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anonymous
 5 years ago
If Theta is the acute angle between al line through the origin and the positive xaxis,the area in cm^2, of part of a rose petal is
A=(9/6).(4Thetasin(4Theta))
If the angle Theta is increasing at a rate of 0.2 radians per minute, at what rate is the area changing when Theta=pi/4 ?
anonymous
 5 years ago
If Theta is the acute angle between al line through the origin and the positive xaxis,the area in cm^2, of part of a rose petal is A=(9/6).(4Thetasin(4Theta)) If the angle Theta is increasing at a rate of 0.2 radians per minute, at what rate is the area changing when Theta=pi/4 ?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2polar coordinates eh...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let me rewrite the question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If \[\theta\] is the acute angle between al line through the origin and the positive xaxis,the area in \[cm^{2}\], of part of a rose petal is \[A=(9/16).(4\theta\sin(4\theta))\] If the angle \[\theta\] is increasing at a rate of 0.2 radians per minute, at what rate is the area changing when \[\theta=\pi/4\] ?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2the normal integral for polar regions is: (1/2) [S] [f(t)]^2 dt ; [a,b]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2but you seems to want derivatives..... so thats no good to you lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this answer involve topic rate of change under diiferentation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2A=(9/6).(4tsin(4t) A' = 0(4t...) + (9/6) D(4tsin(4t))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2A' = (9/6).(4cos(4t)) if i see it right

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2A' = (9/6).(0t' 4cos(4t)) A' = (9/6).(4cos(4t))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2that dt=.2 is throwing me; not sure how it applies in here...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2maybe its my choice of t as a substitute for theta.... A' = (9/6).(4cos(4t)) (.2) is what i wanna say is correct.... do you have an answer sheet to go by?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sadly no.. anyway thanks

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.2i wanna say A' = 1.2 rads/min
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