#### Transcript Where should a pilot start descent?

**Where should a pilot start descent?**

### By: Alex, Chelsea, Gio, Ian, Jake, and Tessa

**The Problem**

### An approach path for an aircraft landing in shown in the figure and satisfies the following conditions…

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

• • • • •

**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