Investigation 2: Space Travel

I hope you're ready for our second investigation.

It connects quadratic functions to space travel.

So fasten your belts and let's take off.

Space travel began in the late 1950s and it's still primarily

reserved for astronauts.

But there's an ongoing race among aerospace companies to

develop the first commercial carrier that will take

tourists into space.

In fact, several prototype spaceships are expected to be

ready for test flights within two years.

So in your lifetime, civilian space travel may become an

everyday reality.

Understanding the mathematics behind it will help you

appreciate it all the more.

Just think, one day you may stare at the Earth the way

today you stare at the moon.

Let's take an imaginary trip aboard a space shuttle to the

International Space Station, or ISS, and investigate where

quadratic functions fit in.

Here is the problem, at take-off, a space shuttle has

0 altitude and it's projected upward by a

powerful initial velocity.

At about 120 kilometers above the earth's surface, it cuts

off its main engine.

At that point, the shuttle's vertical velocity is 2,110

meters per second, or 2.11 kilometers per second.

It still must rise to 340 kilometers above sea level,

the minimum orbiting altitude of the ISS.

Here is the question, how long will it take the shuttle to

reach the ISS located at 340 kilometers above the earth

after its main engine is turned off?

Let's first examine the algebraic expression of the

altitude and the function of time.

The quadratic equation, giving the vertical altitude or

height, h, of any projectile, a space shuttle in our case,

as a function of time, t, is h of t equals negative one half

gt squared plus b sub 0 t plus h sub 0.

Where negative one half g is the a coefficient, c sub 0 sub

d coefficient, h sub 0 the c coefficient.

g stands for the acceleration due to gravity.

Namely, 9.8 meters per second squared.

Or 0.0098 kilometers per second squared.

p sub 0 is the initial velocity.

We call that the start time for our problem is main engine

cut off time.

So v sub 0 is 2.11 kilometers per second.

And a sub 0 is the initial height.

Again, at main engine cut off time.

So, 8 sub 0 equals 120 kilometers.

Putting all this together, we obtain h of t equals negative

one half times 0.0098 t squared, plus 2.11 t plus 120

or h of t equals negative 0.0049 t squared

plus 2.11 t plus 120.

You're now ready to set up the equation you must solve in

order to answer the original question, which was, how long

will it take the shuttle to reach the ISS after its main

engine is turned off.

The question how long means you're solving for t, the time

in seconds.

The target altitude is 340 kilometers.

So our equation is h of t equals 340.

Or negative 0.0049 t squared plus 2.11 t

plus 120 equals 340.

By solving this equation for t, you will obtain the time

elapsed from main engine cut off to the arrival at the ISS.

Instead of working this out algebraically, let's solve it

graphically, using the TI-Nspire.

Keep in mind that on the handheld, t will be x and h of

t will be f of x.

Such mental gymnastics are good for the mind.

Turn on the TI-Nspire.

Press the home key to open a new document.

Save or delete the previous document.

Select 2 to create a graph in geometry page.

The blinking cursor is on the function entry

line by f 1 of x.

Type in negative 0.0049 x squared, plus 2.11 x plus 120

for our altitude function.

Press enter to graph.

Surprised there's no graph?

It's there, but out of the viewing window.

Press menu 4 and 1 for the window settings.

Set x min to negative 75 in order to see the y-axis.

Press tab to move from one entry to the next.

Set x max to 500.

That's more than eight minutes.

Set y min to negative 50, again to see the x-axis, and y

max to 400.

Well above our target altitude of 340.

Then click OK.

Press escape to move the arrow to the work area.

Now the parabola's in sight.

As expected, it opens downwards because the leading

coefficient, a, is negative.

Since the graph plots height over time, the y-intercept

represents the height y equals 120 kilometers at our starting

time, x equals zero.

So that's our initial height.

Now let's find the points where y equals 340 using a

dynamic feature of the TI-Nspire.

Use the nav pad to move the arrow to a point on the graph.

About halfway between the y-intercept and the vertex.

It becomes a finger pointing hand.

Place a point on the graph.

Click and the pencil will plot it.

Press escape to exit point on mode.

An open hand is now hovering over the point.

Using the nav pad, move the open hand over the points

coordinate.

Click and hold until it changes to a closed hand.

Use the nav pad to drag the coordinate to the center of

the screen.

Press escape to exit grab and drag mode.

Move the pointer back over to that point on the graph until

it becomes an open hand.

Click and hold until it changes to a closed hand.

You're now ready to slide the point to the right along the

graph in search for the points where y equals 340.

Press the right arrow until the

y-coordinate approaches 340.

Notice this first occurs at x equals about 177.

Keep pressing the right arrow until you see an m for

maximum, the highest part of the graph.

Keep sliding to the right until y

approaches 340 once again.

This time, x is about 251.

So we solved our quadratic equation graphically, making

sense of the meaning of the parabola.

The quadratic expression on the left is the altitude.

We needed an altitude of 340 kilometers.

So we traced the graph until we found points

where y equals 340.

The first point yielded a value of x close to 177.

That's our answer.

It takes the space shuttle about 177 seconds, almost

three minutes, to reach the International Space Station

after main engine cut off.

There were two solutions, as for many quadratic equations.

But when solving a real world quadratic problem, one often

discards one of the two theoretical solution as

nonsensical.

That's the case for our second point.

You also could solve the problem algebraically by

subtracting 340 from both sides of the equation and

obtaining the quadratic equation negative 0.0049 t

squared plus 2.11 t minus 220 equals 0.

If you know the quadratic formula, use it to compute the

two exact theoretical solutions.

You would then discard the greater one as we did.

Solving quadratic equations algebraically is the focus of

another video in this same series.

I hope you've enjoyed our investigation.

In closing, let's recap the meaning

of a quadratic function.

Algebraically, quadratic means that the highest exponent of x

or the degree of the polynomial is 2.

Geometrically, quadratic means that the

function's graph is a parabola.

Now that you know that quadratic functions can be

found in all sorts of real world phenomena and objects, I

hope you'll delve deeper into the algebra and geometry of

these relationships with your TI-Nspire.

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