this was in my schools trial paper last year for 2u. its q10a. how come we got the same max volume of 4 but different times?dv/dt=root3sin t - cos t
when t=0, dv/dt=-1 so its emptying
uhh i dont think transformations is 2unit...but
V=2-2sin(t+pi/6) [i think]
so its max. at t=4pi/3
and thats V=4.
ah i see, dont worry about what i didthis was in my schools trial paper last year for 2u. its q10a. how come we got the same max volume of 4 but different times?
i differentiated volume then let it equal 0. there are 2 answers pi/6 and 7pi/6 pi/6 gives v=0, while the other one gives v=4 so ones obviously max and ones min. wats taht transformations stuff ur talkin bout?
k lol i got another 2 questions but ill post them up later after i have another crack at it. they are from the same paperah i see, dont worry about what i did
dv/dt=root3sin t - cos t
dv/dt=0 when root3sint-cost=0
root3tan t=1
tan t=1/root3
t=pi/6, 7pi/6,13pi/6....
now we have to find the max.
we can differentiate again.
v''=root3cos t+sin t
subbing in pi/6, v''>0 therefore its a minimum
subbing in 7pi/6, v''<0 therefore its a max.
so i guess u got it right, i have no idea what i did