Pytels Dynamics 12.25: Plane tracking

[Alexanddros81]

Homework Statement

the plane C is being tracked by radar stations A and B. At the instant shown,
the triangle ABC lies in the vertical plane, and the radar readings are θ_A=30^o,
θ_B = 22^o, θ_A(dot) = 0.026rad/s and θ_Β(dot) = 0.032rad/s. Determine (a) the altitude y; (b) the speed v; and (c) the climb angle α of the plane at this instant

Homework Equations

The Attempt at a Solution

I guess I can find (a) from trogonometry. Any hints for part (b)?

 

[Dr. Courtney]

I'd draw a good picture and think long and hard about related rates.

 

[haruspex]

Any hints for part (b)?

Solve part a) keeping everything symbolic - do not plug any of the given numbers in.
That will give you an equation for y in terms of the two angles and the distance between the stations.
Get another equation for the x coordinate.
With those two equations, how might you get an equation for velocities?

 

[SammyS]

Homework Statement

the plane C is being tracked by radar stations A and B. At the instant shown,
the triangle ABC lies in the vertical plane, and the radar readings are θ_A=30^o,
θ_B = 22^o, θ_A(dot) = 0.026rad/s and θ_Β(dot) = 0.032rad/s. Determine (a) the altitude y; (b) the speed v; and (c) the climb angle α of the plane at this instant

Homework Equations

The Attempt at a Solution

I guess I can find (a) from trogonometry. Any hints for part (b)?

Posting the picture, rather than the thumbnail makes the problem clearer and may encourage more help,

 

[Alexanddros81]

Solve part a) keeping everything symbolic - do not plug any of the given numbers in.
That will give you an equation for y in terms of the two angles and the distance between the stations.
Get another equation for the x coordinate.
With those two equations, how might you get an equation for velocities?

x is the distance from the first station to the second station plus the distance to the point that the vertical line meet the plane? (my english are not that good)

 

[jbriggs444]

x is the distance from the first station to the second station plus the distance to the point that the vertical line meet the plane? (my english are not that good)

You can choose any coordinate system you like. Placing the origin at station A is a good choice.

Yes, if the origin is at A then the x coordinate of the plane is the distance from A to B plus the distance from B to a vertical line from plane to ground.

 

[Alexanddros81]

so I got
$y=\frac {1000} {tanθ_A - tanθ_B}$

and

$x=1000\frac {tanθ_A} {tanθ_A-tanθ_B}$

I guess I deffirentiate to get v_x and v_y
but I don't know how to proceed since tan is confusing me

 

[haruspex]

so I got
$y=\frac {1000} {tanθ_A - tanθ_B}$

and

$x=1000\frac {tanθ_A} {tanθ_A-tanθ_B}$

I guess I deffirentiate to get v_x and v_y
but I don't know how to proceed since tan is confusing me

It's just a matter of applying the chain rule. What is the derivative wrt time of tan(θ)?

 

[Alexanddros81]

It's just a matter of applying the chain rule. What is the derivative wrt time of tan(θ)?

since we apply the chain rule it should be:
if $y=tanθ$ then $\frac {dy} {dt} = \frac {dy} {dθ} \frac {dθ} {dt} = sec^2θ\frac {dθ} {dt} = sec^2θ \dot θ$
Any hint how to proceed with the x and y derivatives?

 

[haruspex]

since we apply the chain rule it should be:
if $y=tanθ$ then $\frac {dy} {dt} = \frac {dy} {dθ} \frac {dθ} {dt} = sec^2θ\frac {dθ} {dt} = sec^2θ \dot θ$
Any hint how to proceed with the x and y derivatives?

In post #7 you have an equation for x as a function of the two angles. So differentiate it.

 

[Alexanddros81]

In post #7 you have an equation for x as a function of the two angles. So differentiate it.

 

[haruspex]

Looks right.
What about $\dot y$?

 

[Alexanddros81]

I will proceed finding $\dot y$ later but
for now I want some clarification with the notation.
In Mathematics for Engineers by Croft 2nd edition in pg. 709 the quotient rule is been described.
According to this:
if
$y(x) = \frac {u(x)} {v(x)}$
then

$\frac {dy} {dx} = \frac { v \frac {du} {dx} - u \frac {dv} {dx}} {v^2}$

an example is given

$y=\frac {sinx} {x} = \frac {u} {v}$ so $u=sinx , v=x$
and so
$\frac {du} {dx} = cosx, \frac {dv} {dx} = 1$
applying the quotient rule gives
$\frac {dy} {dx} = \frac { v \frac {du} {dx} - u \frac {dv} {dx}} {v^2}$
$=\frac {xcosx - sinx(1)} {x^2}$
$=\frac {xcosx-sinx} {x^2}$

According to the above example and in relation to this post can we say:
$x(θ) = \frac {u(θ)} {v(θ)}$
but I guess this is not what we want. We want $x(t) = \frac {u(t)} {v(t)}$ so $\frac {dy} {dt} = \frac { v \frac {du} {dt} - u \frac {dv} {dt}} {v^2}$
We can say then that u is a function of θ and θ is a function of t or $u = u(θ) = u(θ(t))$ (chain rule)
So the first line of my attached solution of post #11 should be $x=\frac {u(t)} {v(t)}$ and not $x=\frac {u(θ)} {v(θ)}$

Also I forgot to multiply the solution at post #11 with 1000

Please verify/comment the above

Alexandros

 

[haruspex]

$x=\frac{u(\theta)}{v(\theta)}$ is fine. $\theta$ is allowed to be a function of t. $\dot x=\frac{d}{dt}\frac{u(\theta)}{v(\theta)}= \dot\theta\frac {d}{d\theta}\frac{u(\theta)}{v(\theta)}$.

