數學問題高手幫幫忙preCalculus

[匿名]

2010-05-06 7:13 pm

convert the polar equation into rectangular equation

6/(2cosx-3sinx)

教我convert拜託拜託

上面符號x應該要是theta

更新1:

my bad, the original problem was supposed to be r=6/(2cosx-3sinx) i understand ur concept, but can you check the way i did it since i got something really different from yours.

更新2:

i time (2cosx-3sinx) to both side r(2cos@-3sin@)=6 2rcos@-3rsin@=6 by factoring from polar coordinates x=rcos@ : y=rsin@ so (2x-3y)=6 (by substitution) thus 2x-3y-6=0 thank you so much !

更新3:

nvm i get it thank u

回答 (1)

Prof. Physics

2010-05-06 7:42 pm

✔ 最佳答案

From polar coordinates,

x = rcos@, y = rsin@

So, tan@ = y/x

And from trigo identity, cos@ = x/sqrt(x^2 + y^2)

sin@ = y/sqrt(x^2 + y^2)

So, the equation: 6/(2cos@ - 3sin@)

= 6/[2(x/sqrt(x^2 + y^2)) - 3(y/sqrt(x^2 + y^2))]

= 6[sqrt(x^2 + y^2)] / (2x - 3y)

參考: Physics king

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