate the equation y = sin(x) to find the slope of the tangent at any point.
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the tangent.
    • Use the point-slope form of a line to find the equation of the normal.
  • Example 10: Find the equation of the normal to the curve y = cos(x) at the point (π/2, 0).
    • Differentiate the equation y = cos(x) to find the slope of the tangent at any point.
    • Find the negative reciprocal of the slope to find the slope of the line perpendicular to the tangent.
    • Use the point-slope form of a line to find the equation of the normal.
  1. Problems on one parameter family of curves (contd.)
  • Example 11: Find the condition on parameter c for which the line y = cx is tangent to the curve y = x^2.
    • Substitute y = cx into the equation of the curve y = x^2 to form a quadratic equation in terms of x.
    • Find the discriminant of the quadratic equation for the line to be tangent.
    • Set the discriminant equal to zero and solve for c to find the condition.
  • Example 12: Find the equation of the tangent to the curve y = e^x + 2x - 3 at the point (-1, -1).
    • Differentiate the equation y = e^x + 2x -