quattro wrote,...
So according to the iterative approach and my old mate pythagoras, it's 11
Hello Richard,
Yes indeed,..you are spot on,..very well done indeed...
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.