Presentation on theme: "Where should a pilot start descent?"— Presentation transcript:
1Where should a pilot start descent? By: Alex, Chelsea, Gio, Ian, Jake, and Tessa
2The ProblemAn approach path for an aircraft landing in shown in the figure and satisfies the following conditions…
3ConditionsThe cruising altitude is h when descent starts at a horizontal distance l from touch-down at the originThe pilot must maintain a constant horizontal speed v through the descentThe absolute value of the vertical acceleration should not exceed a constant k (which is much less than the acceleration due to gravity)
4Question #1Find a cubic polynomial p(x)= ax3+bx2+cx+d that satisfies condition 1 , by imposing suitable conditions on p(x) and p’(x) at the start of descent and at touchdown. Objective Find constants a, b, c, and d that justify an equation for the plane’s flight path.
5Question #1 Work Rates Distance: p(x)= ax3+bx2+cx+d Velocity: p’(x)= 3ax2+bx+cAcceleration: p’’(x)= 6ax+2bWe found…d=0, by plugging 0 into the distance equationc = 0, by setting the velocity equation (the tangent line)a in terms of b and LKnowing that p(L)=h, we plugged a back into the distance equation to find b in terms of h and LWe had an equation for a in terms of b and L, so we plugged our value for b back in to find a in terms of h and L12. Slide 1213. Slide 13Solutionp(x) = -2hx3/L3 + 3hx2/L2
6Use the conditions 2 and 3 to show that Question #2Use the conditions 2 and 3 to show that6hv2/L2 < kGivena = -2h/L2b = 3h/L3p(x) = ax3+bx2x = L
7Question #2 Work Steps Find the second implicit derivative of p(x) Find the first implicit derivative of p(L)dp/dt = 3aL2(dx/dt)+2b(dx/dt)note: dx/dt = -vFind the second implicit derivative of p(x)d2p/dt2 = -6avL(dx/dt)-2bv(dx/dt)d2p/dt2 = -6avL(-v)-2bv(-v)d2p/dt2 = -6Lav2-2bv2note: d2p/dt2 < kPlug in a and b-6Lav2 -2bv2 < K-6Lv2(-2h/L3) - 2(3h/L2)v2 < K12hLv2/L3 – 6hv2/L2 < K12hv2/L2 - 6hv2/L2 < K6hv2/L2 < k
8Question #3Suppose that an airline decides not to allow vertical acceleration of a plane to exceed k = 860 mi/h2. If the cruising altitude of a plane is 35,000 ft. and the speed is 300 mi. how far away from the airport should the pilot start descent? Objective Find L, which is the point at which the pilot should start the descent
9Question #3 WorkGiven: h=35,000 ft converted to mi v= 300 mi/h k=860 mi/h26hv2/L2 < k we now need to plug in our values6( )(300) 2/ L2 < 860/L2< 860 multiply l2 on both sides and divide by 860 to isolate l2 on the right side./860 =√ < √L L= milesWhen the plane is miles away the pilot should start the descent to the airport
10Question #4Graph the approach path if the conditions stated in Problem 3 are satisfied.In order to graph we use the original equation which was ax3 + bx2 + cx+d. Because in question one we found that c and d equaled 0, the equation is ax3 + bx2.In problem 2 we found what a and b equaled in terms of h and l, now that we have what h equals and l equals from number 3, we plug it in for a and b.
11Question #4 WorkGivena= -2h/l3b=3h/l2Height=6.629Length=64.517StepsA= -2(6.629)/(64.517) B=3(6.629)/(64.517)2A= B=Now you plug in A and B to the original equationSolutionp(x)= x x2Window xmin=0, xmax=64.5(length), ymin=0, ymax=6.6 (height)