# Thread: help with parametric differentiation

1. ## help with parametric differentiation

Seeking help with finishing this problem, from line 6 from bottom (with ???).

Find dy/dx and d^2y/dx^2 for,

y = ln sin^4 t and x = ln cos^8 t

y = ln sin^4 t
dy/dt = 1/sin^4 t X (cos^4 t)
= cos^4 t/sin^4 t
= cot^4 t

x = ln cos^8 t
dx/dt = 1/cos^8 t X (-sin^8 t)
= -sin^8 t/cos^8 t
= -tan^8 t

dy/dx = dy/dt X dt/dx
= cot^4 t X -1/tan^8 t
= cot^4 t X (-cot^8 t)
= -(cot^4 t X cot^8 t)

d^2y/dx^2 = d(dy/dx)/dx
= d(-cot^4 t X cot^8 t)/dx
dy/dt = -[-cot^4 t.(cosec^2)^8 t] + [cot^8 t.cosec^2)^4 t]
= [cot^4 t.cosec^16 t] + [cot^8 t.cosec^8 t]

= cot^4 t.cosec^8 t(cosec^2 t + cot^2 t) ???

d^2y/dx^2 = dy/dt X dt/dx

= [cot^4 t.cosec^8 t(cosec^2 t + cot^2 t)] X -1/tan^8 t
= [cot^4 t.cosec^8 t(cosec^2 t + cot^2 t)] X -cot^8 t

2. Originally Posted by ashura
Seeking help with finishing this problem, from line 6 from bottom (with ???).

Find dy/dx and d^2y/dx^2 for,

y = ln sin^4 t and x = ln cos^8 t

y = ln sin^4 t
dy/dt = 1/sin^4 t X (cos^4 t)
= cos^4 t/sin^4 t
= cot^4 t

x = ln cos^8 t
dx/dt = 1/cos^8 t X (-sin^8 t)
= -sin^8 t/cos^8 t
= -tan^8 t

dy/dx = dy/dt X dt/dx
= cot^4 t X -1/tan^8 t
= cot^4 t X (-cot^8 t)
= -(cot^4 t X cot^8 t)

d^2y/dx^2 = d(dy/dx)/dx
= d(-cot^4 t X cot^8 t)/dx
dy/dt = -[-cot^4 t.(cosec^2)^8 t] + [cot^8 t.cosec^2)^4 t]
= [cot^4 t.cosec^16 t] + [cot^8 t.cosec^8 t]

= cot^4 t.cosec^8 t(cosec^2 t + cot^2 t) ???

d^2y/dx^2 = dy/dt X dt/dx

= [cot^4 t.cosec^8 t(cosec^2 t + cot^2 t)] X -1/tan^8 t
= [cot^4 t.cosec^8 t(cosec^2 t + cot^2 t)] X -cot^8 t

First off, your first derivatives are incorrect. You didn't use the chain rule properly.

$\displaystyle x(t) = ln \left ( cos^8(t) \right )$
$\displaystyle y(t) = ln \left ( sin^4(t) \right )$

(I will do it your way, but note that: $\displaystyle ln \left ( cos^8(t) \right ) = 8 ln(cos(t))$, which makes the derivative easier to think about.)

$\displaystyle \frac{dx}{dt} = \frac{1}{cos^8(t)} \cdot (8cos^7(t)) \cdot (-sin(t)) = \frac{-8sin(t)}{cos(t)} = -8tan(t)$

$\displaystyle \frac{dy}{dt} = \frac{1}{sin^4(t)} \cdot (4sin^3(t)) \cdot (cos(t)) = \frac{4cos(t)}{sin(t)} = \frac{4}{tan(t)}$

$\displaystyle \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} } = \frac{\frac{4}{tan(t)}}{-8 tan(t)} = -\frac{1}{2tan^2(t)}$

There are two ways to proceed from here. I would suspect that the standard thing to do would be two write dy/dx as a function of x, rather than leaving it a function of t. But it depends on what you are shooting for. I would do it that way simply because its a bit shorter to take the next derivative. So we have:
$\displaystyle x(t) = ln \left ( cos^8(t) \right )$

$\displaystyle e^x = cos^8(t)$

$\displaystyle \pm \left ( e^x \right ) ^{1/8} = cos(t)$

$\displaystyle t = acs \left ( \pm \left ( e^{x/8} \right ) \right )$

So
$\displaystyle \frac{dy}{dx} = -\frac{1}{2tan^2 \left ( acs \left ( \pm \left ( e^{x/8} \right ) \right ) \right )}$
and go from there.

Otherwise I believe your method is correct, simply that your derivatives weren't correct.

-Dan

3. Actually I think the derivative should be,
$\displaystyle y=\ln |\cos^8 x|$
Then you can write,
$\displaystyle y=8\ln |\cos x|$
However, when you have parantheses instead of absolute value then,
$\displaystyle y=\ln (\cos^8 x)=8\ln| \cos x|$

Sorry, topsquark. I hope all my formal corrections do not bother ye.

4. Originally Posted by ThePerfectHacker
Actually I think the derivative should be,
$\displaystyle y=\ln |\cos^8 x|$
Then you can write,
$\displaystyle y=8\ln |\cos x|$
However, when you have parantheses instead of absolute value then,
$\displaystyle y=\ln (\cos^8 x)=8\ln| \cos x|$

Sorry, topsquark. I hope all my formal corrections do not bother ye.
Oops. I missed that point. Ah well.

I'm never bothered by corrections. As long as the corrections are correct.

-Dan

5. Here is a direct way to the second derivative giving the answer as a function of t.
$\displaystyle \frac{{d^2 y}}{{dx^2 }} = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$

6. Thanks for the correction of function of a function.
I'm getting dy/dx = -1/2tan^2(t) and not -2tan^2(t) from your working. Could you review this.

dx/dt = -8tan(t)
dy/dt = 4/tan(t)

dy/dx = dy/dt X dt/dx
= 4/tan(t) X -1/8tan(t)
= -1/2tan^2(t)

7. Originally Posted by ashura
Thanks for the correction of function of a function.
I'm getting dy/dx = -1/2tan^2(t) and not -2tan^2(t) from your working. Could you review this.

dx/dt = -8tan(t)
dy/dt = 4/tan(t)

dy/dx = dy/dt X dt/dx
= 4/tan(t) X -1/8tan(t)
= -1/2tan^2(t)
I have corrected this in my original post. Thanks for the catch and sorry about that!

-Dan