I forgot to multiply the solution at post #11 with 1000

True.

 

[Alexanddros81]

Is the following correct?

$\frac {dx} {dt} = \frac {dx} {dθ} \frac {dθ} {dt} = \dot θ \frac {dx} {dθ}$

and since $x(θ) = \frac {u(θ)} {v(θ)}$ where $u(θ) = tanθ_A$ and $v(θ) = tanθ_A - tanθ_B$

then $\frac {dx} {dθ} =1000 \frac {v \frac {du} {dθ} - u \frac {dv} {dθ}} {v^2}$

so $\frac {du} {dθ} = sec^2 θ_A$ and $\frac {dv} {dθ} = sec^2 θ_Α - sec^2 θ_B$

so $\frac {dx} {dθ} =1000 \frac {(tanθ_A - tanθ_B)(sec^2 θ_A) - (tanθ_A)(sec^2 θ_Α - sec^2 θ_B)} {(tanθ_A - tanθ_B)^2}$
$=1000\frac {tanθ_Asec^2 θ_A - tanθ_Bsec^2 θ_A - tanθ_Asec^2 θ_Α + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}$
$=1000\frac {- tanθ_Bsec^2 θ_A + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}$
so
$\frac {dx} {dt} = \dot θ 1000\frac {- tanθ_Bsec^2 θ_A + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}$

 

[haruspex]

Is the following correct?

$\frac {dx} {dt} = \frac {dx} {dθ} \frac {dθ} {dt} = \dot θ \frac {dx} {dθ}$

and since $x(θ) = \frac {u(θ)} {v(θ)}$ where $u(θ) = tanθ_A$ and $v(θ) = tanθ_A - tanθ_B$

then $\frac {dx} {dθ} =1000 \frac {v \frac {du} {dθ} - u \frac {dv} {dθ}} {v^2}$

so $\frac {du} {dθ} = sec^2 θ_A$ and $\frac {dv} {dθ} = sec^2 θ_Α - sec^2 θ_B$

so $\frac {dx} {dθ} =1000 \frac {(tanθ_A - tanθ_B)(sec^2 θ_A) - (tanθ_A)(sec^2 θ_Α - sec^2 θ_B)} {(tanθ_A - tanθ_B)^2}$
$=1000\frac {tanθ_Asec^2 θ_A - tanθ_Bsec^2 θ_A - tanθ_Asec^2 θ_Α + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}$
$=1000\frac {- tanθ_Bsec^2 θ_A + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}$
so
$\frac {dx} {dt} = \dot θ 1000\frac {- tanθ_Bsec^2 θ_A + tanθ_Asec^2 θ_B} {(tanθ_A - tanθ_B)^2}$

I do not understand the direction you have taken. What is this angle θ? Your attachment in post #11 was correct.

 

[Alexanddros81]

I do not understand the direction you have taken. What is this angle θ? Your attachment in post #11 was correct.

ok.

for v_y look the attachement

 

[haruspex]

ok.

for v_y look the attachement

You have not applied the chain rule correctly this time. The derivative of 1/v is not $-(\dot v)^2$.

 

[Alexanddros81]

ok I am totally confused. any hint?
should that be $\frac {dy} {dt} = 1000\frac {d} {dt} \left(\frac {1} {v}\right) = 1000(-1) (\frac {1} {v^2})$

 

[Alexanddros81]

what about this:

$y(v) =1000\frac {1} {v}$ and $v(θ) = tanθ_A-tanθ_B$

$\frac {dy} {dt} = \frac {dy} {dv} \frac {dv} {dt}$

where $\frac {dy} {dv} = 1000(-1) \frac {1} {v^2}$ and $\frac {dv} {dt} = sec^2θ_A \dot θ_A - sec^2θ_B \dot θ_B$

so

$\frac {dy} {dt}= -1000\frac{sec^2θ_A \dot θ_A - sec^2θ_B \dot θ_B} {(tanθ_A-tanθ_B)^2}$

 

[haruspex]

what about this:

$y(v) =1000\frac {1} {v}$ and $v(θ) = tanθ_A-tanθ_B$

$\frac {dy} {dt} = \frac {dy} {dv} \frac {dv} {dt}$

where $\frac {dy} {dv} = 1000(-1) \frac {1} {v^2}$ and $\frac {dv} {dt} = sec^2θ_A \dot θ_A - sec^2θ_B \dot θ_B$

so

$\frac {dy} {dt}= -1000\frac{sec^2θ_A \dot θ_A - sec^2θ_B \dot θ_B} {(tanθ_A-tanθ_B)^2}$

Yes!

 

[Alexanddros81]

the angle α is given in the attachement. please verify

 

[Alexanddros81]

Anyone to verify angle α?

 

[haruspex]

Anyone to verify angle α?

Yes, that looks right.