calculus?!

[Piney.z]

the equation given is

a) find the gradient function,
and soo i got


and then draw the graph of dy/dx.

but thenn laterr it ask to
c) sstate the coordinates of the point on the track for which the magnitude of the gradient is maximum.

and thus.. now i'm stuck    i dont really get the minimum and maximum thing......
help? 

[excal]

I would imagine, to find the points where the gradient is the maximum (or minimum), you would need to take the the derivative of your original function and find the derivative of THAT.

(In other words, the rate of change of your gradient function).

Consider your circular functions sin/cos. When you have the maximum or minimum, the gradient will be zero. Thus, when you have your derivative of the gradient function (or double derivative of the original function), you just need to find the solutions to it.

Of course, you'll need to work out which ones are the maximums and minimums, but you should be able to do that when you graph all three ...

[cara.mel]

Excalibur is right. One way you can do it is take the derivative again. This is creatively called the second derivative and is in specialist maths.

However, his "Consider your circular functions sin/cos" without continuing on the proper way of thinking reminded me on something.
On a normal sine/cosine curve, the gradient is the maximum when it crosses the x axis. So they will be in those spots.
(you can confirm this by taking the derivative and solving for 0, which is solving cos pi x/80=0. You just don't actually need to calculate anything new

[excal]

I'm getting there!

[Piney.z]

mmm. okay.
i think i'm following with what you're saying.

one question though,
if we're doing both methods and specialist..
can we use the knowledge of spesh to apply it in methods exams?
such as the second derivative which isn't learn in methods?

I'm getting there!

Lol! okok 

[cara.mel]

one question though,
if we're doing both methods and specialist..
can we use the knowledge of spesh to apply it in methods exams?
such as the second derivative which isn't learn in methods?

You should generally try to avoid it. Eg on the topic of this, if in methods you have to show something is a min/max, don't find 2nd derivative, do one of them sign diagram things.

And joe, don't be like everyone else and post
-question line here-
*working*
then edit 50 times, so you get glory of first post. xD

[excal]

Alternatively, if you want stick to purely methods techniques, consider that the period is where . Also consider that your amplitude is



given we know that for max and for min (there are no vertical translations to consider)

Now our horizontal dilation factor is . Plug this into our dilation formula . This gives . Half period therefore is 80; quarter period is 40. Thus our first stationary point is at . Because there is reflection in the x-axis due to the negation of your sine function, the first stationary point is a minimum at and your maximum is at

[excal]

So you don't complain about first posting.

[cara.mel]

So you don't complain about first posting.

Thanks. I love you :smitten:

[ed_saifa]

mmm. okay.
i think i'm following with what you're saying.

one question though,
if we're doing both methods and specialist..
can we use the knowledge of spesh to apply it in methods exams?
such as the second derivative which isn't learn in methods?

I'm getting there!

Lol! okok 

It is ok to use specialist techniques in methods providing the question doesn't specifically ask for a particular method to be used. For example, using the second derivative in place of a sign diagram.

[excal]

I think the general consensus I got when I did methods (in a class with lots of spesh kids) was no. I did not do specialist, so I don't know what skillz they would've brought over.

[Piney.z]

Hmm..
is there anywhere which was said by VCAA that we're allowed/ not allowed?
since there's 2 answers.... 

 #######################

ohh. i forgot to mention that
does that make a difference?
because the answer is (40,12)
...

[cara.mel]

Yeah, the answer is (40,12)

Excalibur == anhero (I'm learning!)

He did 160/4 = 45, and then used wrong equation somehow

Simply:
Find that period = 160
Remeber that cos(x) = 0 at 1/4 and 3/4 of period (40, 120, 200 etc)
Sub in 40 into equation and get y=12

[excal]

Sorry, my bad. I never actually answered the question! Haha.

You plug the back into the original function, as we've found the stationary point (the point which the graident is the highest).

So,








therefore, (40,12).

I'm not thinking straight.

[Piney.z]

AHhh!~
gotcha now. =]
tyty

cool beans xD