00:03
Hello, it is given that the barometric pressure p at a altitude of h miles, h is in miles, this we should remember, above sea level satisfies the differential equation dp by dt is equal to minus 0 .2 p and the pressure at sea level, that is pressure at when h is 0 is 29 .92 inch of mercury is 29 .92 inch of mercury, we have to find the barometric pressure at 18 ,000 feet.
01:18
Okay, first let us solve this differential equation and find the equation for p.
01:25
So, dp by dt equal this thing, this implies that dp by p is equal to minus 0 .2 dt and now let us integrate both sides, we get ln p is equal to minus 0 .2 p.
01:44
Now, this thing plus constant we ignore because we are directly going to the definite integral.
01:51
So, ln p evaluated between p0 to p as pressure changes from p0 to p, sorry there is no t here, it is h dp by dh, as h changes from 0 to some value h.
02:33
So, we have that ln p minus ln p0 is equal to minus 0 .2 h minus 0 is h.
02:44
So, this thing is ln p by p0 is equal to minus 0 .2 h, this implies that p by p0 equal to e to the power minus 0 .2 h and this implies that p equal to p0 times e to the power minus 0 .2 h.
03:08
P0 is the pressure when h equal to 0...