A-Level Differential equations: worked solution

Question

The model P=A|293-6.5h|^b applies, with pressures 101300 Pa at sea level and 80000 Pa at h=2. Estimate the pressure at the summit of Mount Everest, altitude 8848 m.

Step 1: turn the two known pressure readings into two equations. Altitude h is in km here, since h = 2 is used with 293 - 6.5h. At sea level, h = 0: 293 - 6.5(0) = 293, so 101300 = A(293)^b At h = 2: 293 - 6.5(2) = 293 - 13 = 280, so 80000 = A(280)^b Step 2: eliminate A by dividing the first equation by the second. The unknown A cancels. 101300/80000 = A(293)^b / [A(280)^b] = (293/280)^b 1.26625 = (1.046428...)^b Step 3: take natural logarithms of both sides to bring b down, then make b the subject. ln(1.26625) = b × ln(1.046428...) 0.236060 = b × 0.045383 b = 0.236060/0.045383 = 5.2015 (4 d.p.) Note b comes out POSITIVE. That is what the physics requires: as h rises, the base 293 - 6.5h falls, and a positive power of a falling base gives falling pressure. Step 4: substitute b back into either original equation to find A. A = 101300/(293)^5.2015 (293)^5.2015 = 6.784×10¹² A = 101300/(6.784×10¹²) = 1.4935×10⁻⁸ Step 5: evaluate the model at Everest's summit. The altitude is 8848 m = 8.848 km, so h = 8.848. 293 - 6.5(8.848) = 293 - 57.512 = 235.488 P = 1.4935×10⁻⁸ × (235.488)^5.2015 P = 1.4935×10⁻⁸ × 2.1767×10¹² P = 32500 Pa (3 s.f.)