VCE Methods Question Thread!

[wunderkind52]

For the worked example question from my textbook (attached below) I don't really understand why the derivative of the function is 2x-4 if x^2-4x >0 and is -2x+4 if x^2-4x < 0.

I don't really understand how the x^2-4x part comes into being part of the domain of the derivative?

Any help would be appreciated!

It is because it is a modulus function. Whenever working with modulus functions, you should split it into a hybrid function. As you can see, the function is multiplied by-1 (reflected in the x axis) where it falls below the x axis.(x^2-4x<0). The hybrid function is thus x^2-4x, x^2-4x>0, and - x^2+4x,x^2-4x<0.

[garytheasian]

facepalm i thought u had to put a in terms of r than diff

[kinslayer]

Thanks kinslayer   

so if i wrote the integral is undefined as my answer due to the asymptote at x=0.

Would i be wrong?

I'm honestly not sure. I am surprised to find that question in a Methods textbook because I'm sure that improper integrals aren't covered.

The technical way to say it would be that it diverges to infinity because of how improper integrals work. But yeah, just saying it's not defined because of the asymptote should be OK.

[knightrider]

I'm honestly not sure. I am surprised to find that question in a Methods textbook because I'm sure that improper integrals aren't covered.

The technical way to say it would be that it diverges to infinity because of how improper integrals work. But yeah, just saying it's not defined because of the asymptote should be OK.

Thanks kinslayer 

[lzxnl]

Thanks kinslayer   

so if i wrote the integral is undefined as my answer due to the asymptote at x=0.

Would i be wrong?

Yes. That's technically not true
You can integrate y = x^(-1/2) from 0 to 1 but there's an asymptote there

[cameotodd]

Hey guys, could someone help me with my question that I posted about 5 or 6 posts ago ?

Thanks

[qwerty101]

volume of prism V = Ah

dV/dt = 4 cm^3/min

dV/dh = A, dh/dV = 1/(A cm^2)

dh/dt = dV/dt * dh/dV = 4/A cm/min

The current volume of sand in the prism is irrelevant, unless that volume is greater than V = Ah in which case the prism is full and dh/dt = 0 aheh

I understand your working out, but i dont understand why the 100 hasnt been subbed in somewhere

[qwerty101]

Also , with this question.

N is (2,0) and the turning point is (4,y(4)), shouldnt the coordinate of M be (6,0) since the turning point is symmetrical of the two points on either side?

thanks

[kinslayer]

I understand your working out, but i dont understand why the 100 hasnt been subbed in somewhere

It's irrelevant. Either it's been put there to throw you off, or the question was changed somehow and that bit was left in there by mistake.

[qwerty101]

i always see questions that ask find dh/dt for example and then find the rate of change of height when the time is 6 for example ?

[knightrider]

Say if i was to factorise the following the following expression



in the 2nd step why can we cancel out one of the (x-4) factors from step 1?

Why does this work?

[kinslayer]

i always see questions that ask find dh/dt for example and then find the rate of change of height when the time is 6 for example ?

Yeah, it all depends on what form the required rate takes. In your first example, You know that 100 = Ah or A = 100/h, but dh/dt doesn't depend on h (actually it depends on either A or h, but not both). So knowing the volume actually doesn't get you anywhere. If they asked you to express dh/dt in terms of h, then you would need to use your information about V.

Alternatively, if you had a situation where for example V = Ah^2, then knowing dV/dh gives you 2Ah, and now knowing what V is does place a restriction on dh/dt, because it specifies a relationship between A & h, and h is going to show up in dh/dt.

Being given irrelevant information is quite rare; this question is an anomaly. I wouldn't worry about it.

Say if i was to factorise the following the following expression



in the 2nd step why can we cancel out one of the (x-4) factors from step 1?

Why does this work?

What they are doing is splitting the cubic into x^3 - 4x^2, and -4x + 16. (x-4) is a common factor to both, so you have x^2(x-4) and -4(x-4). Add them together and you get the second line.

The next line is just taking out the common factor of x-4 again and the last line is using DOPS.

[kinslayer]

Yes. That's technically not true
You can integrate y = x^(-1/2) from 0 to 1 but there's an asymptote there

This is true -- asymptotes can be endpoints, since there is no reason (in general) why the improper integral should not converge.

But if the asymptote is in the domain of integration and not one of the endpoints, then the definite integral doesn't make sense and has to be split up into two improper integrals.

I'm still surprised that a question like that showed up in a Methods book.

[lzxnl]

This is true -- asymptotes can be endpoints, since there is no reason (in general) why the improper integral should not converge.

But if the asymptote is in the domain of integration and not one of the endpoints, then the definite integral doesn't make sense and has to be split up into two improper integrals.

I'm still surprised that a question like that showed up in a Methods book.

Off-handedly, if your function blows up at zero at the rate of 1/x or faster, then the integral from 0 to 1 won't exist. If it blows up any slower, the integral will likely exist.

[Floatzel98]

My class skipped over Linear Approximation when we were doing differentiation in class. I've tried going back over it now, but i don't think I really understand it. Or at least i really have no idea how to do any questions. I don't understand the steps taken in my textbook for this example:

Given that f(x) = x^4 - x^3, find in terms of p the approximate increase in f(x) as x increases from 2 to 2 + p, where p is small.

For the percentage change questions, do you use the percentage change formula from the start, or is it something you use after you find the linear approximation? Is linear approximation ever actually show up in any exams? Is it more likely to show up in exam 1 or exam 2?

Thanks