Respuesta :
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
[tex]\bf (cos(\pi t), sin(\pi t), t)[/tex]
But
[tex]\bf cos^2(\pi t)+sin^2(\pi t)=1[/tex]
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
[tex]\bf z=x^2+y^2[/tex]
at the point (-1, 0, 1) is given by
[tex]\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)[/tex]
where
[tex]\bf f(x,y)=x^2+y^2[/tex]
since
[tex]\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y[/tex]
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
[tex]\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)[/tex]
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
[tex]\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta[/tex]
where
[tex]\bf \theta[/tex]= angle between the tangent vector and its projection onto the tangent plane. So
[tex]\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[/tex]
and
[tex]\bf \theta=arccos(0.8038)=0.6371\;radians[/tex]
The angle of intersection between the helix and the paraboloid is 0.6371 radians.
What is an angle?
Angle is the space between the line or the surface that meets. And the angle is measured in degree. For complete 1 rotation, the angle is 360 degrees.
The helix r(t) = (cos(πt), sin(πt), t) intersect the paraboloid z = x² + y².
The coordinate (x, y, z) of the helix satisfies the equation of the paraboloid.
Then we have
[tex]\rm cos^2 (\pi t ) + sin^2 (\pi t) = 1[/tex]
So, the helix intersects the paraboloid when t = 1. This is the point
(cosπ, sinπ, 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
[tex]\rm r'(t) \ when\ t = 1[/tex]
[tex]\rm r'(t) = (- \pi sin(\pi t ), \pi cos(\pi t ), 1)\\r'(t) = (0, -\pi , 1)[/tex]
A normal vector to the tangent plane of the surface
z = x² + y²
at the point (-1, 0, 1) is given by
[tex]\rm (\dfrac{\partial f}{\partial x} (-1, 0) , \dfrac{\partial f}{\partial y} (-1, 0) , -1)\\\\\\\dfrac{\partial f}{\partial x} = 2x , \dfrac{\partial f}{\partial y} = 2y[/tex]
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
(0, -π, 1) - ((0, -π, 1)(-2, 0, -1))(-2, 0, 1) = (0, -π, 1) - (-2, 0, 1) = (2, -π, 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.
[tex](2, -\pi , 0)(0, -\pi , 0) = ||(2, -\pi , 0) || \ ||(0, -\pi ,1) || cos \theta[/tex]
where [tex]\theta[/tex] be the angle between the tangent vector and its projection onto the tangent plane. So
[tex]\pi ^{2} = (\sqrt{4 + \pi ^2}\sqrt{\pi ^2 + 1}) cos \theta[/tex]
Then the angle will be
[tex]\rm \theta = \cos ^{-1} 0.8038 = 0.6371 \ radians[/tex]
More about the angles link is given below.
https://brainly.com/question/15767203
