A Question About Delta

[NNNOOOOOO 0]

If you have (delta)x, do they both act as a single variable, or as two separate variables.

[rob86 2]

If you have (delta)x, do they both act as a single variable, or as two separate variables.

Delta means "change in" so Delta x would mean

X2 - X1 = DeltaX

The original number always goes second. For example when calculating change in temperature, the original temperature is X1.

10-7 = 3
7-10 = -3

The negative is most likely important in physics.. unless your instructions tell you to make it an absolute number (no negative signs)

It's used a lot when calculating speeds..so think of it like
car going 100mph slows down to 50mph.. what is the change in velocity?
DeltaV = Vf - Vo
DeltaV = 50 - 100
DeltaV = -50

Change in velocity, DeltaV is -50mph.

It's been a while..so if i'm wrong someone can answer more correctly.

[Bikerman 2]

No need, you nailed it. ΔV = V1-V2 (ie take (subtract) the end value from the start value and that is your delta (difference))You will normally see delta as the capital Greek letter above (an equilateral triangle). If you see the lowercase Greek letter (δ) then that is likely to be a related concept called a differential - then you get into calculus - but don't be scared off by that because it is really useful and the maths gets very pretty.At the intro level then you will be meeting delta as differences in velocity, as stated previously. You might also meet it as a way of calculating the slope of a straight line graph:ΔY/ΔX=slope

Edited August 14, 2010 by Bikerman (see edit history)

[rvalkass 5]

Delta means "change in" in mathematics. It is therefore not a variable but an operator, and cannot be treated as a variable. For example, the differential δx/δt cannot be simplified by "dividing" top and bottom by δ, as mathematically that makes no sense. However, you can treat δx as a whole as a variable. For example:

 

δx/δt = 15t + 7

δx = ( 15t + 7 ) δt

x = 7.5t² + 7t + c

 

The δt can be used as one thing if it makes mathematical and physical sense at the time. However, you shouldn't split the δ and t (or x or whatever variable) as you would be changing the meaning and value.

[Bikerman 2]

Now I think you might have frightened the poster by jumping into calculus before they got difference sorted, which is why I left it out....just a thought (I hate to see people scared of maths - it is fun as well as a vital tool/skill)

[BuffaloHelp 24]

I just realized how long it's been since my last calculus... thanks for making me count my age :(ha ha

[Bikerman 2]

I just realized how long it's been since my last calculus... thanks for making me count my age
ha ha

If you are older than 48 then whoops sorry dad, but if you are younger then stop whinging sonny :-)

[NNNOOOOOO 0]

So delta just means "change in". Thanks. It was really confusing me.

[Bikerman 2]

So delta just means "change in". Thanks. It was really confusing me.

Yep that's all it is. As has been said, don't think of it as part of the variable, it is just something you do to it. It's not quite the same as a multiply or add because it doesn't actually operate on the variable directly - it acts on two values of the variable - an old one and the current one. It is assumed that the variable is doing what the name implies - varying - changing. Obviously there would be no point trying to measure the delta for the height of a car - you wouldn't expect it to change (much anyway), so why measure it? Velocity - ah that changes. Faster - more velocity (in a paricular direction) which is what we get for pressing the accelerator pedal. So acceleration is change in velocity and velocity is change in distance. We can now use our new symbol to write it properly in maths
Velocity = Δdistance/time
acceleration = Δvelocity/time

Now, if we travel at velocity 10m/s for 10 seconds we can do some sums:
10=Δdistance/10
Δdistance = 10*10=100 metres so we travel 100 metres.
And
if we start at 10m/s and then after 5 seconds we are doing 20m/s how much acceleration?
acceleration = 10/5 = 2 m/s^2 or 2 metres per second per second. So every second we get another 2 metres per second quicker...

With me?
Edited August 14, 2010 by Bikerman (see edit history)

[NNNOOOOOO 0]

Yep that's all it is. As has been said, don't think of it as part of the variable, it is just something you do to it. It's not quite the same as a multiply or add because it doesn't actually operate on the variable directly - it acts on two values of the variable - an old one and the current one. It is assumed that the variable is doing what the name implies - varying - changing. Obviously there would be no point trying to measure the delta for the height of a car - you wouldn't expect it to change (much anyway), so why measure it? Velocity - ah that changes. Faster - more velocity (in a paricular direction) which is what we get for pressing the accelerator pedal. So acceleration is change in velocity and velocity is change in distance. We can now use our new symbol to write it properly in mathsVelocity = Δdistance/time
acceleration = Δvelocity/time

Now, if we travel at velocity 10m/s for 10 seconds we can do some sums:
10=Δdistance/10
Δdistance = 10*10=100 metres so we travel 100 metres.
And
if we start at 10m/s and then after 5 seconds we are doing 20m/s how much acceleration?
acceleration = 10/5 = 2 m/s^2 or 2 metres per second per second. So every second we get another 2 metres per second quicker...

