Question

(a) Find a vector-parametric equation r⃗ 1(t)=⟨x(t),y(t),z(t)⟩r→1(t)=⟨x(t),y(t),z(t)⟩ for the shadow of the circular cylinder x2+z2=5x2+z2=5 in the xzxz-plane. Shadow: r⃗ 1(t)=r→1(t)= for 0≤t≤2π0≤t≤2π. (b) Find a vector-parametric equation for intersection of the circular cylinder x2+z2=5x2+z2=5 and the plane 3x+2y+8z=13x+2y+8z=1. Intersection: r⃗ 2(t)=r→2(t)= for 0≤t≤2π0≤t≤2π.

Answer 1

Answer: (a) r1(t) = <2cost , 0 , 2sint>

(b) <2cost , (1 - 12sint - 10cost)/8 , 2sint>

Step-by-step explanation:

x2+z2=4

a)

Now, in the xz plane, we know that y = 0...

So, x^2 + z^2 = 4 will simply be a circle centered at (0,0)..

This can be easily parameterized as

x = 2cos(t)

z = 2sin(t)

So, the required parameterization is :

r1(t) = <2cost , 0 , 2sint>

b)

Cylinder : x^2 + z^2 = 4

Plane : 5x+8y+6z=1

Easily enough, the x^2 + z^2 = 4 can again be parameterized as

x = 2cost , z = 2sint

With this, we can find y using plane equation...

5x+8y+6z=1

5(2cost) + 8y + 6(2sint) = 1

8y = 1 - 12sint - 10cost

y = (1 - 12sint - 10cost)/8

So, the parameterization is :

<2cost , (1 - 12sint - 10cost)/8 , 2sint>