https://isaacphysics.org/questions/maths_ch7_3_q6?board=a8bb5f89-49e4-4045-abe0-d0c9070dbc69&stage=a_level

For part c of this question, I dont really understand how to set up this equation.

I initially said dV/dt = ap-r

so dh/dt = a*rho*g*h-r/A

h = a(e^bt - 1) - since when t = 0, h=0 and when t=T, h=h_eq and since the derivative of the function includes the original function

then got h = -r/(a*rho*h) * [e^(a*rho*g/A *t) -1]

But im really not sure what to do from here or if its even correct.

Any help greatly appreciated.

For part c of this question, I dont really understand how to set up this equation.

I initially said dV/dt = ap-r

so dh/dt = a*rho*g*h-r/A

h = a(e^bt - 1) - since when t = 0, h=0 and when t=T, h=h_eq and since the derivative of the function includes the original function

then got h = -r/(a*rho*h) * [e^(a*rho*g/A *t) -1]

But im really not sure what to do from here or if its even correct.

Any help greatly appreciated.

Original post by mosaurlodon

https://isaacphysics.org/questions/maths_ch7_3_q6?board=a8bb5f89-49e4-4045-abe0-d0c9070dbc69&stage=a_level

For part c of this question, I dont really understand how to set up this equation.

I initially said dV/dt = ap-r

so dh/dt = a*rho*g*h-r/A

h = a(e^bt - 1) - since when t = 0, h=0 and when t=T, h=h_eq and since the derivative of the function includes the original function

then got h = -r/(a*rho*h) * [e^(a*rho*g/A *t) -1]

But im really not sure what to do from here or if its even correct.

Any help greatly appreciated.

For part c of this question, I dont really understand how to set up this equation.

I initially said dV/dt = ap-r

so dh/dt = a*rho*g*h-r/A

h = a(e^bt - 1) - since when t = 0, h=0 and when t=T, h=h_eq and since the derivative of the function includes the original function

then got h = -r/(a*rho*h) * [e^(a*rho*g/A *t) -1]

But im really not sure what to do from here or if its even correct.

Any help greatly appreciated.

Youre along the right lines, but height is positive up so r will increase the height/volume so youd have

dV/dt = r - a*rho*g*h

and V = Ah so

dh/dt = (r - a*rho*g*h)/A

Then its similar to what youve done but get the signs etc right and make sure the sign of the exponential is correct, does it grow/decay with time?

Note that even though you work out the h_eq expression in part b) they seem to want you to just put in "h_eq" rather than the expression. No idea why this is the case.

(edited 3 months ago)

mqb2766

Note that even though you work out the h_eq expression in part b) they seem to want you to just put in "h_eq" rather than the expression. No idea why this is the case.

I've not even looked at the variables in detail, but I would expect a transform $y = h - h_{eq}$ causes a fair amount of stuff to cancel.

Ok so I reworked it out and I got the same thing as before except the exponent for e^ is now negative so

h = r/(a*rho*h) * [1 - e^(-a*rho*g/A *t)]

which I think makes sense as when t increases the e part gets smaller until h=r/(a*rho*h) which is what I got for part b as h_eq

Ohh I see what you mean by the h_eq part now.

Thank you very much I managed to get the right answer and complete the last part as well.

h = r/(a*rho*h) * [1 - e^(-a*rho*g/A *t)]

which I think makes sense as when t increases the e part gets smaller until h=r/(a*rho*h) which is what I got for part b as h_eq

Ohh I see what you mean by the h_eq part now.

Thank you very much I managed to get the right answer and complete the last part as well.

Original post by DFranklin

Probably to encouraging writing the answer in a form highlighting the difference between the current height and the equilibrium.

I've not even looked at the variables in detail, but I would expect a transform $y = h - h_{eq}$ causes a fair amount of stuff to cancel.

I've not even looked at the variables in detail, but I would expect a transform $y = h - h_{eq}$ causes a fair amount of stuff to cancel.

Dont think so, they want

h_eq(1-e^#)

rather than

expression(1-e^#)

It seems weird as part b) is about working out that expression and its just "incorrect" (as usual) there isnt even a hint to say write in an equivalent way.

Original post by mqb2766

Dont think so, they want

h_eq(1-e^#)

rather than

expression(1-e^#)

It seems weird as part b) is about working out that expression and its just "incorrect" (as usual) there isnt even a hint to say write in an equivalent way.

h_eq(1-e^#)

rather than

expression(1-e^#)

It seems weird as part b) is about working out that expression and its just "incorrect" (as usual) there isnt even a hint to say write in an equivalent way.

I mean to me, the form they want is the most "natural", but obviously mileage may vary. I *have* seen IsaacPhysics give hints about what form of answer is expected when you give something "correct but not the right form"; it would certainly help if they'd done so here.

Original post by DFranklin

I mean to me, the form they want is the most "natural", but obviously mileage may vary. I *have* seen IsaacPhysics give hints about what form of answer is expected when you give something "correct but not the right form"; it would certainly help if they'd done so here.

I do understand what you mean and its not really a big deal. As you say though, a hint could have helped or a bit of feedback from the solution checker that they were looking for a different form. Guess its the usual frustrations with isaac.

Well fortunately you wont have to deal with many of them anymore since the deadline ended about 16 minutes ago

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