The next Brain Teaser,....the Balloon and the P6

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
Here is the next thought provoking problem to test the brain cells.

A balloon is rising vertically above a level, straight road at a constant speed of 1 metre per second. Just when the balloon is 65 metes above the ground, a P6 passes under it, travelling 17 metres per second. How fast is the distance between the P6 and the balloon increasing 3 seconds later? Give your answer in metres per second.

Ron.
 
Your teaser is not as complicated as it sounds. As the ascent of the balloon is uniform, as is the forward movement of the Rover is expressed in the same units, we end up with a right angled triangle. As many years have elapsed since I last solved such problems, all I can say is we are dealing with the principles of Pythagorus. All you have to calculate is the length of the adjacent side. Good luck!
 
Hello Classicus,

You are partly correct in that the question does involve some use of Pythagoras, but that is not the whole story... :wink: You might like to have a shot, see how you go.

Ron.
 
Well, as far as I remember Calculus back when I was an Engineering student (long time ago!) it would be necessary to use infinitesimal calculation in order to get the exact answer. I had known how to do it... 25 years ago. Not now, sorry.

But I think that we can use a simple method just for the real life.

At t=0 the distance between them is 65 metres. Three seconds later the vertical height of the balloon is 68 metres and the horizontal distance is 17x3=51 metres, so the distance between balloon and car is 85 metres.

So, in three seconds the distance increased 20 metres. Therefore the speed is 20/3 metres per second.


Regards
 
Hello Shazzbat,

You are correct in that you can use derivatives in order to find the solution, and equally there are alternative methods that can be employed to obtain the answer. I'll give a clue,..the correct answer is an integer. If you would like to try again,..please do.

Each teaser I post will be harder than the one that preceeded it.

Ron.
 
Ohhh, not now, Ron. I was very very good at Calculus when I was a student but after 20 years working for a car's manufacturer I have only dark remembrances about Maths.

You can believe me. Car's Industry shrinks the brains, kills neurons and in thirty years time (more or less) you become a supermarket bag.


Regs
 
Any topic with "Brain Teaaser" in the title I'm going to skip
Too old , too tired and not clever enough to participate
 
Hello Dave,

Don't sell yourself short nor ever under estimate your capabilities. As for being too old or not clever enough,...never!

Does anyone have an answer to the problem at this stage?

Ron.
 
quattro wrote,..
10
Click to expand...

Hello Richard,

Sorry...its not 10 but you are very close.

If you use Pythogoras you need to take an iterative approach as you are after how fast the distance is changing between the P6 and the balloon 3 seconds later. I'll post two solutions, one formal and one from iterations over the weekend.

Ron.
 
I am just reminding myself that I left school 32 years ago. :shock:

at t=0 they are 65M apart.

3.15M in the first second

t=1 68.15M

6.98M in the second second

t=2 75.13

9.87M in the third second

t=3 85M

11.88M in the fourth second

t=4 96.88

If it's not 10, which looking at it now it obviously isn't, it is 11

8)

I don't have a square root button on my keyboard so will use ^ for it.

^((65+x)2 + Y2) - Where x = 1 per second and y = 17 per second. This gives the various distances but not the increase in speed.

I have forgotten how to calculate acceleration so will have to leave it there.

Richard
 
Ok - I have worked out that after the following times, it is moving away at the following speeds.

2.6 sec - 10.02
2.7 sec - 10.25
2.8 sec - 10.47
2.9 sec - 10.69
3 sec - 10.9
3.1 sec - 11.1
3.2 sec - 11.29
3.3 sec - 11.47
3.4 sec - 11.65

So according to the iterative approach and my old mate pythagoras, it's 11
 
quattro wrote,...
So according to the iterative approach and my old mate pythagoras, it's 11
Click to expand...

Hello Richard,

Yes indeed,..you are spot on,..very well done indeed... :D You have quite an analytical brain! You have succesfully solved both or my brain teasers now, so I'll post another that won't involve any Calculus, but none the less challenging.

You have explained the iterative solution precisely, so I'll just add the formal solution. ( oh just one thing,.....when wishing to represent square root without using the radical sign, the equivilant is ^1/2. That is raising an index of 1/2, or it can be said..raising the expression to the power of 1/2. An example.....The square root of 64 .....so using index notation 64^1/2 ).

So to the formal solution.

Information we know....
dy/dt = 1m/s....that is the rate of change of elevation of the balloon with respect to time.
y(initial) = 65m
dx/dt = 17m/s...the rate that the distance covered by the P6 changes with time.

At t = 3 seconds the P6 has covered 17 X 3 = 51m. We shall call that distance x.
At t = 3 seconds, the balloon has risen by another 3 metres to be 68m above the ground. This distance shall be called y.

The problem asks us to find how fast the distance between the P6 and the balloon is increasing 3 seconds later,..so with a Pythogorean triangle with sides x and y, the hypotenuse will be z. The change in distance with respect to time is dz/dt. This is what we need to find.

