Question

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 ?

Answer 1

polar coordinates eh...

Answer 2

let me rewrite the question

Answer 3

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 \[\theta=\pi/4\] ?

Answer 4

the normal integral for polar regions is:

(1/2) [S] [f(t)]^2 dt ; [a,b]

Answer 5

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

Answer 6

this answer involve topic rate of change under diiferentation

Answer 7

A=(9/6).(4t-sin(4t)

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

Answer 8

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

Answer 9

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

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

Answer 10

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

Answer 11

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?

Answer 12

sadly no..
anyway thanks

Answer 13

i wanna say A' = 1.2 rads/min

Answer 14

at t=pi/4