In the project on page 344 we expressed the power needed by bird during its flapping mode as P ( v , x , m ) A v 3 + B ( m g / x ) 2 v where A and B are constants specific to a species of bird, v is the velocity of the bird, m is the mass of the bird, and x is the fraction of the flying time spent in flapping mode. Calculate ∂ P /∂ v , ∂ P /∂ x , and ∂ P /∂ m and interpret them.
In the project on page 344 we expressed the power needed by bird during its flapping mode as P ( v , x , m ) A v 3 + B ( m g / x ) 2 v where A and B are constants specific to a species of bird, v is the velocity of the bird, m is the mass of the bird, and x is the fraction of the flying time spent in flapping mode. Calculate ∂ P /∂ v , ∂ P /∂ x , and ∂ P /∂ m and interpret them.
Solution Summary: The author calculates the values of partial v, if P(v,x,m)=Av
In the project on page 344 we expressed the power needed by bird during its flapping mode as
P
(
v
,
x
,
m
)
A
v
3
+
B
(
m
g
/
x
)
2
v
where A and B are constants specific to a species of bird, v is the velocity of the bird, m is the mass of the bird, and x is the fraction of the flying time spent in flapping mode. Calculate ∂P/∂v, ∂P/∂x, and ∂P/∂m and interpret them.
A seasoned parachutist went for a skydiving trip where he performed freefall before deploying
the parachute. According to Newton's Second Law of Motion, there are two forcës acting on
the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa)
as shown in Figure 1.
Fa = -cv
ITM EUTM FUTM
* UTM TM
Fg= -mg
x(t)
UTM UT
UTM /IM LTM
UTM UTM TUIM
UTM F UT
GROUND
Figure 1: Force acting on body of free-fall
where x(t) is the position of the parachutist from the ground at given time, t is the time of fall
calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration,
v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the
position is given by the equations below:
EUTM PUT
v(t) =
mg
-et/m – 1)
(Eq. 1.1)
x(t) = x(0) –
Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the
critical position to deploy the parachutes is at 762 m from the ground…
(u x v) · v=,
Chapter 14 Solutions
James Stewart Calculus for MAT 127/128/229 8th edition
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