Question

Find dy/dx , if
x=cos^-1(2t/1+t^2), y=sec^-1(√1+t^2)
(Derivartive of Parametric Functions)
Class 12 HSC

Answer 1

Given :  x=cos^-1(2t/1+t^2), y=sec^-1(√1+t^2)

To find :  dy/dx

Solution:

x=cos⁻¹(2t/(1+t²))

=> Cosx  = 2t/(1+t²)

=> -Sinx dx/dt  = 2t ( -1/(1 + t²)²)(2t)  + 2/(1+t²)

=> -Sinx dx/dt  =   (-4t²  + 2 + 2t²)/(1 + t²)²

=>  -Sinx dx/dt  =   ( 2 - 2t²)/(1 + t²)²

=> dx/dt  =  - 2 ( 1 - t²)/(1 + t²)²Sinx

Sinx = √(1 - Cos²x) =  √(1 - (2t/(1+t²))²)   = (1-t²)/(1 + t²)

=> dx/dt  = -2/(1+t²)

y=sec⁻¹(√1+t²)

secy  = √1+t²

secytany dy/dt  =   t/√1+t²

=> tany dy/dt  = t/(1+t²)

sec²y  = 1 + tan²y

=> 1 + tan²y  = 1 + t²

=> tany  = t

t dy/dt  = t/(1+t²)

=> dy/dt  = 1/(1+t²)

dy/dx  =  (dy/dt)/(dx/dt)

=> dy/dx  =( 1/(1+t²))/ ( -2/(1+t²))

=> dy/dx  = -1/2

Another method :

x=cos⁻¹(2t/(1+t²))    y=sec⁻¹(√1+t²)

put  t  = tanα

=> x = cos⁻¹(2tanα/(1+tan²α)) = cos⁻¹(2tanα/(Sec²α)) = cos⁻¹(2SinαCosα)) = cos⁻¹(Sin2α) = cos⁻¹(Cos(π/2 -2α))

=π/2 -2α

dx/dα  = -2

y=sec⁻¹(√1+t²) = sec⁻¹(√1+tan²α) =  sec⁻¹(√sec²α) = sec⁻¹(secα)  = α

dy/dα  = 1

dy/dx = (dy/dα)/(dx/dα) = -1/2

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