VCE Methods Question Thread!

[pi]

I meant in the sense that we haven't learnt it in 3/4 methods yet. My first SAC will be on chs. 1-4, which doesn't cover calculus, so if I use a calculus technique to find an answer a different way to what I was taught from chs. 1-4 I might be penalised.

That's so strange. I'd clarify that with your teacher (in general for this and future SACs), given everything in 1+2 would/should be assumed knowledge, unlike say knowledge from Spesh which isn't appropriate.

[SE_JM]

Hello,

I have a quick question regarding finding the derivative of absolute value log functions.

if y=log_e|f(x)| then dy/dx= f'(x)/f(x)

Is this general rule right? In my textbook, there isn't a rule specifically for the absolute value log functions (there is one for log functions) but this is what I found works as I did more problems.

for example, if I had a question:
find derivative of y= -3 log_e(|(x/5)-3|)

I would first identify that f(x) is (x/5)-3,
and f'(x)= 1/5
then

I would say dy/dx= -3* (1/5)/((x/5)-3)
which is -3/(x-15)

Thanks!

[keltingmeith]

Hello, this is perhaps not relevant but I was wondering if anyone had differentiation of logarithmic and exponential functions problems (difficult ones)
I already went on to do the itute ones and the ones in the textbook, but feel as I need more practice (it's kind of confusing at times! )

Thank you

Differentiate the following:



Those should keep you busy for a while. I don't think you'll need any more practice after those. (mainly chain rules, but my theory is if you can do a bunch of different chain rules, then you only need to practice a couple of product/quotient rules)

Solutions

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[SE_JM]

Thanks! I edited the last post because I found something to do online but this would definitely help also!
Is there answers to this?

[keltingmeith]

Thanks! I edited the last post because I found something to do online but this would definitely help also!
Is there answers to this?

Gimme a bit.

And oki, I'll answer that, then:

Hello,

I have a quick question regarding finding the derivative of absolute value log functions.

if y=log_e|f(x)| then dy/dx= f'(x)/f(x)

Is this general rule right? In my textbook, there isn't a rule specifically for the absolute value log functions (there is one for log functions) but this is what I found works as I did more problems.

for example, if I had a question:
find derivative of y= -3 log_e(|(x/5)-3|)

I would first identify that f(x) is (x/5)-3,
and f'(x)= 1/5
then

I would say dy/dx= -3* (1/5)/((x/5)-3)
which is -3/(x-15)

Thanks!

The thing with logs is that their derivative is only defined for one half of the Cartesian plane. So, a general rule for logs is that:



Which we can show by the chain rule.

Now, let's consider the other side - when f(x)<0. If we want this to work in the log, we have to take the negative, because the logarithm is only defined for positive numbers:



You'll notice I applied the formula from above, but those negatives go away. Now, we consider the scary case - the ABSOLUTE VALUE of f(x):



And we see that on either side of the branch, we actually get the same gradient. So, we can say that:

[SE_JM]

Thanks for the detailed explanations!
It really helps me to understand this thing:P

[keltingmeith]

Solutions now in the post~

[SE_JM]

Thank you!
Time to do some serious work

[SE_JM]

Hello~It's me. again...

This is the explanation of population growth in my textbook and I don't get it...
I don't know what n represents (the textbook doesn't specify)
and because I don't know that, I don't know what I'm looking at right now

I also attached example 9 they were talking about if that will help

Could someone please help me?

[Zues]

in circles and eclippsed etc your not supposed to have a coefficient in front of the x^2, y^2 / x-h, y- h correct?

so if i have 4x^2 + (y+2)^2 = 1, this cannot be a circle, even by expansion of the second term (brackets)

so would it be correct to say

x^2/(1/4) + (y+2)^1 = 1? now an elipse? why cannot you have a coefficient infront of the varables?

[brightsky]

in circles and eclippsed etc your not supposed to have a coefficient in front of the x^2, y^2 / x-h, y- h correct?

so if i have 4x^2 + (y+2)^2 = 1, this cannot be a circle, even by expansion of the second term (brackets)

so would it be correct to say

x^2/(1/4) + (y+2)^1 = 1? now an elipse? why cannot you have a coefficient infront of the varables?

The general equation of an ellipse is (x - h)^2/a^2 + (y - k)^2/b^2 = 1. If a = b, then the ellipse becomes a circle.

Hence, we can generalise and say that if the coefficients in front of x^2 and y^2 are the same, then we have a circle, and if they are different, then we have an ellipse. Of course, you need to ensure that the coefficient in front of y^2 is positive so that you don't get a hyperbola instead.

[keltingmeith]

Hello~It's me. again...

This is the explanation of population growth in my textbook and I don't get it...
I don't know what n represents (the textbook doesn't specify)
and because I don't know that, I don't know what I'm looking at right now

I also attached example 9 they were talking about if that will help

Could someone please help me?

n represents the amount of compounds each year.

Think of it this way - in each example, you could add up each percentage over the year to get 24%. However, you got different results because each time you compounded more frequently. In this case, the population growth over one year is given by:



Let's try this formula out on the three examples:

n=1:
n=2:
n=12:

And so we see that the formula works!

However, only compounding occasionally for population growth is incredibly stupid. I mean, think about it - are you going to just say to people, "okay, you can only have sex every 2 months, so we can get this inconsequential mathematical model done"? No, that's stupid - people will continue to reproduce at all times. And so, we need to be compounding constantly - think of it as compounding an infinite amount of times within a year.

The best way to represent this is to take our amount of compounds (n), and see what happens as it goes to infinity, which is where your book kicks in.

[SE_JM]

Thank you so much! 

[aaziz17]

Question how do you type all the math stuff what app/website do you use?

[SE_JM]

Hello!
I'm currently learning the derivative of circular functions. There is a question i don't get. I was wondering if you could help me?


Find the derivative of each of the following with respect to Ө
sin²2Ө

The answer said:
let y= sin²2Ө and u = sin2Ө
then y=u² and applying the chain rule:

dy/dӨ = 2u*2cos2Ө
           = 4sin 2Ө cos 2Ө
           = 2sin4Ө, as sin4Ө = 2sin2Өcos2Ө
the bold bit is what i don't understand. why is sin4Ө = 2sin2Өcos2Ө?

similarly, in another question, the textbook says:
4sin(2Ө+1)cos(2Ө+1)
=2sin 2(2Ө+1), as sin2(2Ө+1)=2sin(2Ө+1)cos(2Ө+1)

thank you in advance!