+0

+5
726
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### Calculus with Analytic Geometry II

problem1: an airplane @ an altitude of 440m is flying horizontally directly away from an observer. At the instant when the angle of elevation is 45° or /4 radians, the angle is decreasing @ the rate of 0.05 rad/sec. How fast is the airplane flying @ that distant?

and

problem 2: a steel girder 27m long is moved horizontally along a corridor 8m wide and around corner into a hall @ right angles to the corridor. How wide must the hall be to permit this? neglect the width of the girder.

note: i already tried aswering problem 2 but im not sure if my solution is correct, but i arrive at the right answer, without seeing it before i answered.

Jun 24, 2015

#4
+27342
+20

tan(45°) = 8/p  Hence p = 8/tan(45°)  or p = 8 m

cos(45°) = (w+p)/27  or  1/√2 = (w + 8)/27 so  w = 27/√2 - 8 m

$${\mathtt{w}} = {\frac{{\mathtt{27}}}{{\sqrt{{\mathtt{2}}}}}}{\mathtt{\,-\,}}{\mathtt{8}} \Rightarrow {\mathtt{w}} = {\mathtt{11.091\: \!883\: \!092\: \!036\: \!783\: \!2}}$$

or w ≈ 11.1 m

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Jun 25, 2015

#1
+27342
+15

problem1:

Let angle of elevation at any time be θ and the horizontal distance to the plane be h.  Then:

tanθ = 440/h

Differentiate both sides with respect to time, t.

(tan2θ + 1)dθ/dt = -440/h2*dh/dt

The plane's speed is just dh/dt (let's call it v), so v = -(tan2θ + 1)h2*dθ/dt/440

We are told that when θ = 45°  dθ/dt = -0.05radians/sec.  and, since tan(45°) = 1, then h = 440m, so it's just a case of plugging in the numbers:

$${\mathtt{v}} = {\frac{\left({{\mathtt{1}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right){\mathtt{\,\times\,}}{{\mathtt{440}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{\mathtt{0.05}}}{{\mathtt{440}}}} \Rightarrow {\mathtt{v}} = {\mathtt{44}}$$

v = 44 m/s

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Jun 25, 2015
#2
+27342
0

Why not show us what you did for problem 2 and then we can comment on it?

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Jun 25, 2015
#3
+423
+10

For question 2, consider this:

The girder must rotate a full 90 degrees (Or pi/4 if you like your radians), so at one point or another it's going to be at 45° to the corridor walls. Seeing as we ignore width of the beam we can forget about it 'catching' on the walls when it rotates.

So we have a nice situation of all we need is the corridor corner to support a three-quarter circle of diameter 27 metres. This is a radius of a 27/2 = 13.5 metres. So the corridor needs to be 13.5 metres wide.

If my explanation is a bit confusing, just say and I will try to mock up some diagrams to better explain it.

Jun 25, 2015
#4
+27342
+20

tan(45°) = 8/p  Hence p = 8/tan(45°)  or p = 8 m

cos(45°) = (w+p)/27  or  1/√2 = (w + 8)/27 so  w = 27/√2 - 8 m

$${\mathtt{w}} = {\frac{{\mathtt{27}}}{{\sqrt{{\mathtt{2}}}}}}{\mathtt{\,-\,}}{\mathtt{8}} \Rightarrow {\mathtt{w}} = {\mathtt{11.091\: \!883\: \!092\: \!036\: \!783\: \!2}}$$

or w ≈ 11.1 m

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Alan Jun 25, 2015
#5
+95177
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Thank you Alan and Sir-Emo,

Jun 25, 2015
#6
+423
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I completely skimmed over the detail the first corridor is only 8m wide -_- I thought they were the same width. Alan has the right idea.

Jun 25, 2015
#7
+94526
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Here's another way to do this based on Alan's drawing...

The hypotenuse of the the small triangle at the bottom left = √[8^2 + 8^2] = 8√2

So, the length of the hypotenuse of larger triangle at the top right is just 27 - 8√2

And using similar triangles, the length of each side of this larger triangle = w =  [27 - 8√2]/ √2  = about 11.092 ft.  ≈  11.1 ft

Jun 25, 2015
#8
+35
0

@Alan @Sir Emo @CPhill,guys sorry for the late reply, and yeah i also arrive at the same aswer, mine is 11.098 to be exact, my solution is a little bit longer(though it's pretty much the same as your's) and i dont have a pc to write it down like you guys (im using my mom's ipad) so it might confuse you guys if i wright it here manually

Jun 26, 2015
#9
+95177
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Hi Alan,

I have finally gotten back to this one.

I have been looking at the aeroplane one.  I can certainly see how you got an airspeed of 44m/s

I am wondering though.

If you had been asked what is the speed at which the plane is moving away from the observer at that point, what would be the answer?

Jun 27, 2015
#10
+27342
+5

I don't understand what you are asking Melody; that is the speed at which the plane is moving away from the observer!

Or perhaps you are asking how it would be done if the plane were immediately above the observer.  In this case, take a small interval of time δt during which the angle changes by δθ.  Now we have  tan(δθ) = vδt/440.

When δθ is small, tan(δθ) = δθ, so δθ = vδt/440  or v = 440*δθ/δt.

Given the rate of change of angle, δθ/δt, we get the plane's speed.

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Jun 27, 2015
#11
+95177
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Thanks Alan,

Mmm, I did wonder if my question made sense :/

The plane is flying at a horizontal speed of 44m/s.  And a vertical speed of 0m/s

But the observer is not at the same altitude as the plane so I thought the speed from the observers view point would be different  ://

Umm.   The observer sees this 'speed' in radians per second.  I want the change in 'site' distance (hypotenuse) per second.

Does that make sense or am I makeing no sense at all ?

Jun 27, 2015
#12
+27342
+5

Ok.  In that case just replace tanθ = 440/h  by sinθ = 440/h where h is now the line of sight distance.

cosθ*dθ/dt = -440/h2dh/dt    when θ = 45°  then cosθ = 1/√2,  h = 440*√2 so

dh/dt = 4402*2*(1/√2)*0.05/440 = (22*√2) m/s ≈ 31.1 m/s

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Jun 27, 2015
#13
+95177
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Of course, thanks Alan, that is what I was asking :)

Jun 27, 2015