Y'all wanna see a witch?
Yes
Ganeshie > Tis witch
It actually is.
Did you see the image?
polar curve of the italian witch , the only curve was discoverd by a woman , cool i love it also wait ill uploud Gif to how its work
I though it was an actually witch. I got scared for a sec there.
Smh lol.
@ikram002p explained it.
This is called Witch of Agnesi.
\(\large y = e^{-x^2/2}\)
It is actually cartesian. It depends upon two variables: a and b.
well , it have charastaristic equation but still one of famouse 15 special curves i guess
looks ikram did phd in polar curves :o
When a=.5 the witch becomes \( {1 \over {1+x^2}}\) better known as the derivative of arctan x.
The general equation \[ \Large y = {{(2a)^3} \over {x^2 + (2a)^2}} \] is the way the witch is usually given.
Is the derivation easy ? Lres, in your svg file, W should be \(\large(b, a+\sin\theta)\) right ?
@ganeshie8 Yes. This will only take a second to patch up.
i hope so, but the derivation looks hard to me still... we need to find the locus of W hmm
u wanna derevative of y ?
its like 1/x^2 formulla mm
Revised label
sorry i don't mean derivative/differentiation.. i mean, find the equation of curve using the given constraints...
ohh u wanna construct the equation from the curve itself
well note how the change of x,y according to a,b and theta then convert to one line equation
the gif file well help you to note how is that happening
x = b b = OB^2 - (2a)^2
ok now find OB , u wanna convert to a sin theta ?
OB = OS + SB OS = \(\large \sqrt{a^2 + a^2 - 2(a)(a)\cos (90-\theta)}\) SB = \(\large \sqrt{a^2 + a^2 - 2(a)(a)\cos (\theta)}\)
ima google this :) thats better lol
haha xD
these things woud be easy to follow the definition mm
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