VCE Specialist 3/4 Question Thread!

[Phy124]

Is there are way to find all the posts that you've made in this thread? I need help with a question but I'm pretty sure I already asked about it a while back 

You can look through posts you've made by going to your profile and clicking the Show Posts link. You can then find the post in context by clicking the "Re: Specialist Question Thread!" or whatever the relevant link is.

[kinslayer]

Is there are way to find all the posts that you've made in this thread? I need help with a question but I'm pretty sure I already asked about it a while back 

I don't know if there is a button anywhere that does it, but try this link:

Specialist Question Thread!

That will show your own posts in this topic (128233) enjoy

[vintagea]

how do i sketch a slope field diagram?
for example? dy/dx = x for the x values -2 to 2
and state the equation for the general solution curve?

thank you

[keltingmeith]

Slope fields look like this. To sketch it, find what the gradient is at any point (a,b), then draw that gradient at that point.

So, at (0,0), dy/dx = 0, so draw a line with gradient zero at that point. At (1,0), dy/dx = 1, so draw a line with gradient 1 at that point. Do this for all the points in the region they've given you.

A slope field gives the shape of the curve when in the form . So, look at what the shape looks like, then see if you can guess what the general solution curve should be.

[hyunah]

i just wanted to double check
y = tan(2x - pi/8) has a maximal domain of -3pi/16<x<5pi/16 right?

[keltingmeith]

Nope, sorry - both tan and arctan have a maximal domain of R. However, if you did mean arctan, and the domain was changed to -pi/2<x<pi/2, you are right.

[hyunah]

oh, cause the question was find the particular solution, y, to dy/dx = 2+2y^2 if x=pi/4 when y = 1 and state the maximal domain of the particular solution?
did i get it right?

[keltingmeith]

Nope, sorry - the solution is , and its maximal domain is .

[M_BONG]

Need help!

Can anyone spot my mistake?

Question (from Maths Quest)

"A body moves in a line so that its acceleration is (x+1) m/s^2 and when x=2, v=3

Find where the body is when the velocity is 0m/s

My working is in attachment.. I apologise if it's a stupid mistake (I got home at 10PM today and starting maths now lol so probs really stupid mistake cuz I am tired).
Thanks!

[keltingmeith]

I see that everyone's starting their specialist holiday homework now.

Your mistake is in the second line - dv/dt does not mean you anti-diff the RHS with respect to x, you have to do it with respect to t.

On your formula sheet, you will see this:

The reason for this is that you need to know which of these many differential equations to use. In this case, I'd go with the last one. This means we have:



Recognising which fraction of differentials to use comes with practice, so don't stress about not getting it at first.

[M_BONG]

I see that everyone's starting their specialist holiday homework now.

Your mistake is in the second line - dv/dt does not mean you anti-diff the RHS with respect to x, you have to do it with respect to t.

On your formula sheet, you will see this:

The reason for this is that you need to know which of these many differential equations to use. In this case, I'd go with the last one. This means we have:



Recognising which fraction of differentials to use comes with practice, so don't stress about not getting it at first.

Thank you so much for the prompt reply!
Much appreciated

[lzxnl]

In short, if you have dv/dt = f(x), for instance, the dt means integrating the left hand side to get v means you're integrating the right hand side with respect to t. Which, here, is of no help
You want to turn the dt into a dx. If you had dv/dx = f(x), you'd then be able to directly integrate both sides
Hence, you write dv/dt = dv/dx * dx/dt = v dv/dx = f(x) by chain rule
Now you can integrate

[keltingmeith]

In short, if you have dv/dt = f(x), for instance, the dt means integrating the left hand side to get v means you're integrating the right hand side with respect to t. Which, here, is of no help
You want to turn the dt into a dx. If you had dv/dx = f(x), you'd then be able to directly integrate both sides
Hence, you write dv/dt = dv/dx * dx/dt = v dv/dx = f(x) by chain rule
Now you can integrate

Spec doesn't cover separation of variables (yet), so he needs the extra step to get it to , else the differential equation would be "unsolvable".

[keltingmeith]

If you draw its acceleration graph, you'll notice that the object has a positive acceleration for all x > 0 - so, it can't be moving backwards, and we discount the negative velocity, as this implies backwards motion.

[tiff_tiff]

i have an euler's method question
how would i use the euler's method with a step size of h = 0.5, calculate the approx. value of sol. of dy/dx = (3-y)/x at x = 2.5, with initial conditions y=0 and x= 1?