Conditions

1) The cruising altitude is h when descent starts at a

horizontal distance l from touch-down at the origin

1) The pilot must maintain a constant horizontal

speed v through the descent

1) The absolute value of the vertical acceleration

should not exceed a constant k (which is much less than the acceleration due to gravity)

Question #1

Find a cubic polynomial p(x)= ax 3 +bx 2 +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.

Question #1 Work

Rates

Distance: p(x)= ax 3 +bx 2 +cx+d Velocity: p’(x)= 3ax 2 +bx+c Acceleration: p’’(x)= 6ax+2b

12. Slide 12

13. Slide 13

• • • • •

We found…

d=0, by plugging 0 into the distance equation c = 0, by setting the velocity equation (the tangent line) a in terms of b and L Knowing that p(L)=h, we plugged a back into the distance equation to find b in terms of h and L We 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 L

Solution

p(x) = -2hx 3 /L 3 + 3hx 2 /L 2

Question #2

Use the conditions 2 and 3 to show that

6hv 2 /L 2 < k • • • •

Given

a = -2h/L 2 b = 3h/L 3 p(x) = ax 3 +bx 2 x = L

3.

2.

Question #2 Work

Steps

1.

Find the first implicit derivative of p(L) dp/dt = 3aL 2 (dx/dt)+2b(dx/dt) note: dx/dt = -v Find the second implicit derivative of p(x) d 2 p/dt 2 = -6avL(dx/dt)-2bv(dx/dt) d 2 p/dt 2 = -6avL(-v)-2bv(-v) d 2 p/dt 2 = -6Lav 2 -2bv 2 note: d 2 p/dt 2 < k Plug in a and b -6Lav 2 -2bv 2 < K -6Lv 2 (-2h/L 3 ) - 2(3h/L 2 )v 2 12hLv 2 /L 3 12hv 2 /L 2 – 6hv 2 /L - 6hv 2 /L 2 2 < K < K 6hv 2 /L 2 < k < K

Question #3

Suppose that an airline decides not to allow vertical acceleration of a plane to exceed k = 860 mi/h descent?

2 . 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

Objective

Find L, which is the point at which the pilot should start the descent

Question #3 Work

• • • • • • • Given: h=35,000 ft converted to 6.62878 mi v= 300 mi/h k=860 mi/h 2 6hv 2 /L 2 < k we now need to plug in our values 6(6.62878)(300) 2 / L 2 < 860 3579541.2 /L 2 < 860 multiply l 2 and divide by 860 to isolate l 2 on both sides on the right side. 33579541.2/860 = 4162.257209

√4162.257209 < √L 2

L=64.515 miles When the plane is 64.515 miles away the pilot should start the descent to the airport

Question #4

• Graph the approach path if the conditions stated in Problem 3 are satisfied.

In order to graph we use the original equation which was ax one we found that c and d equaled 0, the equation is ax 3 3 + bx + bx 2 2.

+ cx+d. Because in question • 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.

Question #4 Work

Given

a= -2h/l 3 b=3h/l 2 Height=6.629

Length=64.517

Steps

A= -2(6.629)/(64.517) 3 A=-.00004937

B=.004777

Now you plug in A and B to the original equation B=3(6.629)/(64.517) 2

Solution

p(x)=-.00004937x

3 + .004777x

2 Window  xmin=0, xmax=64.5(length), ymin=0, ymax=6.6 (height)

Walk away…

P(0) = 0 P’(0) = 0 P’(L) = 0 P(L) = h

Question

5. Question