anonymous 5 years ago If Theta is the acute angle between al line through the origin and the positive x-axis,the area in cm^2, of part of a rose petal is A=(9/6).(4Theta-sin(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 ?

1. amistre64

polar coordinates eh...

2. anonymous

let me rewrite the question

3. anonymous

If $\theta$ is the acute angle between al line through the origin and the positive x-axis,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 $\theta=\pi/4$ ?

4. amistre64

the normal integral for polar regions is: (1/2) [S] [f(t)]^2 dt ; [a,b]

5. amistre64

but you seems to want derivatives..... so thats no good to you lol

6. anonymous

this answer involve topic rate of change under diiferentation

7. amistre64

A=(9/6).(4t-sin(4t) A' = 0(4t-...) + (9/6) D(4t-sin(4t))

8. amistre64

A' = (9/6).(-4cos(4t)) if i see it right

9. amistre64

A' = (9/6).(0t' -4cos(4t)) A' = (9/6).(-4cos(4t))

10. amistre64

that dt=.2 is throwing me; not sure how it applies in here...

11. amistre64

maybe 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?

12. anonymous

13. amistre64

i wanna say A' = 1.2 rads/min

14. amistre64

at t=pi/4