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May 15, 2026, 04:12:50 pm

Author Topic: VCE Methods Question Thread!  (Read 6090796 times)  Share 

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Special At Specialist

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Re: VCE Methods Question Thread!
« Reply #9315 on: March 16, 2015, 12:04:40 am »
+1
Why does the domain of two functions added together become the intersection of the two?

It's easier to show by example:
Let's say f(x) = sqrt(x) and g(x) = sqrt(4 - x)
We can see that f(x) + g(x) = sqrt(x) + sqrt(4 - x)
But what is the domain of f(x) + g(x)?

For sqrt(x) to be defined, we need x to be 0 or more.
For sqrt(4 - x) to be defined, we need x to be 2 or less.
We need to satisfy both conditions for f(x) + g(x) to be defined.
So x must be between 0 and 2 (inclusive).
Thus the domain of f(x) + g(x) is [0, 2]

More generally, let's say f(x) has condition A imposed on it. And let's say g(x) has condition B imposed on it.
What conditions must h(x) have to be defined if h(x) = f(x) + g(x)?
It must have both conditions: A and B
ie. h(x) is only defined for the intersection of A and B (where both conditions are met simultaneously)
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kinslayer

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Re: VCE Methods Question Thread!
« Reply #9316 on: March 16, 2015, 01:56:09 am »
+1
Why does the domain of two functions added together become the intersection of the two?

If h(x) = f(x) + g(x) then h(x) is defined only when both f(x) and g(x) are. The maximal domain for h(x), then, is all x for which both f(x) and g(x) are defined -- but this is nothing more than the intersection of the domains of f and g.

(An element is in the intersection of two sets if and only if it is in both sets).
« Last Edit: March 16, 2015, 02:00:39 am by kinslayer »

cosine

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Re: VCE Methods Question Thread!
« Reply #9317 on: March 16, 2015, 06:34:21 am »
0
Thank you guys, makes sense now!

Just wondering is the range the similar process? Or would I just have to sketch the new graph h(x) and restrict the domain and then determine the range?
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Apink!

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Re: VCE Methods Question Thread!
« Reply #9318 on: March 16, 2015, 09:00:39 am »
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Hi,
just a quick question.
Are we expected to know integration by parts in the methods course?
Thanks!
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #9319 on: March 16, 2015, 09:09:41 am »
0
Hi,
just a quick question.
Are we expected to know integration by parts in the methods course?
Thanks!
Integration by parts is not in VCE, and as far as we can tell it won't be for a while (grrr, VCAA). However, a very similar technique often dubbed "integration by recognition" is in the methods curriculum, which is when they simply ask you to diff something, and use that result to integrate something. Eg, diff x*ln(x), and hence integrate ln(x).

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Re: VCE Methods Question Thread!
« Reply #9320 on: March 16, 2015, 10:06:49 am »
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Oh thanks! I'm glad.
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Special At Specialist

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Re: VCE Methods Question Thread!
« Reply #9321 on: March 16, 2015, 11:59:59 am »
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Hi,
just a quick question.
Are we expected to know integration by parts in the methods course?
Thanks!

No. Not even VCE Specialist Maths teaches integration by parts.
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #9322 on: March 16, 2015, 06:55:04 pm »
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Technically you shouldn't see anything as you can't 'see' a point in 2D :P

That was sorta the point. :P

cosine

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Re: VCE Methods Question Thread!
« Reply #9323 on: March 16, 2015, 06:58:42 pm »
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What the hell is the point of integration? Like I see how derivatives can provide the gradient of certain functions but what does doing the opposite yield? Thanks
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keltingmeith

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Re: VCE Methods Question Thread!
« Reply #9324 on: March 16, 2015, 07:06:10 pm »
+1
What the hell is the point of integration? Like I see how derivatives can provide the gradient of certain functions but what does doing the opposite yield? Thanks

You'll later discover that the integral has many different applications to functions, such finding the area under a curve or the average value a function takes.

However, one really important use of the integral is its ability to solve what we call "differential equations". For example, you're often told that all objects fall at the same rate (more particularly, about 9.8 m/s^2). Well, we can use this to find the velocity and acceleration of an object as it descends. From our knowledge of derivatives, we know that velocity (m/s) is the rate of change of displacement (m), and so fits the differential equation:



Similarly, velocity is the rate of change of acceleration, and so:



So, we can use integration to solve this differential equation, and so we can then find where an object is at any particular time. This is great, because acceleration is easy to measure - but displacement isn't always very easy at all.

This particular example was quite simple, however they can get quite complicated very quickly.

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Re: VCE Methods Question Thread!
« Reply #9325 on: March 16, 2015, 07:30:52 pm »
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What the hell is the point of integration? Like I see how derivatives can provide the gradient of certain functions but what does doing the opposite yield? Thanks
isn't there something cool like acceleration is the derivative of velocity? If this is true then velocity is the antiderivative of acceleration. Right?

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Re: VCE Methods Question Thread!
« Reply #9326 on: March 16, 2015, 07:50:10 pm »
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isn't there something cool like acceleration is the derivative of velocity? If this is true then velocity is the antiderivative of acceleration. Right?

Position - velocity - acceleration - jerk - snap - crackle - pop!
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Re: VCE Methods Question Thread!
« Reply #9327 on: March 16, 2015, 07:55:43 pm »
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isn't there something cool like acceleration is the derivative of velocity? If this is true then velocity is the antiderivative of acceleration. Right?
Refer to above your post.

TheAspiringDoc

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Re: VCE Methods Question Thread!
« Reply #9328 on: March 16, 2015, 08:01:12 pm »
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Refer to above your post.
Fair point :P
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Snap, Crackle, and Pop are the cartoon mascots of Kellogg's crisped-rice breakfast cereal Rice Krispies, known in Australia as Rice Bubbles.
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"Snap", "Crackle", and "Pop" are terms sometimes facetiously used for the fourth, fifth, and sixth derivatives of position.







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Re: VCE Methods Question Thread!
« Reply #9329 on: March 16, 2015, 08:26:35 pm »
+1
Thank you guys, makes sense now!

Just wondering is the range the similar process? Or would I just have to sketch the new graph h(x) and restrict the domain and then determine the range?

Anyone?