Calculus Question

Question

Atmospheric pressure decays exponentially as altitude increases. With pressure, P , in inches of mercury and altitude, h , in feet above sea level, we have

P=30e^{-3.23E-5 h}

(a) At what altitude is the atmospheric pressure 26 inches of mercury?

(b) A glider measures the pressure to be 26 inches of mercury and experiences a pressure increase of 0.1 inches of mercury per minute. At what rate is it changing altitude?

Answer 1

a) you have the pressure as 26 inches of mercury, plug that into the equation and find h

P = 30 e^{-3.23.10**-5 h} , 26 = 30 e^{-3.23.10**-5 h} , 0.866 = e^{-3.23.10**-5 h} , take Ln of both sides

Ln 0.866 = Ln e^{-3.23.10**-5 h} , then Ln 0.866 = -3.23.10^-5 h Ln e, but Ln e =1, and h = Ln 0.866 / -3.23.10^-5

h = 4432 ft

b) the change in pressure is given as ΔP/Δt = 0.1 in/min , as the glider drops in altitude the pressure increases, therefore the change in altitude must be decreasing wrt time Δh /Δt = -ve

using partial differentiation Δp/Δt = Δp /Δh . Δh/Δt , and Δh/Δt = Δp/Δt / Δp/Δh

we have Δp/Δt so we need Δp/Δh

Δp/Δh = 30 . -3.23 .10^-5 . e^{-3.23.10**-5 h} = -0.000969 e^{-3.23.10**-5 h} _,

and Δh/Δt = 0.1 in/min / -0.000969 e^{-3.23.10**-5 h} = -103.2 e^{-3.23.10**-5 h} , substituting h = 4432

then Δh/Δt = -119 ft /min , and the altitude is dropping at a rate of 119 ft/min

Answer 2

a) We need to find h when P = 26 in. of Hg. So,

26 = 30 e^{-3.23E-5 h}

Divide both sides by 30

0.867 = e^{-3.23E-5 h}

Take natural logarithm of both sides

ln(0.867) = -3.23E-5 h

-0.143 = -3.23E-5 h

Divide both sides by -3.23E-5 (that is, -0.0000323)

h = 4427 ft

b) Differentiating P with respect to time, we get

dP/dt = 30(-3.23E-5) e^{-3.23E-5 h} dh/dt

dP/dt = -9.69E-4 e^{-3.23E-5 h} dh/dt

It is given that dP/dt is 0.1 in. of Hg per minute. Also, we know that h = 4427 ft when P = 26 in of Hg. Therefore,

0.1 = -9.69E-4 e^{-3.23E-5 (4427)} dh/dt

0.1 = -9.69E-4 e^-0.143 dh/dt

0.1 = -9.69E-4 (0.867) dh/dt

0.1 = -8.4E-4 dh/dt

Divide both sides by -8.4E-4 (that is, -0.00084)

dh/dt = -119

So the altitude is DECREASING at the rate 119 ft/min.