With me?

Yes, I'm with you, as this is on my notes packet. Also on the packet of notes:
Geometry stuff
displacement, and distance

I'm not looking at the notes right now, so I can't name the rest of the stuff on it.

[Bikerman 2]

Well that is delta covered.

Obviously the delta value will give you an average. So the two examples I gave (velocity and acceleration) would give the result as an overal average velocity or acceleration for the time taken (t).

 

Sometimes you want an instant reading whilst you move. Then you have to make the value of t very small.

 

Heres a graph of the car moving:

You can see it is a straight line when we plot distance in km against time in hours.You should be able to tell, from the graph, what the velocity is, in km/h. How? Well the velocity is given by the slope of the line.

 

 

If we divide d by t then distance/time = velocity.

or as we now know

Δd/Δt = v

(difference in distance divided by different in time = velocity)

 

But what about if we choose something else - like a ball? Throwing a ball up in the air and catching it produces the following graph....

 

 

Now you can see the ball is not straight-lined but a curve - actually it is a special curve called a parabola.

 

Can you see what the graph shows? It shows the ball starting off at a fast velocity then slowing down continually until it reaches the top of the graph and stops for an instant, then slowly at first starts to come back down, speeding up all the time.

 

So how could we get a value for the slope of this line (the velocity)?

 

Clearly it is constantly changing.

 

 

This is where the lower case greek delta comes in. It means a very very tiny change.

 

If you imagine ddrawing the same triangle on this graph as we did for the car, then we would get:

 

 

Trouble is that the hypotenuse of the triangle (the curve) is not a straight line.

 

If we make our triangle smaller and smaller, then the line gets straighter and straighter.

 

Newton worked out a mathematical dodge. He made the sides of the triangle infinitessimally small (almost infinitely small) and then the side would be as near to straight as made no difference.

 

So instead of using big delta for change, we use little delta for tiny tiny change.

 

now we get:

δh/δt = v

(tiny change in height divided by tiny time period is the velocity at that point.

 

Welcome to calculus. The tiny change in h is correctly called the derivative of height. Likewise for the derivative of time. So we can now work out this sum to get the velocity at any point on the graph.

 

No point progressing past this until your course catches up - they may teach it differently....

good luck..

Edited August 15, 2010 by Bikerman (see edit history)

[NNNOOOOOO 0]

Displacement

Δx=x_f(x final)-t_i(t initial)

 

Average speed

average speed=total distance/total time

 

average velocity

average velocity=Δx/Δt=(x_f-x_i)/(t_f-t_i)

 

instantaneous velocity

Δ=approaching zero (Δx/Δt)

 

average acceleration

average acceleration=Δv/Δt=(v_f-v_i)/(t_f-t_i)

 

instantaneous acceleration

instantaneous acceleration=approaching zero (Δv/Δt)

[Bikerman 2]

Displacement

Δx=x_f(x final)-t_i(t initial)

 

Average speed

average speed=total distance/total time

 

average velocity - total distance in a single direction / total time

average velocity=Δx/Δt=(x_f-x_i)/(t_f-t_i)

 

instantaneous velocity - velocity at a particular instant in time

Δ=approaching zero (Δx/Δt) = δx/δt

 

average acceleration - total velocity / total time

average acceleration=Δv/Δt=(v_f-v_i)/(t_f-t_i)

 

instantaneous acceleration - acceleration at one instant of time

instantaneous acceleration=approaching zero (Δv/Δt) = δv/δt

 

Very good. I've just added a few things for completion. You have it sussed, nice one.

[sheepdog 10]

Well shoot! Here was a question I thought I could answer! You guys have done me in again. I used to really like North West Airlines, but they were bought out by Delta, and now shipping with Delta is a real a real pain in the hind end. Delta charges almost $50 more to ship a puppy than North West or American. So if you have to ship a puppy, don't use Delta if you can avoid it. Not really what you wanted to know is it???

[NNNOOOOOO 0]

Well shoot! Here was a question I thought I could answer! You guys have done me in again. I used to really like North West Airlines, but they were bought out by Delta, and now shipping with Delta is a real a real pain in the hind end. Delta charges almost $50 more to ship a puppy than North West or American. So if you have to ship a puppy, don't use Delta if you can avoid it.
Not really what you wanted to know is it???

*laughing* no....