What did you guys get on 1d
What did you guys get for 4D)?
What did you guys get for 2D?
30+ {{{{{integral from t=0 to t=w of R(t)-D(t) dt}}}}}=50
@zach1198 – I don’t know if I’m right, but I got that both m and b could = 1.
Does anyone know how to do 6b??
@Kylemcgrogan If I used time=8 but did the amount of water in the tube at t=8 instead of t=0, will i get credit still?
Ok, i got 30+ {{{{{integral from t=0 to t=8 of R(t)-D(t) dt}}}}} + {{{{{{{ integral from t=8 to t=w of R(t)-D(t)}}}}}}}=50
So thats pretty much the same thing right?
@eyrar99 I got that too I believe, but I don’t know if it is correct either. My friend got some totally different thing with X’s and stuff. Not sure.
@APScholar18 – 2D???
@eyrar99 there was no 2d
Im not sure, i read it as a new condition that R(t) and D(t) describe the volume rates of change after t=8, since it said in the description of the problem they only work for 0<t<8
Sorry 2C and 5D
I think i remeber them saying that the rates are still constant when t>8
@Kylemcgrogan I got the same thing for 1D.
@Namoree I think that would work…
@Kylemcgrogan did you also to 8 to w for limits of integration?
@Namoree @Kylemcgrogan it says that “for t>8, water continues to flow in and out of the pipe at the given rates until the pipe begins to overflow”
For 2c, i think you take the derivative of f(x) - g(x) at x=3, and you get -74.044 . I could be wrong. I don’t remember what i did on the exam.
@APScholar18 is that what you got?
f(-2)+[[[[integral from -2 to 1 of f’)= f(3)=1, rearrange to solve for f(-2)
f(1)+ [[[[integral from 1 to 4 of f’]]]]=f(4)
@Kylemcgrogan what question is that for?