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Hey I need help with mathematics p3
november 2008 question 8 differential equation.
november 2008 question 8 differential equation.
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Math p3 help
- Thread starter usamah2702
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these are basically just general guidelines... if you want the whole step by step solution, i can do that too... especially the integration part and all... it's a lot to type out 
but i think these can suffice...
i) we know from the question that dV/dt = 20 and also dV/dt = -kh^2, so combining both, we have dV/dt = 20 - kh^2.
we also know V = (4/3)h^3, so dV/dh = 4h^2
you can find out dh/dt using the chain rule dV/dt = dV/dh x dh/dt
this will give you dh/dt = 5/(h^2) - k/4
substitute the value of dh/dt at h = 1 given in the question, which'll give you k= 0.2
substitute the value of k in the equation above, and tada!
ii) take the more complicated-looking side, in this case, RHS.
(10+h)(10-h)=100-h^2 [remember a^2 - b^2 = (a+b)(a-b)]
substitute that, and take the lcm, giving you (-2000 +20h^2 +2000)/(100-h^2) the 2000s cancel off to give the LHS. proven.
iii) we know from (i) that dh/dt = 5/h^2 - 1/20. if we take the lcm, it turns into (100 - h^2)/(20h^2)
thus we get [(20h^2)/(100 - h^2)] dh = dt
substitute the identity from (ii), then integrate both sides. this will give you -20h + 100 ln [(10+h)/(10-h)] + c = t
substitute h=0 when t=0 (given in question) to find c, which'll turn out to be 0.
thus you have t = -20h + 100 ln [(10+h)/(10-h)]
cheers!
but i think these can suffice...i) we know from the question that dV/dt = 20 and also dV/dt = -kh^2, so combining both, we have dV/dt = 20 - kh^2.
we also know V = (4/3)h^3, so dV/dh = 4h^2
you can find out dh/dt using the chain rule dV/dt = dV/dh x dh/dt
this will give you dh/dt = 5/(h^2) - k/4
substitute the value of dh/dt at h = 1 given in the question, which'll give you k= 0.2
substitute the value of k in the equation above, and tada!
ii) take the more complicated-looking side, in this case, RHS.
(10+h)(10-h)=100-h^2 [remember a^2 - b^2 = (a+b)(a-b)]
substitute that, and take the lcm, giving you (-2000 +20h^2 +2000)/(100-h^2) the 2000s cancel off to give the LHS. proven.
iii) we know from (i) that dh/dt = 5/h^2 - 1/20. if we take the lcm, it turns into (100 - h^2)/(20h^2)
thus we get [(20h^2)/(100 - h^2)] dh = dt
substitute the identity from (ii), then integrate both sides. this will give you -20h + 100 ln [(10+h)/(10-h)] + c = t
substitute h=0 when t=0 (given in question) to find c, which'll turn out to be 0.
thus you have t = -20h + 100 ln [(10+h)/(10-h)]
cheers!
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thankks!!
It helped a lot
It helped a lot