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Wire of Varying Area

For this q, https://isaacphysics.org/questions/wire_of_varying_area_sym?board=f289dda7-5340-4dd8-9796-66fb6ebd06bd&stage=a_level

I tried using hint 4 to try and work out the resistance of a small length of the wire and adding these little bits up by integrating but im stuck on how to continue from here - is there a way to integrate this or have I set up the wrong equations.
As always, thank you for your time helping.
Reply 1
Working:

(as for notation - i used x as an increase in the length of the wire)
(edited 2 months ago)
Reply 2
Original post by mosaurlodon
Working:

(as for notation - i used x as an increase in the length of the wire)
The resistor slices are in series so the resistance dR of a slice of width dx would be
dR = (rho / (pi r^2)) dx
and the radius varies linearly so assuming x goes from 0 to l, then
r = sqrt(A1/pi)*(l-x)/l + sqrt(A2/pi)x/l
so
dr = ... dx
....

The actual integral is fairly simple, the only real complexity is the number of constants kicking around. But it goes without saying that really you need to sketch a cone and think about what the integration variable (x) is and the fact you need to get a dr = ... dx relationship to transform the integral.
(edited 2 months ago)
Reply 3
Ok thank you - so using those equations I managed to get this result - is this correct so far?

EDIT: I reread your response and initially I thought you were saying that the fact its a cone just means that theres a linear relationship, but upon rereading I feel you mean there is another important detail that stems from this?
(edited 2 months ago)
Reply 4
Original post by mosaurlodon
Ok thank you - so using those equations I managed to get this result - is this correct so far?

EDIT: I reread your response and initially I thought you were saying that the fact its a cone just means that theres a linear relationship, but upon rereading I feel you mean there is another important detail that stems from this?
Not carefully checked but looks about right when you sub the limits for r into the definite integral. You integrate dR from 0 to R and dx from 0 to l and dr from ... You could argue that youve missed the +c for the indefinite, but there are no integral signs anywhere.

Note Id usually do a sketch and if necessary lump constant terms together until you simplify near the end.
dx = Z dr
...

The linear relationship between r and x is important (truncated cone) as youre transforming the integral. You can obv do it for other relationships, but you need to be clear about what youre trying to do as you sketch the problem.
(edited 2 months ago)
Reply 5
Thank you so much :smile: I had some long complicated answer by the end of it and I had little hope it would be right - but it was!

I do have a question about how you made the original equation for r though ie r = sqrt(A1/pi)*(l-x)/l + sqrt(A2/pi)x/l
My thinking was since it varies linearly with distance along the wire
its form is r=kx+c
then I just tried to equate stuff
so when r = sqrt(a_1/pi), x=0, so c=sqrt(a_1/pi)
and worked out k when r=sqrt(a_2/pi), and x=L
the equation I got was r=(sqrt(a_2/pi)-sqrt(a_1/pi))*x/L +sqrt(A_1/pi)
(edited 2 months ago)
Reply 6
Original post by mosaurlodon
Thank you so much :smile: I had some long complicated answer by the end of it and I had little hope it would be right - but it was!
I do have a question about how you made the original equation for r though ie r = sqrt(A1/pi)*(l-x)/l + sqrt(A2/pi)x/l
My thinking was since it varies linearly with distance along the wire
its form is r=kx+c
then I just tried to equate stuff
so when r = sqrt(a_1/pi), x=0, so c=sqrt(a_1/pi)
and worked out k when r=sqrt(a_2/pi), and x=L
the equation I got was r=(sqrt(a_2/pi)-sqrt(a_1/pi))*x/L +sqrt(A_1/pi)
so what was your process and could you please tell me the error(s) in my method?
I just linearly interpolated between A and B (say) which are a distance l apart, assuming x starts at 0. So
r = A(l-x)/l + Bx/l
both terms are linear so the equation is linear and it equals A and B at x=0 and l so it must be what you want. Its one way to "join the two dots".

Theres nothing really wrong with what you did, its just (arguably) more complicated working than it needs to be. You could write it as
r = A + (B-A)x/l
Rearranging will give the previous. Note the previous hint about lumping constant terms together.
Reply 7
Oh yeah I never knew you could do the 'join the 2 dots' method - ive always learn it as a strict 'straight line equation', that actually makes a lot more sense seeing that.
Thanks for all the help!
Reply 8
Original post by mosaurlodon
Oh yeah I never knew you could do the 'join the 2 dots' method - ive always learn it as a strict 'straight line equation', that actually makes a lot more sense seeing that.
Thanks for all the help!
Whichever one you do, dont overcomplicate it. As in the previous post, writing the usual equation of a line doesnt require any working. You could have simply noted that really you want dr/dx so all you needed was (B-A)/l and nothing more.

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