# A small aircraft starts its descent from an altitude of h = 4/3 mile, 4 miles west of the runway....

## Question:

A small aircraft starts its descent from an altitude of h = 4/3 mile, 4 miles west of the runway.

a. Find the cubic {eq}f(x) = ax^3 + bx^2 + cx + d {/eq} on the interval {eq}(-4, 0] {/eq} that describes a smooth glide path for the landing.

b. When would the plane be descending at the greatest rate?

## Hermite Approximations to Functions:

Suppose {eq}f(x) {/eq} is a function, and we're given the values {eq}f(a),f'(a),f(b),f'(b) {/eq} of {eq}f {/eq}. Then there is a unique polynomial {eq}P(x) {/eq} whose degree is 3 or less, such that

{eq}\begin{align*} P(a)&=f(a)\\ P'(a)&=f'(a)\\ P(b)&=f(b)\\ P'(b)&=f'(b) \, . \end{align*} {/eq}

This polynomial {eq}P {/eq} is called the (cubic) Hermite approximation of {eq}f {/eq} on the interval {eq}[a,b] {/eq}.

If {eq}x {/eq} is the east-west distance from the runway in miles, with positive distance to the east, then the cubic {eq}f(x)=ax^3+bx^2+cx+d {/eq} must satisfy the following conditions:

{eq}\begin{align*} f(-4)&=\frac{4}{3}&&\text{(because the altitude of the plane is }\frac{4}{3}\text{ miles when the plane is }4\text{ miles to the west)}\\ f'(-4)&=0&&\text{(because the plane is starting its descent at this location, so it is not already descending at this location)}\\ f(0)&=0&&\text{(because the plane should land on the runway)}\\ f'(0)&=0&&\text{(so that the landing on the runway will be smooth).} \end{align*} {/eq}

Differentiating {eq}f {/eq} gives {eq}f'(x)=3ax^2+2bx+c {/eq}.

Since {eq}f(0)=0 {/eq}, we have {eq}a(0^3)+b(0^2)+c(0)+d=0 {/eq}, and so {eq}d=0 {/eq}. Since {eq}f'(0)=0 {/eq}, we have {eq}3a(0^2)+2b(0)+c=0 {/eq} and so {eq}c=0 {/eq}.

Thus {eq}f {/eq} is of the form {eq}f(x)=ax^3+bx^2 {/eq}, and {eq}f'(x)=3ax^2+2bx {/eq}. Letting {eq}x=-4 {/eq} and applying the above conditions, we have:

{eq}\begin{align*} \frac{4}{3}&=f(-4)\\ &=a(-4)^3+b(-4)^2\\ &=-64a+16b\\ 0&=f'(-4)\\ &=3a(-4)^2+2b(-4)\\ &=48a-8b \, . \end{align*} {/eq}

Since {eq}48a-8b=0 {/eq}, we have {eq}b=6a {/eq}. Substituting this into the first equation gives:

{eq}\begin{align*} -64a+16b&=\frac{4}{3}\\ -64a+16(6a)&=\frac{4}{3}\\ 32a&=\frac{4}{3}\\ a&=\frac{4}{3(32)}\\ &=\frac{1}{3(8)}\\ &=\frac{1}{24} \, . \end{align*} {/eq}

So {eq}b=\dfrac{6}{24}=\dfrac{1}{4} {/eq}. That is, the cubic {eq}f(x) {/eq} is {eq}\boxed{f(x)=\frac{1}{24}x^3+\frac{1}{4}x^2 \, .} {/eq}

b. Using the above expression for {eq}f {/eq}, we have

{eq}\begin{align*} f'(x)&=\frac{1}{8}x^2+\frac{1}{2}x\\ &=\frac{1}{8}x(x+4) \, . \end{align*} {/eq}

The plane will be descending most rapidly when {eq}f'(x) {/eq} is minimized (that is, is as negative as possible). Since {eq}f'(x) {/eq} is quadratic with positive leading coefficient, the minimum of {eq}f'(x) {/eq} will occur halfway between the roots of {eq}f'(x) {/eq}. By the above factored form, these roots occur at {eq}x=0 {/eq} and {eq}x=-4 {/eq}, so the minimum will occur when {eq}x=-2 {/eq}.

That is, the plane will be descending at its greatest rate when it is {eq}\boxed{2\text{ miles west}} {/eq} of the runway. 