Question

Where does the helix r(t) = cos(πt), sin(πt), t intersect the paraboloid z = x2 + y2? (x, y, z) = What is the angle of intersection between the helix and the paraboloid? (This is the angle between the tangent vector to the curve and the tangent plane to the paraboloid. Round your answer to one decimal place.)

Answer 1

Answer:

Intersection at (-1, 0, 1).

Angle 0.6 radians

Step-by-step explanation:

The helix r(t) = (cos(πt), sin(πt), t) intersects the paraboloid  

z = x2 + y2 when the coordinates (x,y,z)=(cos(πt), sin(πt), t) of the helix satisfy the equation of the paraboloid. That is, when

$\bf (cos(\pi t), sin(\pi t), t)$

But  

$\bf cos^2(\pi t)+sin^2(\pi t)=1$

so, the helix intersects the paraboloid when t=1. This is the point

(cos(π), sin(π), 1) = (-1, 0, 1)

The angle of intersection between the helix and the paraboloid is the angle between the tangent vector to the curve and the tangent plane to the paraboloid.

The tangent vector to the helix in t=1 is

r'(t) when t=1

r'(t) = (-πsin(πt), πcos(πt), 1), hence

r'(1) = (0, -π, 1)

A normal vector to the tangent plane of the surface  

$\bf z=x^2+y^2$

at the point (-1, 0, 1) is given by

$\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)$

where

$\bf f(x,y)=x^2+y^2$

since

$\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y$

so, a normal vector to the tangent plane is

(-2,0,-1)

Hence, a vector in the same direction as the projection of the helix's tangent vector (0, -π, 1) onto the tangent plane is given by

$\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)$

The angle between the tangent vector to the curve and the tangent plane to the paraboloid equals the angle between the tangent vector to the curve and the vector we just found.  

But we now

$\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta$

where  

$\bf \theta$= angle between the tangent vector and its projection onto the tangent plane. So

$\bf \pi^2=(\sqrt{4+\pi^2}\sqrt{\pi^2+1})cos\theta\rightarrow cos\theta=\frac{\pi^2}{\sqrt{4+\pi^2}\sqrt{\pi^2+1}}=0.8038$

and

$\bf \theta=arccos(0.8038)=0.6371\;radians$