The Student Room Group
Do you mean dV/dx there? :s-smilie:

If you mean dV/dt, separate the variables (and if necessary, substitute u = 2t + 1).
that should be dv/dt.
to do part i just integrate both sides of, vdv=1000(2t+1)^(-2) dt
ii) put t=5 into answer part(i) to get V then use V=4/3pi(r)^3 to find r
iii) put t=5 and your answer to part(ii) into your expression for dr/dt which you should have as 1000/[4pi r^2(2t+1)^(2)]
Reply 3
see attached (may be useful?)
Reply 4
syrinx41
question 7 - connected rates of change.

(c) Given that V = 0 when t = 0, solve the differential equation:

dV/dx=1000/(2t+1)2dV/dx = 1000/(2t+1)^2 to obtain V in terms of t.

(d) hence, at t = 5,

(i) find the radius of the ballon, answer to 3 sig fig.

(ii) show that the rate of increase of the radius of the balloon is approximately;

Unparseable latex formula:

2.90X10^-^2 cms^-^1



HELP.



Use a substitution of u=2t+1 for that integral.. i remmeber it came out nicely
Reply 5
evariste
that should be dv/dt.
to do part i just integrate both sides of, vdv=1000(2t+1)^(-2) dt
ii) put t=5 into answer part(i) to get V then use V=4/3pi(r)^3 to find r
iii) put t=5 and your answer to part(ii) into your expression for dr/dt which you should have as 1000/[4pi r^2(2t+1)^(2)]


I did this question today and this is pretty much what I did to get the right answers (although it's not just a case of integrating for part C), you have to rearrange to find C, but that's not hard). I guess you could substitute u=2t+1, but I can't see why it'd make it any easier to do.