ADNAN MUHAMMED

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Angle Between Two Curves

Angle Between Two Curves

Let there be two curves y = f₁(x) and y = f₂(x) which intersect each other at point (x₁, y₁). If we draw tangents to these curves at the intersecting point, the angle between these tangents, is called the angle between two curves.

Let m₁ = (df₁ (x))/dx |_(x=x1) and m₂ = (df₂ (x))/dx |_(x=x1)

And both m₁ and m₂ are finite.

The acute angle between the curves is given by

θ = tan^-1 |(m₁-m₂)/(1+m₁ m₂ )|

Important note Note: (p - q) is also an angle between lines.

If m₁ (or m₂) is infinity the angle is given by θ = |π/2-θ₁ | where

θ₁ = tan^-1 m₂ (or tan^-1 m₁)

In the figure given below, f is the angle between the two curves,which is given by

f = Ψ₁ - Ψ₂

=> tanΦ = tan (Ψ₁ - Ψ₂)

= (tan Ψ1-tan Ψ2)/(1+tan Ψ1 tan Ψ2 ),

where tan Ψ₁ = f'(x₁) and tan Ψ₂ = g'(x₁).

Two curves are said to cut each other orthogonally if the angle between them is a right angle, that is, if f = 90^o, in which case we will have,

tanΨ₁ tanΨ₂ = -1.

Two curves touch each other if the angle between the tangents to the curves at the point of intersection is 0^o, in which case we will have,

tanΨ₁ = tanΨ₂.

Illustration:

Show that the curves ax² + by² = 1 and cx² + dy² = 1 cut each other orthogonally if, 1/a-1/b=1/c-1/d.

Solution:

Let the two curves cut each other at the point (x₁, y₁). Then

ax₁² + by₁² = 1 ...... (1)

and cx₁² + dy₁² = 1. ...... (2)

From (1) and (2), we get

(a - c)x₁² + (b - d)y₁² = 0. ...... (3)

Slope of the tangent to the curve ax² + by² = 1, at (x₁, y₁) is given by

tan Ψ₁ = [dy/dx]_(x1,y1) = -ax₁/by₁ .

Slope of the tangent to the curve cx² + dy² = 1 at (x₁, y₁) is given by

tan Ψ₂ = [dy/dx]_(x1,y1) = -cx₁/dy₁. If the two curves cut orthogonally, we must have,

(-ax₁/by₁ )(-cx₁/dy₁ ) = -1 => acx₁² + bdy₁² = 0. ...... (4)

From (3) and (4) we have

(a-c)/ac=(b-d)/bd => 1/a-1/c=1/c-1/d.

Illustration:

Prove that the tangent lines to the curve y2 = 4ax at points where x = a are at right angles to each other.

Solution:

y² = 4ax

When x = a

y² = 4a²

=> y = +2a

points are (a, 2a) and (a, - 2a)

Slope of the tangent of the curve y² = 4ax is

y dy/dx = 4a

dy/dx = 2a/y

for y = 2a, dy/dx = 1 = m₁

for y = -2a dy/dx = - 1 = m₂

m₁m₂ = -1

To read more, Buy study materials of Applications of Derivatives comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Mathematics here.