Angle between two curves

Angle between two curves

Definition 7.3

Angle between two curves, if they intersect, is defined as the acute angle between the tangent lines to those two curves at the point of intersection.

For the given curves, at the point of intersection using the slopes of the tangents, we can measure the acute angle between the two curves. Suppose y = m₁ x + c₁ and y = m₂ x + c₂ are two lines, then the acute angle θ between these lines is given by,

where m₁ and m₂ are finite.

Remark

(i) If the two curves are parallel at ( x₁ , y₁ ) , then m₁ = m₂ .

(ii) If the two curves are perpendicular at ( x₁ , y₁ ) and if m₁ and m₂ exists and finite then m₁m₂ = −1 .

Example 7.14

Find the angle between y = x² and y = (x − 3)².

Solution

Let us now find the point of intersection of the two given curves. Equating x² = (x − 3)² we get, x = 3/2. Therefore, the point of intersection is ( 3/2 ,9/4). Let θ be the angle between the curves. The slopes of the curves are as follows :

For the curve y = x²,

Example 7.15

Find the angle between the curves y = x² and x = y² at their points of intersection (0,0) and (1,1).

Solution

Let us now find the slopes of the curves.

Let m₁ be the slope of the curve y = x²,

then m = dy/dx = 2x .

Let m₂ be the slope of the curve x = y²,

then m₂ =  dy/dx =1/2y .

Let θ₁ and θ₂ be the angles at (0,0) and (1,1) respectively.

At (0, 0) , we come across the indeterminate form of 0× ∞ in the denominator of tanθ₁ =  and so we follow the limiting process.

Example 7.16

Find the angle of intersection of the curve y = sin x  with the positive x -axis.

Solution

When the curve y = sin x intersects the positive x -axis, y = 0 which gives, x = nÏ€ , n = 1, 2, 3,…. Now, dy/dx = cos x. The slope at x = nÏ€ are cos(nÏ€) = (−1)ⁿ. Hence, the required angle of intersection is

Example 7.17

If the curves ax² + by² = 1 and cx² + dy² = 1 intersect each other orthogonally then, show that 1/a – 1/b = 1/c – 1/d .

Solution

Let the two curves intersect at a point ( x₀ , y₀ ) . This leads to (a − c)x₀² + (b − d ) y₀² = 0 .

Let us now find the slope of the curves at the point of intersection ( x₀ , y₀ ) . The slopes of the curves are as follows :

For the curve ax² + by² = 1, dy/ dx = − ax/by

For the curve cx² + dy² = 1, dy/dx = − cx/dy

Now, if two curves cut orthogonally, then the product of their slopes, at the point of intersection (x₀ , y₀ ) , is −1. Hence, if the above two curves cut orthogonally at ( x₀ , y₀ ) then

Remark

In the above example, the converse is also true. That is assuming the condition 1/a – 1/b = 1/c – 1/d one can easily establish that the curves cut orthogonally.

Example 7.18

Prove that the ellipse x² + 4y² = 8 and the hyperbola x² − 2y² = 4 intersect orthogonally.

Solution

Let the point of intersection of the two curves be (a , b) . Hence,

a² + 4b² = 8 and a² − 2b² = 4         ... (4)

It is enough to show that the product of the slopes of the two curves evaluated at (a , b) is −1.

Differentiation of x² + 4 y² = 8 with respect x , gives

Differentiation of x² − 2y² = 4 with respect to x, gives

Applying the ratio of proportions in (4), we get

Therefore a²/b² = 32/4 = 8 . Substituting in (5), we get m₁ × m₂ = −1. Hence, the curves cut orthogonally.