Question

A small aircraft starts its descent from an altitude of h = 1.4 miles, 4 miles west of the runway

Answer 1

Answer:

Step-by-step explanation:

From the question we are told that

Altitude of Height 1.4miles

4miles west

a) From the question

$f(x)=ax3+bx2+cx+d$ on interval (-4,0)

The descent can be expressed in $\triangle _x =-4,0$

therefore  initial points is (-4 , 1.4).

final co-ordinate is given by (0,0) as its landing

Mathematically solving in above  cubic equation

 $0 = 0 + 0 + 0 + d$

$d = 0$

$1.4 = -64a + 16b -4c + d$

$1 = -45.7a + 11.4b-2.9c+d$

Generally in this case Slope of tangent at point of descent is zero.assuming flight was initially straight before descent

     $f'(x) = 3ax^2 + 2bx+c$

    $f'(-4) = 48a -8b + c = 0$

    $f'(0) = c = 0$

Mathematically solving above equations we have

$a = \frac{1}{40} \ and \ b = \frac{3}{20}$

Therefore having found a and b the Cubic equation is given as

$f(x) = \frac{1}{40}x^3 + \frac{3}{20}x^2$

b)descent at its greatest rate

$f(x)=\frac{1.4}{40} x^3 + \frac{129}{20} x^2$

$f(x)=\frac{1.4}{40} x^3 + \frac{129}{20} x^2$