VCE Methods Question Thread!

[paper-back]

Genuine help needed for e, f and g! Honestly have no idea and do have answer to them this time :p

e) You have to make the equation = 9.5, and then solve for x, in the domain [24h,48h], as it is asking for Monday. Although it asks for the first time, you might as well solve for this domain, so you don't have to solve again for the next questions
f) Find difference between the first 2 solutions
g) 7.09pm

Spoiler

[Alwin]

Genuine help needed for e, f and g! Honestly have no idea and do have answer to them this time :p

hey there

here's the graph I'll be referring to: https://www.desmos.com/calculator/yofmq9juix

Now, for (e) we want h(t) > 9.5
We can easily solve h(t) = 9.5 on the calculator or by hand, and you should be able to obtain a values of t = 1.167, 5.833, 15.167, 19.833, 29.167, 33.833, 43.167, 47.833
These can be converted to time values:
SUNDAY: 1:10 am, 5:50 am, 3:10 pm, 7:50 pm.
MONDAY: 5:10 am, 9:50 am, 7:10 pm, 11:50 pm.

So if we take the first solution on monday, she should drop the line in at 5:10 am

For (f), we refer to the graph (see above). Clearly from t = 29.167 to 33.833 h(t) is bigger than 9.5 m
Thus, the time she can leave the line in is 33.833-29.167 = 4 hours and 40 minutes

For (g) we can see that our solutions on monday night are 7:10 pm to 11:50 pm
This is when the depth is greater than 9.5
So, she should return at 7.10 pm to drop a line in.

Hope it helps + makes sense

Edit: beaten

[Jason12]

i have an answer to a question as 2 + 4loge^(1) but the answer is 2

does this mean that loge^(1) is 0 as 4 X 0 + 2 = 2?

[b^3]

Yeah, . You can just use this fact, but I've added why below.


“Log base of the answer gives the power".

[paper-back]

This may sound like a silly question
But what is the purpose of sketching the gradient function? What does it do relative to the derivative function (from which we can get the gradient at a point)?

Thanks

[soNasty]

i think, in the graph of a gradient function, the x intercepts represent the stationary points of the original function, and visually show us when and where the graph contains a negative and positive gradient. this may not be new information to you, but still, its good to know

Also, i have a question

is the general solution to: equal to ?

[Orb]

This may sound like a silly question
But what is the purpose of sketching the gradient function? What does it do relative to the derivative function (from which we can get the gradient at a point)?

Thanks

In the graph of a gradient function, you can just look towards where the gradient = 0 (x-intercepts) and find what the stationary points of the original graph are.

Simultaneously, when x>0 you know that the gradient is positive.
and when x<0 you know that the gradient is negative.

So you can find out multiple details from the one graph which may be harder to simply deduce from the original graph.



EDIT: beaten by andrew

[soNasty]

can i please get help with the first one or two steps into solving these, dont give me the answers

over



and



for that ^^^ one, does it work if i make it:  ?

[paper-back]

Thanks guys

[paper-back]

can i please get help with the first one or two steps into solving these, dont give me the answers

over



and



for that ^^^ one, does it work if i make it:  ?

Let

Solve for a, then solve for cos(pi x)

[Zealous]

can i please get help with the first one or two steps into solving these, dont give me the answers

over



and



for that ^^^ one, does it work if i make it:  ?

[edit] beaten by paper-back.

1.


3. Yep, you can convert the sin to cos:

Spoiler

I made a slight positive/negative error when I typed this up the first time (making it impossible to factorise by inspection) and hence I used the long method of completing the square. Just view the above as LaTex practice! 
Have fun working from there!

[soNasty]

oh sorry for the second one, its x/3

Thanks zealous, however cant you just factorise from 2a^2-3a+1=0 to get (a-1)(2a-1)?
also, since its asking us to solve within the domain of [0,2pi], would the inside the cos make the new domain of values ? i dont even know, im confused now lol. thanks for the help! its just this question thats annoying me now.

the calculator gives solutions of x=0, 1/3, 5/3, or 2, or 7/3 or 11/3 or 17/3 or 6. idk how the hell its getting that many solutions

[Zealous]

Thanks zealous, however cant you just factorise from 2a^2-3a+1=0 to get (a-1)(2a-1)?

Haha yep! I made a silly positive/negative error when I did it the first time round which made it not possible to factorise by inspection.

[soNasty]

haha no probs, thanks again

[soNasty]

another q lol