Where should a pilot start descent, in order that the path

What have you tried? How far have you gotten?

Please be complete. Thank you! :D

Eliz.
 
i thought i knew what to do but it was nowhere near right. how should i go about this?
 
Please show us what you have tried - even if you know it is not correct.

That will help us determine a starting point for our assistance.

The book looks like a math book for engineers. If you are planning to be one - then you really need to understand these types of problems. This problem requires your understanding of functions, derivatives and physics - that is what engineers do.

So go at it - tell us where you are stuck - exactly .....
 
For those who prefer to work locally, or whose connections might choke on the whole-chapter file, the following is an approximation of the "figure":

Code: 
^
|                   _ 
y                 -  ^
|              /     |
| y = P(x)  /        |
|         /          h
|      /             |
| _ -                |
*--------------------|--x-->
|<---------L-------->
That is, the plane is decending from a height of h through an horizontal distance of L along polynomial path P(x). The somewhat-S-shaped path is smooth when leaving the horizontal flight path and when landing at the origin.

The text is as follows:

An approach path for an aircraft landing is shown in the figure and satisfies the following conditions:

. . . . .(i) The cruising altitude is "h" when the descent starts at a
. . . . .horizontal distance of "L" from touchdown at the origin.

. . . . .(ii) The pilot must maintain a constant horizontal speed "v"
. . . . .thoughout descent.

. . . . .(iii) The absolute value of the vertical acceleration should not
. . . . .exceed a constant "k" (which is much less than the acceleration
. . . . .due to gravity).

1) Find the cubic polynomial P(x) = ax<sup>3</sup> + bx<sup>2</sup> + cx + d that satisfies condition (i) by imposing suitable conditions on P(x) and P'(x) at the start of descent and at touchdown.

2) Use conditions (ii) and (iii) to show that \(\displaystyle \L \, \, \frac{6hv^2}{L^2}\, \leq \,k\)

3) Suppose that an airline decides not to allow vertical acceleration of a plane to exceed k = 860 mi / hr<sup>2</sup>. If the cruising altitude of a plane is 35,000 feet and the speed is 300 mph, how far away from the airport should the pilot start descent?

4) Graph the approach path if the conditions stated in (3) are satisfied.
Click to expand...
 
Re: Where should a pilot start descent, in order that the pa

terry33199 said:
can't figure this out:

"Where should a pilot start descent" p. 232 of the following PDF file (1 megabyte).
Click to expand...

Did you get the first two parts?

If you did, what does y=f(x) look like? -- part (1)
 
I'll start you off.

y = Ax^3 + Bx^2 + Cx + D

Given

y(0) = 0 ---> D = 0

y'(0) = 0 ---> ???

Now investgate y'(L) = 0 and finally y(L) = h

Show us what you get out of these....
 
I am also doing this problem and just want to know if I'm doing it right. I solved P'(L)=0 for A and B and then solved P(L)=h for A and B. Then I had A in terms of L,H and B; and B in terms of L,H and A. So I took the A I had gotten from solving P'(L)=0 and the A I got from solving P(L)=h, set them equal to each other and solved for the b. I did the same with the A.
This is what I got for B and A. Could someone tell me if they are right?
A=(4h)/(L^3)
B=(3h)/(-2(L^3)+3(L^2))

For part 2, I said that x is secretly a function of time, so if I implicitly differentiate P(x) with respect to time I should get the vertical acceleration. And where ever there is a dx/dt, I can substitute in v=horizontal speed. However in this derivative there is also A and B, so I wanted to make sure I had them right before I substituted in and did a lot of work. I found the derivative of P(x) with respect to time to be dP(x)/dt=3a(x^2)v+2bxv.
Is everything I said right? Am I on the right track?
Thanks.[/tex]
 
Calc Student said:
I am also doing this problem and just want to know if I'm doing it right. I solved P'(L)=0 for A and B and then....
Click to expand...
Since, in the posted exercise, the polynomial P is given as being in terms of x, not of L, it is difficult to evaluate what you've done. Are you sure you're doing the same exercise...? :oops:

Eliz.
 
Yes, I'm sure I'm doing the same exercise. The reason I found A and B in terms of L and H is because I evaluated P'(L)=0 and P(L)=H. I know P is given in terms of x, but I plugged L in as my x.
Hope that clears things up a bit.
 
Calc Student said:
...I plugged L in as my x.
Click to expand...
But x is variable, giving you a function for the entire path; L is fixed, and would give you only one point (and the derivative would be zero, since P(L) is just a number).

Eliz.
 
Sorry, you're right Stapel. But then how does P(L)=h and P'(L)=0 help me? And how can I find A and B?[/quote]
 
Calc Student said:
I am also doing this problem and just want to know if I'm doing it right. I solved P'(L)=0 for A and B and then solved P(L)=h for A and B. Then I had A in terms of L,H and B; and B in terms of L,H and A. So I took the A I had gotten from solving P'(L)=0 and the A I got from solving P(L)=h, set them equal to each other and solved for the b. I did the same with the A.
This is what I got for B and A. Could someone tell me if they are right?
A=(4h)/(L^3)
B=(3h)/(-2(L^3)+3(L^2))

For part 2, I said that x is secretly a function of time, so if I implicitly differentiate P(x) with respect to time I should get the vertical acceleration. And where ever there is a dx/dt, I can substitute in v=horizontal speed. However in this derivative there is also A and B, so I wanted to make sure I had them right before I substituted in and did a lot of work. I found the derivative of P(x) with respect to time to be dP(x)/dt=3a(x^2)v+2bxv.
Is everything I said right? Am I on the right track?
Thanks.[/tex]
Click to expand...

As far as I can tell, you are on right path.
 
Calc Student said:
I am? But what Stapel said makes sense. So how can what I did be right?
Click to expand...

P'(L) means P'(x) evaluated at x = L.

for example

P(x) = Ax^3 - Bx^2

P'(x) = 3Ax^2 - 2Bx

P'(L) = 3AL^2 - 2BL

if P'(L) =0 then

3AL = 2B

That is sort of what you did - right ?
 
Yep that's what I did. So now for the second part, I found the derivative of P'(x) with respect to time. If I plug in the A's and B's I found earlier, I should find my equation, right?
Thanks so much for the help.
 
Calc Student said:
Yep that's what I did. So now for the second part, I found the derivative of P'(x) with respect to time.


If I plug in the A's and B's I found earlier, I should find my equation, right? Correct



Thanks so much for the help.
Click to expand...
 
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