VCE Methods Question Thread!

[speedy]

Ahhhh thanks! I should start to use the formula sheet more haha. However the answer is 2 not root(2)?

Yeah - I just edited that in aha, I was just showing what sec(pi/4) is

[Yacoubb]

How do we get 2? I understand that () = 1 thus . How does this equate to 2?
I was thinking to differentiate sec^2 but seems like I'm getting no where haha. Thanks.

sec^2(pi/4) = 1/[cos(pi/4)]^2 = 1/(sqrt2/2)^2 = 1/2/4 = 4/2 = 2.

tan(pi/4) = 1

So, sec^2(pi/4) / tan(pi/4) = 2/1 =2

[Rishi97]

Can someone please help me with 2007 VCAA exam 1 q 2b?
It's a really simple differentiation question but I can't do it (haha don't judge! )

Thanks heaps

[myanacondadont]

Spoiler

Here's my working. The answer is A - I'm not sure how




Now sub them in


Which led me to believe the answer is C?

[speedy]

Spoiler

Here's my working. The answer is A - I'm not sure how




Now sub them in


Which led me to believe the answer is C?

Because you're trying to find the original function, don't transpose for x/y -> sub in x^' and y^'.
Basically, when you're going from f -> x², that would require subbing in x/y (as you found above) -> however, to go backwards, from x² -> f, you sub x^'/y^'.

[IndefatigableLover]

Can someone please help me with 2007 VCAA exam 1 q 2b?
It's a really simple differentiation question but I can't do it (haha don't judge! )

Thanks heaps

Well you know the derivative of log will be the differentiated bit in the inside over the inside so essentially you differentiate tan(x) and that will be over tan(x).

Looking at your formula sheet, you know the derivative of tan(x) is 1/cos(x)^2 which will be all over tan(x). However I would split up tan(x) into sin(x)/cos(x) and make it all into one fraction so you can cancel one of the cos(x) to get:



Here you can sub in pi/4 and you'll get 1/(2/4) which is just 2

[Yacoubb]

Can someone please help me with 2007 VCAA exam 1 q 2b?
It's a really simple differentiation question but I can't do it (haha don't judge! )

Thanks heaps

It's okay

1. Find the derivative.

If y = loge(x), dy/dx = 1/x

So, if g(x) = loge(tan(x)), g'(x) = sec^2(x) / tan(x).

Therefore, g'(x) = sec^2(c) / tan(x)

3. Make the substitution.

Let's dissect what we have.

sec^2(x) = 1/[cos(x)]^2.
So, sec^2(pi/4) = 1/[cos(pi/4)]^2 = 1/(sqrt2/2)^2 = 1/2/4 = 4/2 = 2.

The denominator, tan(x), also requires substitution. So, tan(pi/4) = 1.

Bringing it together, 2/1 = 2.

Hope this helps

[Reus]

Thanks Speedy n Yacoubb.
Another question, how do you do these? I find them soooooooooo hard. It kills me haha.

[speedy]

Thanks Speedy n Yacoubb.
Another question, how do you do these? I find them soooooooooo hard. It kills me haha.

First look at stationary points - the derivative function will cross the x-axis/touch the x-axis here.
Next, look at the types of turning points -> for example: max (gradient) goes +ve 0 -ve. So for the derivative graph, it will start above the x-axis (+ve), cross it (0) and then continue down (-ve). A point of inflection will just touch the x-axis as the gradient stays the same.

Then have open circles where there are sharp points and discontinuities/end points.

Edit: another thing, the derivative graph is always going to be n-1 power. Therefore, for quadratics you will get linear etc. So for your graph, (on the LHS) the curve derivative will be a straight line, and (on the RHS) the straight line will be at a constant value.  The final bit will be at 0 (horizontal line - m = 0)

[Yacoubb]

Thanks Speedy n Yacoubb.
Another question, how do you do these? I find them soooooooooo hard. It kills me haha.

Okay so you need to take some things into consideration here.
(1.) You cannot differentiate the function at points like:
- Endpoints
- Cusps
- Points of discontinuity.

Any points like that, you must place an uncoloured circle (i.e. a white circle denoting the exclusion of that point). Then remember basic principles. If your function is quadratic, your derivative will be linear. Cubic --> quadratic, etc. SO employ these, and sketch the function of your derivative.

Your domain of the derivative is simply going to be everything EXCEPT within the points you're given EXCEPT endpoints, cusps and points of discontinuity.

[myanacondadont]

Because you're trying to find the original function, don't transpose for x/y -> sub in x^' and y^'.
Basically, when you're going from f -> x², that would require subbing in x/y (as you found above) -> however, to go backwards, from x² -> f, you sub x^'/y^'.

Ahhhh I think I get you. So since it says that the matrix maps a function TO x² then the transformations remain untouched? So when it says a function of x² to a new function THEN you do what I did above.

Never knew that - Nice.

[faredcarsking123]

For Q19 in 2013 VCAA Exam 2, in the solution it says solve in the domain [0, 3pi]

Where did they get this domain from?



Also can someone explain VCAA Exam 2 2007 quesiton 5fii please?

Thank you

[prishabal]

Could someone super please show me how to answer this???

[speedy]

Ahhhh I think I get you. So since it says that the matrix maps a function TO x² then the transformations remain untouched? So when it says a function of x² to a new function THEN you do what I did above.

Never knew that - Nice.

Mm kinda, don't memorise it. If the transformation x = x'-5 is applied to f to get x², then the transformation x' = x+5 'undoes' this change to get x² it back to f.

[Reus]

First look at stationary points - the derivative function will cross the x-axis/touch the x-axis here.
Next, look at the types of turning points -> for example: max (gradient) goes +ve 0 -ve. So for the derivative graph, it will start above the x-axis (+ve), cross it (0) and then continue down (-ve). A point of inflection will just touch the x-axis as the gradient stays the same.

Then have open circles where there are sharp points and discontinuities/end points.

Edit: another thing, the derivative graph is always going to be n-1 power. Therefore, for quadratics you will get linear etc. So for your graph, (on the LHS) the curve derivative will be a straight line, and (on the RHS) the straight line will be at a constant value.  The final bit will be at 0 (horizontal line - m = 0)

Okay so you need to take some things into consideration here.
(1.) You cannot differentiate the function at points like:
- Endpoints
- Cusps
- Points of discontinuity.

Any points like that, you must place an uncoloured circle (i.e. a white circle denoting the exclusion of that point). Then remember basic principles. If your function is quadratic, your derivative will be linear. Cubic --> quadratic, etc. SO employ these, and sketch the function of your derivative.

Your domain of the derivative is simply going to be everything EXCEPT within the points you're given EXCEPT endpoints, cusps and points of discontinuity.

Don't think I'll be getting this anytime soon so complicated it hurts LOL.

Could someone super please show me how to answer this???

Use the quotient rule!