Of the 100 students enrolled at the dance studio, 50 take ballet and 24 take tap dancing. Eight of the students who take ballet
2 answers:
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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
But
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 <em>tangent vector</em> 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
at the point (-1, 0, 1) is given by
where
since
so, a normal vector to the tangent plane is
(-2,0,-1)
Hence, <em>a vector in the same direction as the projection of the helix's tangent vector (0, -π, 1) onto the tangent plane </em>is given by
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
where
= angle between the tangent vector and its projection onto the tangent plane. So
and
<span>Point B has coordinates (3,-4) and lies on the circle. Draw the perpendiculars from point B to the x-axis and y-axis. Denote the points of intersection with x-axis A and with y-axis C. Consider the right triangle ABO (O is the origin), by tha conditions data: AB=4 and AO=3, then by Pythagorean theorem:
</span>
<span>
.
</span>
{Note, that BO is a radius of circle and it wasn't necessarily to use Pythagorean theorem to find BO}
<span>The sine of the angle BOA is</span>
Since point B is placed in the IV quadrant, the sine of the angle that is <span> drawn in a standard position with its terminal ray will be </span>
<span /><span>
</span><span>
</span>
.
</span>
<span>
</span>
{Note, that BO is a radius of circle and it wasn't necessarily to use Pythagorean theorem to find BO}
<span>The sine of the angle BOA is</span>
Since point B is placed in the IV quadrant, the sine of the angle that is <span> drawn in a standard position with its terminal ray will be </span>
<span /><span>
</span><span>
</span>