cover up method of doing partial fraction decomposition. its just a simplified way of looking at a pretty normal way of solving for the constants in the numerator (you multiply by the terms of the denominator on both sides of the equation, select values for the unknown variable which would make a particular constant=0). you can work out the math for yourself by solving it algebraically, you'll find the method is doingn exactly that blackpenredpen has a nice video on the cover up method for partial fraction decomposition
So, for part (c), one can disregard the partial fractions bit in (b), and directly integrate dV/dt = − kV to get V/V0 = exp[ −k(t−t0) ], and then substitute for V in terms of h at the end, where h0 = 5 and t0 = 0: [ ⅓πh²(15−h) ] / [ ⅓π5²(15−5) ] = exp(−kt). The ⅓π cancels and this directly gives the result requested. Would this be accepted to get the marks, where you take the "smart" option and disregard the suggested path? (In earlier years, exam papers would have the text "using this result or otherwise, show that...". I notice recent papers do not seem to include such text.)
Yes, this would be accepted. Only if it says "hence", "deduce that", or "using ..." (or similar wording) do you have to use previous results - otherwise, you're free to use any valid approach.
I have to say that this question is such a gift. It's something nice and juicy that is essentially straightforward.
it always works out in the end
4:55 no K would be 3 time smaller right?
No theirs would be 3 times smaller not his
bro did not cut the bit
hokia boogin
I don't understand what he meant when he said, "h-15 = 0," then implied that on the numerator, and made h = 0 in the denominator. Can someone explain?
If you're talking about 8:08 onwards, it's a technique called the "cover-up rule" - you can just Google it and you'll see many explanations of it.
cover up method of doing partial fraction decomposition. its just a simplified way of looking at a pretty normal way of solving for the constants in the numerator (you multiply by the terms of the denominator on both sides of the equation, select values for the unknown variable which would make a particular constant=0). you can work out the math for yourself by solving it algebraically, you'll find the method is doingn exactly that
blackpenredpen has a nice video on the cover up method for partial fraction decomposition
So, for part (c), one can disregard the partial fractions bit in (b), and directly integrate dV/dt = − kV to get V/V0 = exp[ −k(t−t0) ], and then substitute for V in terms of h at the end, where h0 = 5 and t0 = 0: [ ⅓πh²(15−h) ] / [ ⅓π5²(15−5) ] = exp(−kt). The ⅓π cancels and this directly gives the result requested. Would this be accepted to get the marks, where you take the "smart" option and disregard the suggested path? (In earlier years, exam papers would have the text "using this result or otherwise, show that...". I notice recent papers do not seem to include such text.)
bro why are you back here yapping
Shush
yh you are smater than the wasteman examiner so it will work ez
Yes, this would be accepted. Only if it says "hence", "deduce that", or "using ..." (or similar wording) do you have to use previous results - otherwise, you're free to use any valid approach.
@@amritlohia8240 Thanks for the reply.
easy peasy squeeze the 🍋