quadratic equation?

faolan · #1 $Old$ November 17th, 2009, 07:19 PM

I want to create a quadratic equation. I understand the form is Ax^2 + Bx + C = 0. However, the instructor says we can use the form y = +/- x^2.

If I wanted to create a quadratic equation simulating a bird falling out of the sky to the earth...where would I begin? I'm not sure what factors to bring into such a question? The initial height? The speed of flight? How do I translate any question regarding a bird falling to hit the earth in this quadratic equation? This is the first time I've studied this stuff at all, so I could use a little direction in construction a word problem.

Bacterius · #2 $Old$ November 17th, 2009, 07:23 PM

Firstly, you could try to draw a "half-parabola" where the $x$ value is the time in seconds and the $y$ value is the height of the bird. So it would look a bit like the right part of the graph of $y = -x^2$ . Then you try to make an equation that fits the particular curve you get. You can use the y-intercept point (the value of $y$ when $x = 0$ , that is the initial height), the x-intercept point (the value of $x$ when $y = 0$ ) and other features of the parabola to model an equation fitting a bird falling from the sky. You must always consider the $x$ unit (time in seconds) and the $y$ unit (height in ... meters ?).

mr fantastic · #3 $Old$ November 17th, 2009, 07:28 PM

Quote:

Originally Posted by faolan $View Post$

I want to create a quadratic equation. I understand the form is Ax^2 + Bx + C = 0. However, the instructor says we can use the form y = +/- x^2.

If I wanted to create a quadratic equation simulating a bird falling out of the sky to the earth...where would I begin? I'm not sure what factors to bring into such a question? The initial height? The speed of flight? How do I translate any question regarding a bird falling to hit the earth in this quadratic equation? This is the first time I've studied this stuff at all, so I could use a little direction in construction a word problem.

Do you want to simulate the path followed by the falling bird or the height of the bird from the ground at time t? Either way, you will need to make the (unrealistic) assumption that air resistance is negligible.

faolan · #4 $Old$ November 17th, 2009, 07:46 PM

Quote:

Originally Posted by mr fantastic $View Post$

Do you want to simulate the path followed by the falling bird or the height of the bird from the ground at time t? Either way, you will need to make the (unrealistic) assumption that air resistance is negligible.

I think the height of the bird from the ground at time t would be best. Air resistance would complicate this too much for me, I think.

I put y = 100 feet in the air, and x = 10 seconds, just to see what happens, and lost any sense of a parabola, so I knew I was approaching that wrong.
y would therefore be the unknown variable, right?

faolan · #5 $Old$ November 17th, 2009, 07:49 PM

Quote:

Originally Posted by Bacterius $View Post$

Firstly, you could try to draw a "half-parabola" where the $x$ value is the time in seconds and the $y$ value is the height of the bird. So it would look a bit like the right part of the graph of $y = -x^2$ . Then you try to make an equation that fits the particular curve you get. You can use the y-intercept point (the value of $y$ when $x = 0$ , that is the initial height), the x-intercept point (the value of $x$ when $y = 0$ ) and other features of the parabola to model an equation fitting a bird falling from the sky. You must always consider the $x$ unit (time in seconds) and the $y$ unit (height in ... meters ?).

I think I made the mistake of entering in the value of 100 for the value of the height of the bird and 10 for the seconds...I wasn't sure what to put in there, but the nice parabola vanished.

Should it look something like this ? y (height from earth) = 10 (seconds)^2
...I'm not sure I can plug my idea into an equation this small (y = +/- x^2).

Grandad · #6 $Old$ November 17th, 2009, 11:44 PM

Hello faolan

Quote:

Originally Posted by faolan $View Post$

I want to create a quadratic equation. I understand the form is Ax^2 + Bx + C = 0. However, the instructor says we can use the form y = +/- x^2.

If I wanted to create a quadratic equation simulating a bird falling out of the sky to the earth...where would I begin? I'm not sure what factors to bring into such a question? The initial height? The speed of flight? How do I translate any question regarding a bird falling to hit the earth in this quadratic equation? This is the first time I've studied this stuff at all, so I could use a little direction in construction a word problem.

A body falling freely under gravity (that is, without air resistance - which in the case of a bird falling is a gross assumption!) has a height, $h$ , above ground, after time $t$ , given by the equation:

$h=ut-\tfrac12gt^2$ , where $g$ is the acceleration due to gravity and $u$ is its initial velocity, measured vertically upwards.

This is a quadratic expression in terms of $t$ , so you should be able to work out something based on this formula.

Grandad

rowe · #7 $Old$ November 18th, 2009, 03:15 AM

Quote:

Originally Posted by Grandad $View Post$

Hello faolanA body falling freely under gravity (that is, without air resistance - which in the case of a bird falling is a gross assumption!) has a height, $h$ , above ground, after time $t$ , given by the equation:

$h=ut-\tfrac12gt^2$ , where $g$ is the acceleration due to gravity and $u$ is its initial velocity, measured vertically upwards.

This is a quadratic expression in terms of $t$ , so you should be able to work out something based on this formula.

Grandad

Hi, a question on this, if we set initial velocity to 0 and rearrange:

$-\tfrac12gt^2-h=0$

We get an "inverse" parabola for positive h. So if, I set h = 3 meters, we get complex solutions to the polynomial. If I flip the sign to:

$-\tfrac12gt^2+h=0$

Then on a plot, you can get the time from h to hitting the ground at the zero of the function. Why does it have to modified to give this result? I mean, I understand the maths behind it, but why is this the formula when the results give imaginary solutions when solved for different variables (in this case, t)?

EDIT: Although Im(t) = the root of the second equation, interestingly

Grandad · #8 $Old$ November 18th, 2009, 03:39 AM

Hello rowe

Quote:

Originally Posted by rowe $View Post$

Hi, a question on this, if we set initial velocity to 0 and rearrange:

$-\tfrac12gt^2-h=0$

We get an "inverse" parabola for positive h. So if, I set h = 3 meters, we get complex solutions to the polynomial. If I flip the sign to:

$-\tfrac12gt^2+h=0$

Then on a plot, you can get the time from h to hitting the ground at the zero of the function. Why does it have to modified to give this result? I mean, I understand the maths behind it, but why is this the formula when the results give imaginary solutions when solved for different variables (in this case, t)?

Yes, sorry, I should have been a bit more specific.

The equation

$h=ut-\tfrac12gt^2$

assumes that the object in question starts from ground level with an initial speed of $u$ upwards (as you'll see, because $h=0$ when $t = 0$ ).

If the object starts at an initial height $h_0$ , then we need to add on $h_0$ to get:

$h=ut-\tfrac12gt^2 + h_0$

You'll find that will answer your question about real and complex roots.

Grandad