Question

If x=asin2t(1+cos2t) and y=b cos2t(1-cos2t) find dy/dx at t= pi/4

Answer 1

Step-by-step explanation:

x = asin2t ( 1+ cos2t)

we know,

cos2t = 2cos²t - 1

cos2t +1 = 2cos²t

1+cos2t = 2cos²t

x = asin2t ( 1+ cos2t) = asin2t x 2cos²t

x = asin2t x 2cos²t

Differentiation wrt x

dx/dt = d/dt (asin2t x 2cos²t)

we know ,

y= u. v then

dy/dx = u.( dv/dx) + v. (du/dx)

=)) d/dt (asin2t x 2cos²t)

=)) asin2t x 4cost x ( -sint)

+ 2cos²t x (acos2t) (2)

=)) -sint x asin2t x 4cost

+ 4 cos²t x acos2t

=)) 4cos²t x acos2t

- sint x asin2t x 4cost

=)) 4acost ( cost x cos2t

- sint x sin2t )

dx/dt = 4acost ( cost x cos2t - sint x sin2t )

________________________

Now,

y=b cos2t(1-cos2t)

we know,

cos2t = 1 - 2sin²t

take (-) on both the sides

-cos2t = - (1 - 2sin²t)

-cos2t = -1 + 2sin²t

- cos2t + 1 = 2sin²t

1 - cos2t = 2sin²t

y = b cos2t(1-cos2t)=bcos2t x 2sin²t

y = bcos2t x 2sin²t

Differentiation wrt t

dy / dt = d/dt ( bcos2t x 2sin²t )

=)) bcos2t x 4sint x (cost )

+ ( 2sin²t x ( -bsin2t) x (2)

=)) bcos2t x 4sint x cost

- 4 sin²t x bsint

=)) 4bsint (cos2t x cost

- sint x sin2t)

dy /dt = 4bsint (cos2t x cost

- sint x sin2t)

______________________

we know, dy/dx = (dy/dt)/(dx/dt)

therefor,

dy/dx=

=)) (4bsint (cos2t x cost - sint x sin2t) ) / (4acost ( cost x cos2t

- sint x sin2t ))

= )) (4bsint (cos2t x cost - sint x

sin2t)) / (4a cost ( cos2t x cost

-sint x sin2t) )

= )) 4b sint / 4a cost

= )) ( b/a ) tant

here, we have to find dy/dx at t= π/4t = π/4

= )) ( b/a) tan ( π/4 )

= )) (b/a) ( 1) ............

.................(since tan(π/4 ) = 1)

= )) b/a

dy/dx = b/a

Answer 2

Answer:

$\displaystyle \dfrac{dy}{dx}=\dfrac{b}{a}$

Step-by-step explanation:

Given :

$\displaystyle x=a\sin2t(1+\cos2t)\\\\\\\displaystyle y=b\cos2t(1-\cos2t)$

Now

$\displaystyle \text{ \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt}} }$

First for $\dfrac{dy}{dt}$

$\displaystyle \dfrac{dy}{dt} =b\cos2t(1-\cos2t)\\\\\\\displaystyle \dfrac{dy}{dt} =-b\sin2t+2b\sin4t\\\\\\\displaystyle \text{We applied here multiplication rule and we know \dfrac{d}{dt}\cost=-\sin t }$

Now for $\dfrac{dx}{dt}$

$\displaystyle \dfrac{dx}{dt} =a\sin2t(1+\cos2t)\\\\\\\displaystyle \dfrac{dx}{dt}= 2a\cos2t+2a\cos4t\\\\\\\displaystyle \dfrac{d}{dt}(\sin t)=\cos t$

Now for $\dfrac{dy}{dx}$

$\displaystyle \dfrac{dy}{dx}=\dfrac{-b\sin2t+2b\sin4t}{ 2a\cos2t+2a\cos4t}\\\\\\\displaystyle \text{Now put here t=\dfrac{\pi}{4} }\\\\\\\displaystyle \text{ \dfrac{dy}{dx}=\dfrac{-b\sin2\times\dfrac{\pi}{4}+2b\sin4\times\dfrac{\pi}{4}}{ 2a\cos2\times\dfrac{\pi}{4}+2a\cos4\times\dfrac{\pi}{4}} }\\\\\\\displaystyle \text{ \dfrac{dy}{dx}=\dfrac{-b\sin \dfrac{\pi}{2}+2b\sin \pi}{ 2a\cos \dfrac{\pi}{2}+2a\cos \pi} }\\\\\\\displaystyle \dfrac{dy}{dx}=\dfrac{-2b+0}{0-2a}$

Here we put ratios values

$\displaystyle \dfrac{dy}{dx}=\dfrac{-2b}{-2a}\\\\\\\displaystyle \dfrac{dy}{dx}=\dfrac{b}{a}$

Thus we get answer.