Question

Show that the curves r=asec² theta/2 and r= bcosec² theta/2 intersect at right angle

Answer 1

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QUESTION

Show that the curves r=asec² theta/2 and r= bcosec² theta/2 intersect at right angle

ANSWER.

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Answer 2

Answer:

The given two curves intersect at right angle.

Step-by-step explanation:

From the above question,

They have given :

The curve r = $asec^2(theta/2)$ is defined in the interval (0, pi/2) U (3pi/2, 2pi) and the curve r = $bcosec^2(theta/2)$ is defined in the interval (pi/2, 3pi/2).

To show that these two curves intersect at right angle, we can use the property that the angle between two curves is given by the dot product of their respective gradient vectors.

The gradient of the first curve r = $asec^2(theta/2)$ is given by:

∇r = $(d/d(theta)) (asec^2(theta/2)) (cos(theta/2), sin(theta/2))$

=$(a*sec^2(theta/2)*tan(theta/2)cos(theta/2), asec^2(theta/2)*sin(theta/2))$

The gradient of the second curve r = bcosec^2(theta/2) is given by:

∇r = $(d/d(theta)) (bcosec^2(theta/2)) (-sin(theta/2), cos(theta/2))$

$(-b*cosec^2(theta/2)*cot(theta/2)sin(theta/2), bcosec^2(theta/2)*cos(theta/2))$

The dot product of these two gradient vectors is:

∇r1. ∇r2 = $= a*sec^2(theta/2)*tan(theta/2)cos^2(theta/2) + asec^2(theta/2)sin^2(theta/2) - bcosec^2(theta/2)*cot(theta/2)sin^2(theta/2) - bcosec^2(theta/2)*cos^2(theta/2)$

= $asec^2(theta/2) - bcosec^2(theta/2)$

This dot product is equal to zero for all theta, which means that the angle between the two curves is 90 degrees (or pi/2 radians) at all points of intersection, so the two curves intersect at right angle.

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