2000 HSC Q 7a ii (1 Viewer)

[Paroissien]

I can't be bothered to write out the whole question, but I simply don't understand how this question works. If anyone could shed some light on the topic, if would be greatly appreciated.

 

[BillyMak]

Can you do part (i)?

 

[Paroissien]

Yeah no problem, I'm assuming you have the Success one book so the part I don't understand (referring to answers) is the step after you have determined the minimum value for u

 

[BillyMak]

(i) To remain at level flight for greatest possible time, F is a minimum.

F = Au³ + B/u

dF/du = 3Au² - B/u²

d²F/du² = 6Au + 2B/u³

d²F/du² > 0 since u>0, A & B are positive. Therefore min when dF/du = 0

3Au² - B/u² = 0
3Au⁴ = B
u = (B/3A)^1/4

 

[BillyMak]

meh, I had to do that before I could attempt (ii)

 

[BillyMak]

hmm.... I dunno

 

[Archman]

just post the question up.

 

[Paroissien]

I'll do it tomorrow if the question still hasn't been answered

 

[BillyMak]

Question 7
(a) The amount of fuel F in litres required per hour to propel a plane in level flight at constant speed u km/h is given by
F = Au³ + B/u

where A and B are positive constants.

(i) Show that a pilot wishing to remain in level flight for as long a period as possible should fly at
(B/3A)^1/4 km/h.

(ii) Show that a pilot wishing to fly as far as possible in level flight should fly approximately 32% faster than the speed given in part (i).

 

[Paroissien]

Thanks, and the main reason I didn't post it is because I don't know how to do superscript. How do you do it?

 

[Supra]

its easy

say u want x squared

u type x < sup > 2 < / sup> but remove the spaces

x^2=x²

 

[Archman]

say there are K litres on the plane, since fuel consumption is constant so fuel will last
K/f hours.
distance = time * speed
= Ku/f
= Ku/(Au^3+B/u)
= Ku^2/(Au^4 + B)
now differentiate that by quotient rule u get
K(2u(Au^4 + B) - 4Au^5)/(crap)^2
set that to 0:
2u(Au^4 + B) = 4Au^5
u = (B/A)^1/4, now du the routine check to show that its a maximum.
so comparing to the other thing
(B/A)^1/4
-------------- = 3^1/4 = around 1.32
(B/3A)^1/4

so there you go.

 

[Jedi-Nerd]

ok since F represents fuel usage in litres per hour, you divide through by "u" which is the speed to get fuel usage in litres per kilometre
(this works because.. say if you you're using 2 litres per hour at 10 km/h speed, to find how many litres per kilometre you're using you divide 2/10 = 0.2 litres per km)

anyways so therefore let X represent fuels per km:

X = F/U
= (Au³ + B/u) / u
= Au² + B/u²

dX/du = 2Au - 2B/u³

at dX/du = 0

2Au = 2B/u³
2Au⁴ = 2B
u⁴ = B/A
u = (B/A)^1/4

prove this is a minimum using second derivative, to prove minimum fuel used per km for max distance

now: put result from part (ii) and divide it by result in part (i):

B^1/4/A^1/4 divided by B^1/4/(3A)^1/4

= 3^1/4A^1/4 / A^1/4
= 3^1/4
= 1.32 correct to two dec place

therefore plane should fly 32% faster if he wants to maximise distance flown for minimum fuel.

 

[wogboy]

i) F = Au^3 + B/u

You wish to find the value of u that minimises F (the fuel used in a given amount of time), which is equivalent to maximising the time taken given a fixed amount of fuel.

dF/du = 3Au^2 - B/u^2
setting dF/du = 0,
3Au^4 = B,

u = (B/3A)^(1/4) km/h

ii) In this case, you wish to maximise u/F. The quantity u/F is proportional to the distance travelled (it's the distance travelled, multiplied by the amount of fuel used which you take as a fixed constant in this part).

d(u/F)/du = du/du * 1/F + u*d(1/F)/du (using chain rule)
= 1/(Au^3 + B/u) + u * d/du (1/(Au^3 + B/u))
= 1/(Au^3 + B/u) + u*(-1/(Au^3 + B/u)^2 * (3Au^2 - B/u^2))

setting d(u/F)/du = 0,
1/(Au^3 + B/u) = u * (1/(Au^3 + B/u)^2) * (3Au^2 - B/u^2))
Au^2 + B/u^2 = 3Au^2 - B/u^2
2Au^2 = 2B/u^2
u^4 = B/A
u = (B/A)^(1/4)
= (B/3A)^(1/4) * 3^(1/4)
~ 1.32 (B/3A)^(1/4)

So the plane should fly ~ 1.32 times as fast as the answer given in i), so that the distance is maximised.

 

[Archman]

isn't it great when u get ur post in faster than the others

 

[Paroissien]

Thanks very much, and that made me realise how foolish I was not to realise that final step.
Anyway, cheers.