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Equation of Curves

A line which deviate from straightness is called curved line. Like a straight line it is also smooth and continuous. Polynomial curve with degree greater than 1 is made up of curved line; degree for any polynomial curve is highest power of variable in that polynomial. For example if there is any polynomial curve with equation 2x3 + 5x2 + 6x + 3 = 0 then degree for this curve is 3 because highest power of variable is 3. To know the equation of straight line we will need slope of line X-intercept, Y-intercept but in the case of curved line we will need the points where curve line is intersecting the X-axis and that points is also called as roots for that polynomial curve.

How to find equation of curved line in the form of ax2 + bx + c = 0

These types of line will cut the X-axis at two points by help of these two points we will be able to find the equation curved line.

Assume curved line ax2 + bx + c = 0 is intersecting the X-axis at points (α, 0) and (β, 0) then equation of curved line will be:

x2 - (α + β)x + αβ = 0

How to find equation of curved line in the form of ax3 + bx2 + cx + d = 0

These types of line will cut the X-axis at three points by help of these three points we will be able to find the equation of curved line.

Assume curved line ax3 + bx2 + cx + d = 0 is intersecting the X-axis at points (α, 0), (β, 0), (λ, 0) then equation of this curve line will be:

X3 - (α+β+λ) X2 + (αβ+βλ+λα) X - αβλ = 0

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Parametric Form of Curve

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The parametric equation is a way of define a relation using with the help of parameter. For example a curve is given by x = x(t), y = y(t) is said to be a parametrically defined curve and the functions of the curve is x = x(t) and y = y(t) are said to be the parametric equation for the curve. In the below sections, learn this concept of parametric curve equation in detail.

Solve Parametric Curve Equations


Below are some examples on parametric curve equations

Example 1:

The line passes through the line points (4, 2) and (6, 8) when t =0 we are at the point (4, 2) and when t =1 we are at the points (6, 8) find the parametric equations.

Solution:


We write equations symbolically:

(x,y) = (1 - t) (4, 2)+ (t) (6, 8)

= (4 - 4t + 6t, 2 - 2t + 8t) = (4 + 2t, 2 + 6t)

so that

x(t) = 4 + 2t and y(t) = 2 + 6t

Example 2:

The line passes through the line points (5, 3) and (6, 7) when t =0 we are at the point (5, 3) and when t =1 we are at the points (6, 7) find the parametric equations.

Solution:


We write equations symbolically:

(x,y) = (1 - t) (5, 3)+ (t) (6, 7)

= (5 - 5t + 6t, 3 - 3t + 7t) = (5 + t, 3 + 4t)

so that the equation is

x(t) = 5 + t and y(t) = 3 + 4t

Example 3:

The line passes through the line points (6, 3) and (5, 6) when t = 0 we are at the point (6,3) and when t = 1 we are at the points (5,6) find the parametric equations.

Solution:


We write equations symbolically:

(x,y) = (1 - t) (6, 3)+ (t) (5, 6)

= (6 - 6t + 5t, 3 - 3t + 6t) = (6 + t, 3 + 3t)

so that the equation is

x(t) = 6 + t and y(t) = 3 + 3t

Example 4:

How to find the parametric equations for the line through the points (10, 8) and (15, 6) so that when t = 0 we are at the point (10, 8) and when t = 1 we are at the point (15, 6).

Solution:

The parametric equations generally written as,

To find equations

(x, t) = (1-t) (10, 8) + (t) (15, 6)

= (10-10t+15t, 8-8t+6t)

= (10+5t, 8- 2t)

(x, t)= x (t) +y (t)

(x, t) can be written as x (t) =10 +5t and y (t) =8-2t.

Examples on Curve Equation

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Below are few examples on curve line equations explained with steps:

Example 1: what will be equation of curved line which is of degree two and intersecting the X-axis at points (2, 0) and (3, 0)?

Solution: equation of curved line is of degree 2 so their equation will be

x2 - (α + β)x + αβ = 0

Here α =2 and β =3

α + β =2+3=5

αβ =2 x 3 = 6

So equation of curved line is x2 - 5x + 6 = 0

Example 2: what will be equation of curved line which is of degree three and intersecting the X-axis at points (1, 0), (2, 0) and (3, 0)?

Solution: equation of curved line is of degree 3 so their equation will be

X3 - (α+ β + λ) X2 + (αβ+βλ+λα) X - αβλ = 0

Here α =1, β =2, λ =3

α + β + λ = 1+2+3 = 6

αβ+βλ+λα = 1 x 2 + 2x3 + 3 x 1 = 2 + 6 + 3 = 11

αβλ = 1 x 2 x 3 = 6

So equation of curved line is x3 - 6x2 + 11x - 6 = 0

Practice Problems on Equation of Curved Line

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Question 1: what will be equation of curved line which is of degree two and intersecting the X-axis at points (3, 0) and (5, 0)?

Answer: x2 - 8x + 15 = 0

Question 2: what will be equation of curved line which is of degree three and intersecting the X-axis at points (1, 0), (4, 0) and (7, 0)?

Answer: x3- 12x2 + 39x + 28 = 0

Practice Problems on Parametric Equations

Problem:

The line passes through the line points (4, 3) and (6, 4) when t =0 we are at the point (4,3) and when t =1 we are at the points (6,4) find the parametric equations.

Answer:

x(t) = 4 + 2t and y(t) = 3 + t

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