So using the theorem of Pythogoras and implicit differentiation as opposed to explicit differentiation, the following expression is formed.

z^2 dz/dt = x^2 dx/dt + y^2 dy/dt.

Differentiating yields…

2z dz/dt = 2x dx/dt + 2y dy/dt so….

dz/dt = 1/z(x dx/dt + y dy/dt)

Substituting in values

dz/dt = 1/85( 51(17) + 68(1) )

dz/dt = 935/85 which equals…..11m/s.

Ron.
 
I'm shocked! An Aussie knowing calculus! Seeing they only understand beer and barbie's, I think someone else did it for him! :p

Seriously, a very good problem, I enjoyed working through it. Really got the mind ticking.

Yes, I got 11m/s also... and practised my calculus from some 10+ years ago!

Only one nit-pick though...

z^2 dz/dt = x^2 dx/dt + y^2 dy/dt. <== Should be z^2 = x^2 + y^2 (no d/dt yet!)

You can actually do it using explicit differentiation (as opposed to implicit differentiation), but it's messy and very easy to make a mistake. Implicit is much better in this case!

However... here goes!

z = [x^2 + y^2]^0.5

x= 17t (the Rover just under the balloon at t=0) and, y = t+65 (the balloon 65 meters above at t=0)

x^2 = 289t^2 and y^2 = t^2 + 130t + 4225 so z = [289t^2 + t^2 + 130t + 4225]^0.5 = [290t^2 + 130t + 4225]^0.5

dz/dt = 0.5 ([290t^2 + 130t + 4225]^-0.5) x (2 x (290t) + 130) = 0.5 ([290t^2 + 130t + 4225]^-0.5) x (2 x (290t + 65))

130 = 2 x 65, so ((2 x 290t) + 130) = 2 x (290t +65),

dz/dt = (0.5 x 2) x ([290t^2 + 130t + 4225]^-0.5) x (290t + 65)

Seeing 0.5 x 2 = 1 :-

dz/dt = ([290t^2 + 130t + 4225]^-0.5) x (290t + 65)) = (290t + 65)/[290t^2 + 130t + 4225]^0.5

(Really messy! Implicit Differentiation in this case is much cleaner! Ron's implicit differentiation has z as the denominator, which ties in seeing the denominator in my method is actually the expression for z)

at 3 seconds... [290t^2 + 130t + 4225] = (290 x 3^2) + (130 x 3) + 4225 = (290 x 9) + (130 x 3) + 4225 = 2610 + 390 + 4225 = 3000 + 4225 = 7225

[290t^2 + 130t + 4225]^0.5 = 7225^0.5 = ±85. As we are talking distances, we take the +85 answer!

290t + 65 = (290 x 3) + 65 = 870 + 65 = (850 + 20) + 65 = 850 + (20 + 65) = 850 + 85

So...

dz/dt (at t = 3s) = (850 + 85)/85 = 850/85 + 85/85 = 10 + 1 = 11m/s

They both yield the same answer, phew! But it shows sometimes why implicit differentiation is preferable!
 
Hello Darth,

Glad you liked that problem too.. :D Explicit differentiation is normally used where the expression being differentiated is a function (one variable expressed explicitly in terms of another) as opposed to a relation where the variables are defined implicitly. I should expand on that by saying that all functions are relations but not all relations are functions. A circle at the origin is a perfect example of an implicit relation,...x^2 + y^2 = r^2. To differentiate explicitly you need to split the relation into two semi circles, y = f(x) = + or - (r^2 - x^2)^0.5, being just one possible pair, and each then becomes a function of just one variable. (r is the radius, a constant and not a function of x). Much easier though to differentiate implicitly especially where the relation is something like the "folium of Descartes"....x^3 + y^3 = 3xy.

I'll post another problem in a day or two that will take things a step further.

Ron.
 
SydneyRoverP6B said:
Hello Darth,

Glad you liked that problem too.. :D Explicit differentiation is normally used where the expression being differentiated is a function (one variable expressed explicitly in terms of another) as opposed to a relation where the variables are defined implicitly. I should expand on that by saying that all functions are relations but not all relations are functions. A circle at the origin is a perfect example of an implicit relation,...x^2 + y^2 = r^2. To differentiate explicitly you need to split the relation into two semi circles, y = f(x) = + or - (r^2 - x^2)^0.5, being just one possible pair, and each then becomes a function of just one variable. (r is the radius, a constant and not a function of x). Much easier though to differentiate implicitly especially where the relation is something like the "folium of Descartes"....x^3 + y^3 = 3xy.

I'll post another problem in a day or two that will take things a step further.

Ron.
Click to expand...

Yes I agree! I was trying to show that explicit diff was possible to be used, just it's really messy and quite unwieldy!

If x^3 + y^3 = 3xy, then 3x^2 + 3y^2 dy/dx = 3(x dy/dx + y) or dy/dx = (y - x^2) / (y^2 - x) <== wow! :-S

I liked quattro's 'numeric differentiation' method too!
 
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