Given:

Dog run as twice as fast as the Rabbit.

Calculation:

The speed of Dog is twice the speed of Rabbit. Both are travelled in straight line. Let the distance travelled by the dog is then the position of rabbit is ( 0,s/2 ).

The tangent of path joining dog and rabbit has slope is dy dx = s/2−y 0−x

Therefore,

   s=2y−2x dy dx ds dx =2 dy dx −( 2x d 2 y d x 2 +2 dy dx )=−2x d 2 y d x 2 ds dx =2 dy dx −( 2x d 2 y d x 2 +2 dy dx )=−2x d 2 y d x 2 and   ds dx =− 1+ ( dy dx ) 2   then, −2x d 2 y d x 2 =− 1+ ( dy dx ) 2

Let z= dy dx then put this is above equation:

   2x dz dx = 1+ z 2 2dz 1+ z 2 = dx x

Integrating both side then,

   ln x= ∫ 2dz 1+ z 2         = 2ln ( 1+ z 2 +z ) +C

When x=L, z=dy/dx=0, so

   ln L=2 ln 1+C=C ln x=2ln ( 1+ z 2 +z )+ ln L          =ln ( L  ( 1+ z 2 +z ) 2 ) x=L  ( 1+ z 2 +z ) 2 1+ z 2 = x L −z 1+ z 2 = x L + z 2 −2z x L 2z x L = x L −1 dy dx =z= 1 2 x L − 1 2 x L dy dx    = 1 2 L x 1/2 − L 2 x −1/2 y= 1 3 L x 3/2 − L x 1/2 + C 1

When x=L, y=0 , then,

   0= 1 3 L L 3/2 − L L 1/2 + C 1     = L 3 −L+ C 1 C 1 = 2 3 L

Therefore,

   y= x 3/2 3 L − L x 1/2 + 2 3 L

As x=0 then the dog catches the rabbit when rabbit position is ( 0, 2/3L ).

Conclusion:

The Dog catches the rabbit when rabbit position is ( 0, 2/3L ).

Given:

Dog run as half as fast as the Rabbit.

Calculation:

The speed of Dog is half the speed of Rabbit. Both are travelled in straight line.

From the part a

   z= dy dx = x 2 2 L 2 − L 2 2 x 2

Therefore,

   y= x 3 6 L 2 + L 2 2x − 2 3 L

We have to minimize the distance D from the dog ( x,y ) and rabbit ( 0,2sso,

The tangent of path joining dog and rabbit has slope is dy dx = 2s−y 0−x

   s= 1 2 y− 1 2 x dy dx 2s=y−xz y−2s=xz=x( x 2 2 L 2 − L 2 2 x 2 ) y−2s= x 3 2 L 2 − L 2 2x        D= ( x−0 ) 2 + ( y−2s ) 2             = x 2 + ( x 3 2 L 2 − L 2 2x ) 2      = x 6 4 L 4 + L 4 4 x 2 + x 2 2 = ( x 3 2 L 2 + L 2 2x ) 2      =( x 3 2 L 2 + L 2 2x )

For miminum value of D then check value of D ′ =0

   D ′ =0 3 x 2 2 L 2 − L 2 2 x 2 =0 3 x 2 2 L 2 = L 2 2 x 2 x 4 = L 4 3 x= L 3 4  , x>0, L>0

Then

   D ′ ′ ( x )= 3x L 2 + L 2 x 3 >0 for all x>0 so,

   D( L 3 4 )= ( L 3 4 ) 3 2 L 2 + L 2 2( L 3 4 )                  = 2L 3 3/4

   D= 2L 3 3/4 is the minimum distance between dog and rabbit.

The position of dog and rabbit is:

Dog: ( x,y )=( L 3 4 ,( 5 3 7/4 − 2 3 )L )=( L 3 4 ,  5 3 4 −6 9 L )

Rabbit: ( 0,2s )=( 0, 8 3 4 L 9 − 2L 3 )=( 0, 8 3 4 −6 9 L )

Conclusion:

The position of dog is ( L 3 4 ,  5 3 4 −6 9 L ) and rabbit is ( 0, 8 3 4 −6 9 L ).