Two particles move on two different paths which start from a common point 'O'. Particle A follows a path described by f(x) = (3ax² + 2)/(2x + c) and particle B follows a path described by g(x) = (2bx² + 3)/(2x + c). Find the rate at which particle A moves with respect to particle B.
Options:
(1) (3ax² + 3acx - 2)/(2bx² + 2bcx - 3)
(2) (3ax² + 3acx - 2)/(3bx² + 3bcx - 2)
(3) (3bx² + 3bcx - 2)/(2ax² + 2acx - 3)
(4) (2bx² + 2bcx - 3)/(3ax² + 3acx - 2)
Answer (1)
To find the rate at which particle A moves with respect to particle B, we need to calculate the derivative of both functions, f ( x ) and g ( x ) , with respect to x , and then find the ratio of these derivatives.
The path of particle A is described by: f ( x ) = 2 x + c 3 a x 2 + 2
The path of particle B is described by: g ( x ) = 2 x + c 2 b x 2 + 3
Step 1: Differentiate f ( x ) and g ( x )
To find the derivative, apply the quotient rule. The quotient rule states that if you have a function h ( x ) = v ( x ) u ( x ) , then: h ′ ( x ) = [ v ( x ) ] 2 u ′ ( x ) v ( x ) − u ( x ) v ′ ( x )
Differentiate
f ( x )
:
Let u ( x ) = 3 a x 2 + 2 and v ( x ) = 2 x + c .
u ′ ( x ) = 6 a x
v ′ ( x ) = 2
The derivative f ′ ( x ) is: f ′ ( x ) = ( 2 x + c ) 2 ( 6 a x ) ( 2 x + c ) − ( 3 a x 2 + 2 ) ( 2 )
Simplifying: f ′ ( x ) = ( 2 x + c ) 2 12 a x 2 + 6 a c x − 6 a x 2 − 4
Combine terms: f ′ ( x ) = ( 2 x + c ) 2 6 a x 2 + 6 a c x − 4
Differentiate
g ( x )
:
Let u ( x ) = 2 b x 2 + 3 and v ( x ) = 2 x + c .
u ′ ( x ) = 4 b x
v ′ ( x ) = 2
The derivative g ′ ( x ) is: g ′ ( x ) = ( 2 x + c ) 2 ( 4 b x ) ( 2 x + c ) − ( 2 b x 2 + 3 ) ( 2 )
Simplifying: g ′ ( x ) = ( 2 x + c ) 2 8 b x 2 + 4 b c x − 4 b x 2 − 6
Combine terms: g ′ ( x ) = ( 2 x + c ) 2 4 b x 2 + 4 b c x − 6
Step 2: Determine the Rate of Particle A Relative to Particle B
The rate at which particle A moves with respect to particle B is given by the ratio g ′ ( x ) f ′ ( x ) :
g ′ ( x ) f ′ ( x ) = ( 2 x + c ) 2 4 b x 2 + 4 b c x − 6 ( 2 x + c ) 2 6 a x 2 + 6 a c x − 4
Cancel out ( 2 x + c ) 2 from both the numerator and denominator:
g ′ ( x ) f ′ ( x ) = 4 b x 2 + 4 b c x − 6 6 a x 2 + 6 a c x − 4
After simplification, this corresponds to option (1): 2 b x 2 + 2 b c x − 3 3 a x 2 + 3 a c x − 2
