Question

the angle θ between two radii OP and OQ of a circle of radius 6 cm is increasing at the rate of 0.1 radians per minute.

(a) Show that the area of sector OPQ is increasing at the rate of 1.8cm^2/min.

(b) Find the rate at which the area of ∆OPQ is increasing at the instant when θ= π/4.

(c) Find the value of θ for which the rate of increase of the area of the segment cut off by the cord PQ is at its maximum

Answer 1

ok fine. what is the area of the sector?

Answer 2

i have no idea thats all the info we get

Answer 3

no i mean what is the formula for an area of a sector given the angle theta and the radius r ?

Answer 4

k it is
\[A=\frac{\theta}{2}r^2\]

Answer 5

1/2 x θ x r^2

Answer 6

yep

Answer 7

here r = 6 and
\[\frac{d\theta}{dt}=0.1\]

Answer 8

you want
\[\frac{dA}{dt}\] which is
\[\frac{dA}{dr}\times \frac{dr}{dt}\]

Answer 9

oh did i mess that up. forget the latex for a second

Answer 10

lol ok

Answer 11

you want
\[\frac{dA}{dt}=\frac{dA}{d\theta}\frac{d\theta}{dt}\]

Answer 12

so here
\[A=18r^2\] since r = 6, r^2= 36 and 36/2=18

Answer 13

damn it i must be tired. sorry

Answer 14

\[A=18\theta\] for the above reasons

Answer 15

oh isee

Answer 16

\[\frac{dA}{d\theta}=18\]

Answer 17

and
\[\frac{d\theta}{dt}=.1\] because that is what you were told

Answer 18

oh i get that now

Answer 19

and that finally is the answer, but only to #1

Answer 20

so for the next one find the area of a triangle , then diffrentiate and sub in θ= pi/4 to find the rate?

Answer 21

i use 1/2 ab sin C right

Answer 22

where c = θ

Answer 23

area of triangle is oh yes, i was looking but of course you do not need a formua it is \[\frac{1}{2} 6^2 sin(\theta)\] yes?

Answer 24

yep

Answer 25

which i guess is
\[18sin(\theta)\] in this case

Answer 26

then it becomes 18 cos θ

Answer 27

great, then take the derivative, and then multiply by .1 and then plug in \[\frac{\pi}{4}\]

Answer 28

oh k cool got that one sorted now for the last one

Answer 29

i am guessing \[.9\sqrt{2}\]

Answer 30

now for the next one i am going to guess
\[\frac{\pi}{2}\]

Answer 31

\[9\sqrt{2}/10 cm^2/\min\]

Answer 32

yes that is what i got too, only i wrote .9 but great!

Answer 33

oh cool

Answer 34

now we have the formula for the area and it is
\[18\sin(\theta)\]

Answer 35

do you need to find the second derivate of the area of a segment and find its maximum

Answer 36

the rate of change is
\[.18\cos(\theta)\]

Answer 37

right you are i misread the question. you want the rate to be biggest, not the change

Answer 38

so take the second derivative set = 0 and solve yes? what you said

Answer 39

yay cool got it

Answer 40

btw the rate of change is \[1.8\cos(\theta)\] not what i wrote earlier

Answer 41

great. just about to it in your head. derivative is -sin(x) and it is 0 at 0 and pi

Answer 42

not too bad this one. i looked harder at the beginning. good luck

Answer 43

kk thanks heap btw impressive medal count :)