Bucko
New Member
I got nothing when it comes to these. I know the basic stuff, but is there anything I'm missing, I'll do one question, then not be able to do the next. Please help!
Cheers
Cheers
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Rates of change... (1 Viewer)
- Thread starter Bucko
- Start date
what basic studd do you know? 
maybe some of your errors come from the exclusion of the constant?
say (basic question)..
cubical block of ice of edge 10cm melts so that its volume decreases by 25cm^3/s.
find when it will melt completely.
here you are given the rate of change (the derivative).
so you need to integrate it.
v' = -25 (decreases - pay close attention to the word of the question)
v = -25t + C (many people forget about the constant)
to find C, get other values.
so initially, at t = 0, v = 1000 (cubical, 10 cm edges .'. 10^3)
1000 = -25(0) + C
C = 1000
V = -25t + 1000
when v = 0 (thats when it melts completely)
0 = -25t + 1000
t = 40 secs.
this is a general question... it doesn't get TOO much harder than this. hope this is helpful - if not, i can make harder/easier spinoffs of the question...
maybe some of your errors come from the exclusion of the constant?
say (basic question)..
cubical block of ice of edge 10cm melts so that its volume decreases by 25cm^3/s.
find when it will melt completely.
here you are given the rate of change (the derivative).
so you need to integrate it.
v' = -25 (decreases - pay close attention to the word of the question)
v = -25t + C (many people forget about the constant)
to find C, get other values.
so initially, at t = 0, v = 1000 (cubical, 10 cm edges .'. 10^3)
1000 = -25(0) + C
C = 1000
V = -25t + 1000
when v = 0 (thats when it melts completely)
0 = -25t + 1000
t = 40 secs.
this is a general question... it doesn't get TOO much harder than this. hope this is helpful - if not, i can make harder/easier spinoffs of the question...
kazerati
Member
hey, slyball, thanx for that, ey!!
sometimes all ya need is someone different explainin the exact same thing in a slightly different way, n it all makes sense
sigh.. if only the hsc was as easy as the stuff in the 'basic' part of the revise in a month..
not to worry! the hsc is too much of a nervous breakdown waiting to happen
all the best everyone!
kazza
sometimes all ya need is someone different explainin the exact same thing in a slightly different way, n it all makes sense
sigh.. if only the hsc was as easy as the stuff in the 'basic' part of the revise in a month..
not to worry! the hsc is too much of a nervous breakdown waiting to happen
all the best everyone!
kazza
Bucko
New Member
yeah cheers mate, that's good
Can anyone please help me with this question?
Sand is tipped from a truck onto a pile. The rate R kg/s, at which the sand is flowing at R = 100t - t^3, for 0<t<T, where t is the time in seconds after the sands starts to flow.
i)What is the largest value of T for which expression for R is phsyically reasonable.
ii)When the sand start to flow, the pile already contains 300 kg of sand. Find expression for amount of sand in the pile at time t.
Lots of thanks!!
Note: it's from 1998 HSC paper by the way
Sand is tipped from a truck onto a pile. The rate R kg/s, at which the sand is flowing at R = 100t - t^3, for 0<t<T, where t is the time in seconds after the sands starts to flow.
i)What is the largest value of T for which expression for R is phsyically reasonable.
ii)When the sand start to flow, the pile already contains 300 kg of sand. Find expression for amount of sand in the pile at time t.
Lots of thanks!!
Note: it's from 1998 HSC paper by the way
N
ND
Guest
i) It's not possible to have a negative amount of sand. So R>=0.
ii) Think about it, you know the initial amount and the rate at which it's flowing in. Amount is the integral of rate.
ii) Think about it, you know the initial amount and the rate at which it's flowing in. Amount is the integral of rate.
