vertical speed = 100 mph = (440/3) ft/sec = dh/dt
evaluate dP/dt at h = 0 and dh/dt = (440/3)
Im stuck on this problem.
At an altitude of h feet above sea level, the air pressure, P, in inches of mercury (in Hg), is given by P=30e^(-.0000323h)
An unpressurized seaplane takes off at an angle of 30 degrees to the horizontal and a speed of 200 mph.
What is the rate of change of pressure in the plane with respect to time at take-off, in inches of mercury per second?
Converted 200 miles/hours to 3520 inches/second
Converted the vertical rate of change to 3520sin(30)=1760
Derived P to equal 30e^(-.0000323h)(-.0000323)
What should I do next?
Got it. The function of pressure is dependent on feet above sea level, and not inches. You pointed out my mistake of converting the height to inches. The inches relate only to the content of mercury in the air. So multiplying dP/dh with dh/dt gives the answer, which is -.1421154096.
Thanks for quick reply!