Question

Solve for dy/dx

Answer 1

\[\LARGE x=\frac{\sin^3t}{\sqrt{\cos2t}}\]
\[\LARGE y=\frac{\cos^3t}{\sqrt{\cos2t}}\]

Answer 2

@hartnn

Answer 3

@amistre64 @terenzreignz @zepdrix

Answer 4

Found it... it's on your original post :D

Answer 5

:O

Answer 6

But seriously...
\[\Large \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]

Answer 7

#inb4youtellMeToSolveItDirectly

Answer 8

that is gonna be a heptic calculation,can we do something to simplify things first?

Answer 9

you are never going to answer if you chose to solve it directly

Answer 10

\[\frac{dy}{dt}=\large (\sqrt{\cos2t} \times \sin^2t \times cost)-((\sin^3t) \times \frac{1}{\sqrt{\cos2t}} \times 2\sin2t)\]

Answer 11

we are gonna land no where
how about something else?

Answer 12

That's somewhere all right, unless you wanted dy/dx in terms of x?

Answer 13

how about squaring and adding them?

Answer 14

somehow I think I'd rather subtract.
\[\large y^2 - x^2 = \frac{\cos^6t-\sin^6t}{\cos(2t)}\]

Answer 15

adding leads to a dead end that I don't know how to factor...

Answer 16

alright :O

Answer 17

So, this is a difference of two cubes...
\[\Large = \frac{[\cos^2(t) - \sin^2(t)][\cos^4(t)+2\sin^2(t)\cos^2(t)+\sin^4(t)]}{\cos(2t)}\]

Answer 18

Does this really help?
The best I could hope for is to cancel out the denominator..
\[\Large = \cos^4(t)+2\sin^2(t)\cos^2(t) + \sin^4(t) \]

Answer 19

hmm..should be easy now

Answer 20

yeah, but that's y^2 - x^2

Answer 21

yeah i know

Answer 22

just tighten your belt and do it directly :D

Answer 23

lol

Answer 24

would you do this directly too? :O
\[Prove ~That: \frac{d}{dx}(\frac{1}{4 \sqrt2}\log(\frac{x^2+\sqrt2x+1}{x^2-\sqrt2x+1}(+\frac{1}{2\sqrt2}\tan^{-1}(\frac{\sqrt2x}{1-x^2})\]
=1/1+x^4

Answer 25

If all else fails, maybe :D

Answer 26

hmm \m/

Answer 27

very brave of you,master!

Answer 28

brave? No
desperate? maybe...