Question

Question
$find \: d= \frac{dy}{dx} if \: x = \frac{sin {}^{3}t}{ \sqrt{cos2t} } \\ y = \frac{ {cos}^{3} t}{ \sqrt{cos2t} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$
Can anyone help me with this question????​

Answer 1

By the chain rule,

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

So first differentiate both x and y with respect to t.

• dx/dt :

$x = \dfrac{\sin^3(t)}{\sqrt{\cos(2t)}} = \sin^3(t) \left(\cos(2t)\right)^{-\frac12}$

By the product rule,

$\dfrac{dx}{dt} = \dfrac{d(\sin^3(t))}{dt} \left(\cos(2t)\right)^{-\frac12} + \sin^3(t) \dfrac{d\left(\cos(2t)\right)^{-\frac12}}{dt}$

By the power and chain rules,

$\dfrac{dx}{dt} = \dfrac{d(\sin(t))}{dt} 3\sin^2(t) \left(\cos(2t)\right)^{-\frac12} -\dfrac12 \sin^3(t) \left(\cos(2t)\right)^{-\frac32} \dfrac{d(\cos(2t))}{dt}$

$\dfrac{dx}{dt} = 3\sin^2(t) \cos(t) \left(\cos(2t)\right)^{-\frac12} -\dfrac12 \sin^3(t) \left(\cos(2t)\right)^{-\frac32} (-2 \sin(2t))$

We can clean this up a bit and write

$\dfrac{dx}{dt} = \dfrac{3\sin^2(t) \cos(t)}{\sqrt{\cos(2t)}} + \dfrac{\sin^3(t) \sin(2t)}{\sqrt{\cos^3(2t)}}$

$\dfrac{dx}{dt} = \dfrac{\sin^2(t)}{\sqrt{\cos^3(2t)}} \left(3 \cos(t) \cos(2t) + \sin(t) \sin(2t)\right)$

• dy/dt : (I'll skip some steps here, the derivative is nearly the same)

$y = \cos^3(t) \left(\cos(2t)\right)^{-\frac12}$

$\dfrac{dy}{dt} = 3\cos^2(t) (-\sin(t)) \left(\cos(2t)\right)^{-\frac12} + \cos^3(t) \left(-\dfrac12\right) \left(\cos(2t)\right)^{-\frac32} (-2 \sin(2t))$

$\dfrac{dy}{dt} = -\dfrac{3\cos^2(t) \sin(t)}{\sqrt{\cos(2t)}} + \dfrac{\cos^3(t) \sin(2t)}{\sqrt{\cos^3(2t)}}$

$\dfrac{dy}{dt} = -\dfrac{\cos^2(t)}{\sqrt{\cos^3(2t)}} \left(3 \sin(t) \cos(2t) - \cos(t) \sin(2t)\right)$

Finally, we get dy/dx by dividing the two derivatives above:

$\dfrac{dy}{dx} = \dfrac{-\frac{\cos^2(t)}{\sqrt{\cos^3(2t)}} \left(3 \sin(t) \cos(2t) - \cos(t) \sin(2t)\right)}{\frac{\sin^2(t)}{\sqrt{\cos^3(2t)}} \left(3 \cos(t) \cos(2t) + \sin(t) \sin(2t)\right)}$

$\dfrac{dy}{dx} = -\cot^2(t) \dfrac{3\sin(t)\cos(2t) - \cos(t)\sin(2t)}{3\cos(t)\cos(2t) + \sin(t)\sin(2t)}$

and by applying some trig identities,

$\dfrac{dy}{dx} = -\cot^2(t) \dfrac{\sin(3t) - 2\sin(t)}{\cos(3t) + 2\cos(t)}$

$\boxed{\dfrac{dy}{dx} = -\dfrac{\cos(3t)}{\sin(t) (2\cos(2t) + 1)}}$

Answer 2

Answer:

$\frac{dy}{dx}=-\cot (3 t)$

Very important note!

Both answer are correct, the thing is: the expression can be more simplified  as

$\sin(t)(2\cos(2t)+1) = \sin(3t)$

that will lead the the answer I just gave.

Step-by-step explanation:

The question is looking for $\dfrac{dy}{dx}$

We have

$x = \dfrac{\sin^3 (t)}{\sqrt{\cos(2t)}} \quad \text{ and } \quad y = \dfrac{\cos^3 (t)}{\sqrt{\cos(2t)}}$

One approach would be calculating $\dot{x}$ and $\dot{y}$ considering

$\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}$

We would have

$\frac{dx}{dt}=\frac{3 \sin ^2(t) \cos(t)}{\sqrt{\cos (2 t)}} + \frac{\sin (2 t) \sin ^3(t)}{\cos ^{3/2}(2 t)}$

$= \frac{\sin^{2} t}{\cos ^{3/2}(2 t)}(3\cos (t) \cos(2t)+2\sin^{2} (t) \cos(t))$

$=\boxed{ \frac{\sin ^2(t) (2 \cos (t)+\cos (3 t))}{\cos ^{3/2}(2 t)}}$

===============================================

$\frac{dy}{dt}=\frac{\sin (2 t) \cos ^3(t)}{\cos ^{\frac{3}{2}}(2 t)}-\frac{3 \sin (t) \cos ^2(t)}{\sqrt{\cos (2 t)}}$

$= \frac{\cos^{2} t}{(\cos ^{3/2}(2 t)} (-3\sin (t) \cos (2t)+2\cos^{2} (t) \sin (t))$

$=\boxed{ \frac{\sin (t)-\sin (5 t)}{4 \cos ^{3/2}(2 t)}}$

Finally,

$\frac{dy}{dx}= \frac{\sin (t)-\sin (5 t)}{4 \cos ^{3/2}(2 t)} \cdot \frac{\cos ^{3/2}(2t) }{ \sin ^2(t) (2 \cos (t)+\cos (3 t))}$

$=\frac{\cos(t)(2\cos^{2} t-3(2\cos^{2} t-1))} {\sin(t)(2\sin^{2} t+3(1-2\sin^{2} t))}$

$=-\frac{4\cos^{3} (t)-3\cos (t)}{3\sin (t)-4\sin^{3} (t)}$

$=-\frac{\cos (3t)}{\sin (3t)}$

$=\boxed{-\cot (3t)}$