Question

Two particles travel along the space curves r1(t) = t, t2, t3 r2(t) = 1 + 4t, 1 + 16t, 1 + 52t . Find the points at which their paths intersect. (If an answer does not exist, enter DNE.) (x, y, z) = (smaller x-value) (x, y, z) = (larger x-value) Find the time(s) when the particles collide. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) t =

Answer 1

Answer:

Thus the path intersect twice at(3,9,27) when $s=\frac{1}{2}$ and t=3  and (1,1,1)when

s=0 and t=1.

The particles do not collide.

Step-by-step explanation:

r₁(t)=(t,t²,t³)    and r₂(t)=(1+4t,1+6t,1+52t)

If the particles collide provided r₁(t)=r₂(t)

(t,t²,t³)=(1+4t,1+6t,1+52t)

Equating the equation,

t=1+4t

$\Rightarrow t-4t=1$

$\Rightarrow t=-\frac{1}{3}$

and  t² = 1+6t

$\Rightarrow (-\frac{1}{3})^2= 1+6\times(- \frac{1}{3} )$

so this equation does not satisfy $t=-\frac{1}{3}$

So the particles do not collide.

For the path intersection  r₁(t) = r₂(s)

⇒(t,t²,t³)= (1+4s,1+16s,1+52s)

Therefore

t=1+4s..........(1)                t²=1+16s..........(2)            t³=1+52s..........(3)

Putting t=1+4s in equation (2)

(1+4s)²=1+16s

⇒1+8s+16s²=1+16s

⇒16s²-8s=0

⇒8s(2s-1)=0

$\Rightarrow s =0\ and \ \frac{1}{2}$

When s=0  ,  t =1+4.0=1

when $s=\frac{1}{2}$,  t=1+2=3

Putting s=0 and t=1 in equation (3)

1³=1+52.0

⇒1=1

Again putting $s=\frac{1}{2}$ and t=3

3³= 1+26

⇒27=27

Therefore s=0 and t=1  satisfies the third equation and  $s=\frac{1}{2}$ and t=3 also satisfies the third equation.

Thus the path intersect twice at(3,9,27) when $s=\frac{1}{2}$ and t=3  and (1,1,1)when

s=0 and t=1.