Slope Fields

[Zebra]

Can someone please help me with Slope fields/curves?
I don't even know how to approach these questions.....

is it subbing in points or is there a way to do it?
argh!

for eg. 2006 exam 2 slope field question (short answer question that wants you to draw the slope curve)....

HOW? HOW? HOW do you do it?

thanks!

[jane1234]

Essentially, you join the dots.

Slope fields give the gradients of particular points for a FAMILY of curves. Hence why they are little dashes - the co-ordinates of the dash gives the x and y values for a particular point, and the gradient of the dash gives you the gradient of that point on the actual curve.
So slope fields, in a very very general sense, just kind of show a broken-up version of a family of curves. If they want a particular curve, they will give you a point and all you have to do is trace over the little dashes that follow that point ONLY (like in VCAA 2006 2).

It's kind of a weird concept, but practice drawing them and practice trying to find the equation from them. Also, if it asks you to match a differential equation to a slope field, first antidifferentiate the differential equation to get the general idea of the family of curves.

Hope that helped, best way to understand this is just to practice. Also, just draw random differential equation slope fields on your calculator, you'll get the general idea.

[Zebra]

:S so say you are given a slope field and nothing. can you work out the general graph equation?

I am so lost. SOrry!

Hey jane when you say family of curves you mean all the curves that undergo vertical translation right? y=x^2 +c (c is the constant)...

Thanks

[jane1234]

:S so say you are given a slope field and nothing. can you work out the general graph equation?

I am so lost. SOrry!

Hey jane when you say family of curves you mean all the curves that undergo vertical translation right? y=x^2 +c (c is the constant)...

Thanks

Yeah, family of curves is just like y = x^2 + c

Okay so for this:

|        |
 \      /
  \ _ /

You can kind of see it is a parabola yeah (if it wasn't a crappy diagram)?
So what if we had the vertical translations?

It would look like a bunch of dotted parabolas on top of each other:

| |  |   | | |
 \ \ \ _/ / /
  \ \ _/ / /

^ See how each is just a vertical translation (well, if it wasn't a terrible representation)?

So if a question asks you to sketch the curve that passes through point (0,0) on the slope field, all you have to do is trace the parabola that passes through this point.

VCAA 2006 exam 2 makes you work out dy/dx, and then gives you the slope field. So all you had to do was solve  the differential equation to get y in terms of x, and this will give you a "family" of curves, which should resemble the slope field. You are also given a boundary condition, so trace over the curve that passes through the specific point.

Make sense?

[HarveyD]

how would you do this one :S

[jane1234]

Haha I think I did that exam...

Well sub each x-point into the differential equation to get the gradient of the dash, then draw identical dashes for each corresponding y-value.

For example:

x = 0

dy/dx = sqrt(0) = 0

This means that the dash will have a gradient of 0 and will look like this: _

So, for every x = 0, put a _

This means (0,0) and (0,1) and (0,2) and (0,3) and (0,4) and (0,5) will all have a little dash that looks like this: _

Another example:

Let's take x = 4

dy/dx = sqrt(x) = sqrt(4) = 2

So the gradient of the little dash at EVERY x = 1 will have a gradient of 2, or look something like this: /

So for points (4,0) (4,1) (4,2) (4,3) and (4,4), you will put a little dash that looks like this: /

Repeat for the other x values (1,2,3)

See how each dash is vertically translated?

[HarveyD]

ahhh yes i see now
thanks!

is that also how you can identify them in the mc part?
i.e. find if its dependent on x or y and then matching with the equation?

[jane1234]

ahhh yes i see now
thanks!

is that also how you can identify them in the mc part?
i.e. find if its dependent on x or y and then matching with the equation?

For MC, with the "match the equation", I'll plot each equation given on the calculator and see which one looks like the one in the question. Simple!

[HarveyD]

oh okay, that'd probably be easier
thanks again

[costa94]

given a graph of a slope field, join the dots and that is the shape of your function
if it asks for you to choose what dy/dx is equal to (like a multichoice) then either
1. work out y as above and diff it
2. look at various x co-ords on the graph and the gradient there (eg. find x where dy/dx = 0 or undefined(vertical)) and then find the dy/dx from choices that fits

[Zebra]

I know this is going to be a big ask. but can someone upload their working out for 2006 slope field question?!?!!?!?!?!?!??!?!?!?!?!??!?!?!? WTAEKTJRHGLKAR HGARLKARGHRAJGHLKARG argh...

[Natters]

these, the more difficult vector proofs and some of the complex number stuff are the only things im not confident with =(

[HarveyD]

do VCAA generally put vector proof questions?
I've done a few exams and havent seen many on the company trials...
is it false hope?

[costa94]

its always easy stuff, proving magnitudes are the same, proving right angles, proving a = k*b (parallel), etc
they never include any of the more difficult stuff from the textbook (which, if you can do them, means youre set)

[Natters]

i can't do a lot of the one's in the text book, though i havent tried in a while