Show that the curves r=acos(è) and r=cos(è) intersect at right angles.

can it be shown that the derivative of one is the negative reciprocal of the derivative of the other?

To show that the curves intersect at right angles, we need to verify if the derivative of one curve is the negative reciprocal of the derivative of the other curve at the point of intersection.

Let's start by finding the derivatives of the curves:

For the curve r = acos(θ):
1. Differentiate both sides of the equation with respect to θ:
dr/dθ = d(acos(θ))/dθ

Using the chain rule, we get:
dr/dθ = -asin(θ)

For the curve r = cos(θ):
1. Differentiate both sides of the equation with respect to θ:
dr/dθ = d(cos(θ))/dθ

Using the chain rule, we get:
dr/dθ = -sin(θ)

Now, let's check if the derivative of one curve is the negative reciprocal of the derivative of the other curve at the point of intersection.

At the point of intersection, the curves r = acos(θ) and r = cos(θ) share the same value of r and θ. Therefore, we can set them equal to each other:

acos(θ) = cos(θ)

To find the point(s) of intersection, we need to solve this equation for θ. One such solution is when θ = π/4.

Now, let's evaluate the derivatives at θ = π/4:

For dr/dθ of the curve r = acos(θ):
dr/dθ = -asin(π/4) = -(1/√2)

For dr/dθ of the curve r = cos(θ):
dr/dθ = -sin(π/4) = -(1/√2)

Now, let's check if the derivative of one curve is the negative reciprocal of the derivative of the other curve at θ = π/4:

(-1/√2) = -(1/√2)

Since the derivatives of both curves are equal at the point of intersection, we can conclude that the curves r = acos(θ) and r = cos(θ) intersect at a right angle.