#I need help to solve the full answer. also D is supposed to be. infinity or mostly a number that's

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thin plinthBOT
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quasi meadow
polar spade
# quasi meadow

doubt this is the intended way of solving, could you send the original problem statement

quasi meadow
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D is an input

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0 to infinity

quasi meadow
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Ok

polar spade
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or are you just writing random expressions for fun and trying to simplify them

quasi meadow
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I am mostly just trying to find the perimeter of the function

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Let me get jt

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Here

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This is the function that I'm trying to solve

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Only this frame of this perimeter

polar spade
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you can rephrase this to find length of the curve from 0 to D

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(1-t)cos(pi/2 t), (1-t)sin(pi/2 t)

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try polar coordinates

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do you know polar arc length formula

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@quasi meadow

quasi meadow
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A bit

polar spade
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try that

quasi meadow
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Ok

quasi meadow
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Thank you

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r=2theta/pi

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1.32365235051302140488742854200618380945320639547052945195151453540264931019607717540563165049421959222360706687376958640359897031961700216892658733309887800025703841088547659708277308425119733267516388677439029216714145919008637352378272646640242605878878861430930629695216288154541546460064980080304439781320967228504590648704472231659041876566135158811504830259956007648126390413324909915327469002750211528830924336694072341066857003717243435617726193242859190429161806494237110326610954167199422637919875388962449426466716040913753503356988494975546588710291463890410695201052219496452727514188296428718959791750955820430828307524855940001291835998186114366224393192480556210201285803530902621991798199759811096453822270740312539939355147737844385605829530996150186514875516429989883691748677827519704238190977693454643336935044492414312473998436266773513714574153808515212274203751349587577635540043552346293189290849009393932504724920310256076750507994549016094009293325960453200991662762428547632272418361060647148823084270351666701...

polar spade
quasi meadow
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sorry for not putting context , I used polar arc length formula

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On r=2theta/pi

polar spade
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still unsure why its not r = 1 - 2theta/π

quasi meadow
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its both

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your correct

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when finding the same shape you get the same answer

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its just mine is up at the y and yours was at the x

polar spade
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if D =/= 1 the shapes are different