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.)
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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
bro did not cut the bit
4:55 no K would be 3 time smaller right?
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?
easy peasy squeeze the 🍋