Question

if y=cos^2t and x=sin2t then d2y/dx2 at t= π/6​

Answer 1

Given,

$\displaystyle\small\longrightarrow y=\cos^2t$

Differentiating wrt t we get,

$\displaystyle\small\longrightarrow\dfrac{dy}{dt}=-\sin(2t)$

Given,

$\displaystyle\small\longrightarrow x=\sin(2t)$

Differentiating wrt t,

$\displaystyle\small\longrightarrow\dfrac{dx}{dt}=2\cos(2t)$

Then,

$\displaystyle\small\text{ \longrightarrow\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)} }$

$\displaystyle\small\longrightarrow\dfrac{dy}{dx}=-\dfrac{1}{2}\tan(2t)$

And,

$\displaystyle\small\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$

$\displaystyle\small\text{ \longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{d\left(\dfrac{dy}{dx}\right)}{dx} }$

$\displaystyle\small\text{ \longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}} }$

$\displaystyle\small\text{ \longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(-\dfrac{1}{2}\tan(2t)\right)}{2\cos(2t)} }$

$\displaystyle\small\longrightarrow\dfrac{d^2y}{dx^2}=-\dfrac{1}{2}\sec^3(2t)$

At $\smallt=\dfrac{\pi}{6},$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{\dfrac{d^2y}{dx^2}=-4}} }$

Answer 2

Step-by-step explanation:

Given,

\displaystyle\small\text{$\longrightarrow y=\cos^2t$}⟶y=cos

2

t

Differentiating wrt t we get,

\displaystyle\small\text{$\longrightarrow\dfrac{dy}{dt}=-\sin(2t)$}⟶

dt

dy

=−sin(2t)

Given,

\displaystyle\small\text{$\longrightarrow x=\sin(2t)$}⟶x=sin(2t)

Differentiating wrt t,

\displaystyle\small\text{$\longrightarrow\dfrac{dx}{dt}=2\cos(2t)$}⟶

dt

dx

=2cos(2t)

Then,

\displaystyle\small\text{$\longrightarrow\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}$}⟶

dx

dy

=

(

dt

dx

)

(

dt

dy

)

\displaystyle\small\text{$\longrightarrow\dfrac{dy}{dx}=-\dfrac{1}{2}\tan(2t)$}⟶

dx

dy

=−

2

1

tan(2t)

And,

\displaystyle\small\text{$\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)$}⟶

dx

2

d

2

y

=

dx

d

(

dx

dy

)

\displaystyle\small\text{$\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{d\left(\dfrac{dy}{dx}\right)}{dx}$}⟶

dx

2

d

2

y

=

dx

d(

dx

dy

)

\displaystyle\small\text{$\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}$}⟶

dx

2

d

2

y

=

dt

dx

dt

d

(

dx

dy

)

\displaystyle\small\text{$\longrightarrow\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(-\dfrac{1}{2}\tan(2t)\right)}{2\cos(2t)}$}⟶

dx

2

d

2

y

=

2cos(2t)

dt

d

(−

2

1

tan(2t))

\displaystyle\small\text{$\longrightarrow\dfrac{d^2y}{dx^2}=-\dfrac{1}{2}\sec^3(2t)$}⟶

dx

2

d

2

y

=−

2

1

sec

3

(2t)

At \small\text{$t=\dfrac{\pi}{6},$}t=

6

π

,

\displaystyle\small\text{$\longrightarrow\underline{\underline{\dfrac{d^2y}{dx^2}=-4}}$}⟶

dx

2

d

2

y

=−4