Show that the following curves are intersect eachother orthoganally.r-(1+coso) r-(1-coso)

To determine if two curves intersect orthogonally, we can check their slopes or gradients. If the product of their gradients is -1 at the point of intersection, then the curves are orthogonal.

Let's start by finding the gradients of the curves r = 1 + cos(Θ) and r = 1 - cos(Θ). To do this, we'll differentiate both equations with respect to Θ.

For the first curve, r = 1 + cos(Θ):
Differentiating both sides with respect to Θ:
dr/dΘ = -sin(Θ)

For the second curve, r = 1 - cos(Θ):
Differentiating both sides with respect to Θ:
dr/dΘ = sin(Θ)

Now, we need to find the angles (Θ) at which the gradients are equal. Setting the two derivatives equal to each other:
-sin(Θ) = sin(Θ)

Simplifying this equation:
2sin(Θ) = 0

This equation is satisfied when sin(Θ) = 0. The values of Θ that satisfy this condition are 0, π, 2π, etc. Essentially, the values of Θ where the curves intersect.

To determine if the curves are orthogonal at these points, we need to check the product of their gradients at these angles:

For Θ = 0:
Gradient of the first curve: dr/dΘ (r = 1 + cos(Θ)) = -sin(0) = 0
Gradient of the second curve: dr/dΘ (r = 1 - cos(Θ)) = sin(0) = 0
The product of the gradients is 0 * 0 = 0.

For Θ = π:
Gradient of the first curve: dr/dΘ (r = 1 + cos(Θ)) = -sin(π) = 0
Gradient of the second curve: dr/dΘ (r = 1 - cos(Θ)) = sin(π) = 0
The product of the gradients is 0 * 0 = 0.

We can repeat this process for Θ = 2π, 3π, and so on.

In all cases, the product of the gradients is 0, which means the gradients are perpendicular at these angles. Hence, the curves r = 1 + cos(Θ) and r = 1 - cos(Θ) intersect each other orthogonally at the angles Θ = 0, π, 2π, etc.