1. ## parametric equation

could any one help me solving this parametric equation.

1, if y=cos^3 t and x=sin^3t calculate d^2 y/dx^2
2,the parametric equation of a function are given as x=3sin2t and y= 1-e^cos2t calculate dy/dx and d^2 y/dx^2
3, partial differetiation- determine d^2 z/dxdy if z=3xy^2+5x^2 y^4-y^5
4, if v=2loge^(4y-3x) calculate d^2 v/dxdy
5,if u=e^-x(xcosy) determine the value of
i,du/dx
ii,d^2 u/dydx

could any one help me solving this parametric equation.

1, if y=cos^3 t and x=sin^3t calculate d^2 y/dx^2
$\displaystyle \frac{dy}{dt}=-3\sin t \cos^2 t=-3\sin t \cos t\cos t=-\frac{3}{2}\sin 2t \cos t$
$\displaystyle \frac{dx}{dx}=3\cos t \sin^2 t= 3\cos t\sin t \sin t=\frac{3}{2}\sin t \sin 2t$

Thus,
$\displaystyle f(t)=\frac{dy}{dx}=\frac{-\frac{3}{2}\sin 2t \cos t}{\frac{3}{2}\sin t \sin 2t}=-\cot t$

Then,
$\displaystyle f'(t)=\csc^2 t$
Then,
$\displaystyle \frac{d^2 y}{dx^2}= \frac{3\csc^2 t}{2\sin 2t \sin t}$

2,the parametric equation of a function are given as x=3sin2t and y= 1-e^cos2t calculate dy/dx and d^2 y/dx^2
$\displaystyle x=3\sin 2t$
$\displaystyle y=1-e^{\cos 2t}$

Thus,
$\displaystyle x'_t=6\cos 2t$
$\displaystyle y'_t=2\sin 2t e^{\cos 2t}$

Thus,
$\displaystyle f(t)=\frac{dy}{dx}=\frac{2\sin 2t e^{\cos 2t}}{6\cos 2t}=\frac{1}{3}\tan 2t \cdot e^{\cos 2t}$

Thus, (product rule),
$\displaystyle f'(t)=\frac{2}{3}\sec^2 2t\cdot e^{\cos 2t} - \frac{2}{3}\tan 2t \sin 2t\cdot e^{\cos 2t}$
Divide by $\displaystyle x'_t$,
$\displaystyle \frac{d^2y}{dx^2}=\frac{1}{9}\sec^3 2t \cdot e^{\cos 2t}- \frac{1}{9} \tan^2 2t \cdot e^{\cos 2t}$

3, partial differetiation- determine d^2 z/dxdy if z=3xy^2+5x^2 y^4-y^5
Since $\displaystyle z\,$ is continous everywhere you can take the partials in any order.

$\displaystyle z_x=3y^2+10x y^4$
$\displaystyle z_{xy}=6y+40xy^3$

4, if v=2loge^(4y-3x) calculate d^2 v/dxdy
Same thing,
$\displaystyle v=2\ln (4y-3x)$
$\displaystyle v_x=\frac{-6}{4y-3x}$
$\displaystyle v_{xy}=\frac{24}{(4y-3x)^2}$

We have two formulas for parametric differentiation.
. . The second one is commonly misunderstood.

. . . $\displaystyle (1)\;\frac{dy}{dx} \;=\;\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \qquad\qquad (2)\;\frac{d^2y}{dx^2} \;=\;\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)} {\frac{dx}{dt}}$

For the second derivative:
. . (a) Differentiate the result in part (1).
. . (b) Divide by another $\displaystyle \frac{dx}{dt}$

$\displaystyle y \:=\:\cos^3t,\;x = \sin^3t$

We have:
. . $\displaystyle \frac{dy}{dt} \:=\:\text{-}3\cos^2t\sin t$
. . $\displaystyle \frac{dx}{dt} \:=\:3\sin^2t\cos t$

Therefore: .$\displaystyle \frac{dy}{dx} \:=\:\frac{\text{-}3\cos^2t\sin t}{3\sin^2t\cos t} \:=\:-\frac{\cos t}{\sin t} \:=\:-\cot t$

We have: .$\displaystyle \frac{d}{dt}\left(\frac{dy}{dx}\right) \:=\:\csc^2x$

Therefore: .$\displaystyle \frac{d^2y}{dx^2} \:=\:\frac{\csc^2x}{3\sin^2x\cos t} \:=\:\frac{1}{3}\csc^4t\sec t$

### parametric differetiation

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