2000 HSC Question 7 a ii (1 Viewer)

DVDVDVDV

I did a past paper today, 100% except for this question. I loooked at the solutions and I still don't get it. Part i) is fine. Part ii) screws me up big time. Can someone help?

"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 = Au^3 + B/u

where A and B are constants."

i) "Show that a pilot wishing to remain in level flight for as long a period as possible should fly at

(B/3A)^0.25 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)"

Could someone solve this for me and explain the steps for me please?

nightweaver066

Well-Known Member
i)

$\bg_white F = Au^3 + \frac{B}{u}$
$\bg_white \frac{dF}{du} = 3Au^2 - \frac{B}{u^2}$
$\bg_white \text{For minimum/maximum fuel, } \frac{dF}{du} = 0$
$\bg_white 3Au^2 - \frac{B}{u^2}= 0$
$\bg_white 3Au^4 - B = 0$
$\bg_white u^4 = \frac{B}{3A}$
$\bg_white u = (\frac{B}{3A})^{0.25}$
$\bg_white \frac{d^2F}{du^2} = 6Au + \frac{2B}{u^3}$
$\bg_white \text{At } u = (\frac{B}{3A})^{0.25}, \frac{d^2F}{du^2} = 6A(\frac{B}{3A})^{0.25} + \frac{2B}{(\frac{B}{3A})^{0.25}} > 0 \text{ as }(\frac{B}{3A})^{0.25} > 0$
$\bg_white \therefore \text{ for minimum fuel to be consumed, the pilot should fly at } (\frac{B}{3A})^{0.25}$

ii)

$\bg_white \text{Let s be the distance travelled and t be the time taken}$
$\bg_white s = ut$
$\bg_white \text{Since the amount of fuel F is in litres required per hour (from the question),}$
$\bg_white F = \frac{L}{t}$
$\bg_white t = \frac{L}{F}$
$\bg_white \therefore s = \frac{uL}{F}$
$\bg_white = \frac{uL}{Au^3 + \frac{B}{u}}$
$\bg_white \frac{ds}{du} = \frac{L(Au^3 + \frac{B}{u}) - uL(3Au^2 - \frac{B}{u^2})}{(Au^3 + \frac{B}{u})^2}$
$\bg_white \text{For minimum or maximum distance, } \frac{ds}{du} = 0$
$\bg_white L(Au^3 + \frac{B}{u}) - uL(3Au^2 - \frac{B}{u^2}) = 0$
$\bg_white Au^3 + \frac{B}{u} - 3Au^3 + \frac{B}{u} = 0$
$\bg_white \frac{B}{u} - Au^3 = 0$
$\bg_white Au^4 = B$
$\bg_white u^4 = \frac{B}{A}$
$\bg_white \therefore u = (\frac{B}{A})^{0.25}, u > 0$
$\bg_white \text{Test for minimum/maximum.. too much latex for me..}$
$\bg_white \frac{(\frac{B}{A})^{0.25}}{(\frac{B}{3A})^{0.25}}$
$\bg_white = 3^{0.25}$
$\bg_white = 1.32$
$\bg_white \therefore \text{the speed at which the pilot flies as far as possible is } 32$%$\bg_white \text{ faster than the speed given in part (i)}$

Last edited:

DVDVDVDV

I did say the first bit wasn't a problem...

math man

Member
for max distance we need to incorporate distance into this expression using:

$\bg_white D = ut$ distance = speed(u) x time

now we know F is the rate of fuel per hour in litres, therefore:

$\bg_white F = \frac{dl}{dt}$ which reads fuel is rate of change of litres with respect to time. If we invert and integrate we find that:

$\bg_white t = \frac{l}{F}$ the constant goes to 0 as t= 0 means l =0.

now we have:

$\bg_white D = u\frac{l}{F}$

subbing in F and simplyfying gives:

$\bg_white D = \frac{lu^{2}}{Au^{4}+B}$

differentiating gives:

$\bg_white \frac{dD}{du} = \frac{2lu(Au^{4}+B)-4Au^{3}(lu^{2})}{(Au^{4}+B)^{2}}$

setting to zero to find max/min gives:

$\bg_white 0 = 2Alu^{5}+2Blu-4Alu^{5}$

which gives:

$\bg_white 0 = 2lu(B-Au^{4})$

and we get:

$\bg_white u = (\frac{B}{A})^{\frac{1}{4}}$

now if we compare this speed to the original speed from i):

$\bg_white u = \frac{(\frac{B}{A})^{\frac{1}{4}}}{(\frac{B}{3A})^{\frac{1}{4}}}$

we get the ratio approx equal to:

$\bg_white u = 3^{\frac{1}{4}}}$

which is approx 1.32, therefore the new speed is 32% greater