VCE Methods Question Thread!

[achre]

First, our equation is going to be in the form of , where k is the additional vertical translation.

We want the gradient of the two curves to be equal, as then a tangent will occur.

Gradient of y=2x will be 2.



Make these equal to find the x value when the gradient is 2.




Anyone mind finishing this?
Had a complete mind blank.

Sub 3 into 2x to get the coordinate (3, 6). Sub values back into original function plus a constant for the translation and solve for the constant term, k=5.

[Stevensmay]

Achre,
Turns out I fluked the answer, thank you.

[ahat]

First, our equation is going to be in the form of , where k is the additional vertical translation.

shorten quote

We want the gradient of the two curves to be equal, as then a tangent will occur.

Gradient of y=2x will be 2.



Make these equal to find the x value when the gradient is 2.




Anyone mind finishing this?
Had a complete mind blank.

Discriminant method also works, if you had a mind blank in the exam. It also ensures that you don'y sub x = 3 into the wrong function.

[Stevensmay]

f(x)=e^2x - e^-2x.
Find the inverse.

We find the inverse by replacing all our x's with y's and the y's with x's.
Inverse is

This is really as  simple as it gets.

[Lejn]

f(x)=e^2x - e^-2x.
Find the inverse.

Sub your y for x, naturally.

Times by e^2y to get rid of the fraction, then transpose to make zero the subject.

e^4y-xe^2y+1=0

Let u=e^2y

u^2-xu-1=o

Solve for u, log it, divide by two, and you'll have x in terms of y.

[shadows]

How do you diff this?

log₂(x+2)     

[Damoz.G]

How do you diff this?

log₂(x+2)   

When I did it on the CAS, it comes up with (log₂(e))/x+2

With a bit of trial and error, it seems whatever is in the bracket is used as the denominator, and the numerator is always the same (log₂(e)).

[Lejn]

How do you diff this?

log₂(x+2)   

Change of base rule.

Log₂(x+2) = Ln(x+2)/Ln(2)

Where 1/Ln(2) is just a coefficient.

[achre]

How do you diff this?

log₂(x+2)   

Change of base to e => log₂(x+2)=ln(x+2)/ln(2)
then just proceed as normal

[Damoz.G]

Oh yeah, oops. Forgot about that Log rule. =/

[b^3]

We can't differentiate it straight away as the base of the log is not , so we'll turn it into two logs with base using . In other words we'll use a change of base for the log.


Now the bottom is a constant, so we can just differentiate the log normally.


EDIT: Beaten a few times
EDIT2: Just to show that it's the same thing as the CAS gives,

[shadows]

OH thank you!
It's funny how I've actually never come across a question like that o.o

[Damoz.G]

OH thank you!
It's funny how I've actually never come across a question like that o.o

Same here! I've never seen one in an Exam question.

The last time I saw a question like that was in the Essentials Textbook when we covered that chapter.

[shadows]

It was in the MAV 2013 Exam 1

MAV and NEAP exam papers have gotten really crazy these couple of years!
Their exam 1s are sometimes a lot harder than Kilbaha...

[Damoz.G]

Its was in the MAV 2013 Exam 1

MAV and NEAP exam papers have gotten really crazy these couple of years!
Their exam 1s are sometimes a lot harder than Kilbaha...

Ohh....I'm doing MAV 2013 Exam 1 tomorrow.

Yeah, I do agree with you there about difficulty. I guess its good though, just in case VCAA give us a hard paper (Means that we are fully prepared). Even the 2013 Heffernan paper was pretty hard to be honest.