Related Rates

• December 18th 2005, 11:06 AM
ASE2
Related Rates
• December 18th 2005, 01:17 PM
CaptainBlack
Clearly we would be doing you no favours in just solving them
all for you, so I will do one example and if you have specific
problems with the others you can post the question, what you have
done and what the difficulty is.

<usual warning about checking the working yourself>

<another sugestion-it might be worthwhile converting the units to
metres (m), kilograms (kg) and seconds (s)>

Quote:

2.An airplane of mass 10^5 kg flies in a straight line while ice builds up
on the leading edges of its wings at a constant rate of 30kg/h.

a) At what rate is the planes’s momentum changing if it
is flying at a constant rate of 800 km/h?
(Momentum (p) is the product of an object’s mass and
velocity: p = mv)

b) At what rate is the momentum of the plane changing
at t = 1h if at that instant its velocity is 750 km/h and
is increasing at a rate of 20 km/h/h?
The planes momentum is:

$p=m.v$,

so the rate of change of momentum (using the product rule) is:

$\frac{dp}{dt}=\frac{dm}{dt}.v + m. \frac{dv}{dt}$

Part (a) we have the speed is constant:

$\frac{dv}{dt}=0$,

so:

$\frac{dp}{dt}= \frac{dm}{dt}.v$.

We are told: $\frac{dm}{dt}=30\ \mbox{kg/hr}$, and $v=800\ \mbox{km/hr}$,
plugging these in (and leaving the units as they are):

$\frac{dp}{dt}= 30.800=24000\ \mbox{km.kg.hr^{-2}}$

Part (b) at 1hr we have the mass of the aircraft and ice is $100,030 \mbox{kg}$, the speed is $750 \mbox{km/hr}$,
$\frac{dm}{dt}=30\ \mbox{kg/hr}$, and $\frac{dv}{dt}=20 \mbox{km/hr^2}$, so:

$\frac{dp}{dt}= 30.750+100030.20=2023100\ \mbox{km.kg.hr^{-2}}$

RonL