Find the curvature of the curve of intersection?

Question

Find the curvature of the curve of intersection of the cylinder x^2+y^2=16 and the plane x+z=5 at (4, 0, 1).

 

Answer 1

We must first parameterize the curve.

First of all, solving the second equation for z gives z = 5 - x.

Given the first equation, x²+y² = 16, we can take x = 4 cos t, y = 4 sin t so that z = 5 - 4 cos t.

Thus r = <4 cos t, 4 sin t, 5 - 4 cos t>

We now need v = dr/dt, a = dv/dt, and v×a.

v = <-4 sin t, 4 cos t, 4 sin t>

a = <-4 cos t, -4 sin t, 4 cos t>

v×a = <(4 cos t)(4 cos t) - (4 sin t)(-4 sin t) ,

             (4 sin t)(-4 cos t) - (-4 cos t)(4 sin t) ,

             (-4 sin t)(-4 sin t) - (4 cos t)(-4 cos t)>

       = <16, 0, 16>

Now ||v|| = √[(-4 sin t)² + (4 cos t)² + (4 sin t)²] = √(16 + 16 sin² t) = 4√(1+sin² t)

and ||v×a|| = 16√2

So the curvature will be

κ = ||v×a|| / ||v||³ = √2 / [4(1+sin² t)^3/2]

r = <4,0,1> occurs when cos t = 1 so that we can take t = 0.

We get κ = √2 / [4(1+sin² 0)^3/2] = √2 / 4.

Answer 2

I will outline the method.  To find the intersection, you first solve the plane equation for x.   

x = 5-z;  So x^2 = (5-z)^2 or (z-5)^2.

Now substitute this expression for x^2 in the first equation.

(z-5)^2 + y^2 = 16.   This is a circle of radius 4.  A circle has a constant curvature of 1/r or 1/4.  But you can simply use calculus to find the curvature using the formula using the first and second derivatives.  Use implicit differentiation to find dz/dy and d2z/dy2.