VCE Methods Question Thread!

[Shadowxo]

Can someone please help me with the following question?
Find the values of x for which the graph of y=f(x) has a stationary point and state the nature of each stationary point. Consider 0≤x≤2π only.
f(x)=2cosx+2sinxcosx
Thanks

First you need to find the derivative. For the second part, you need to apply the product rule
f'(x) = -2sinx -2sin²x + 2cos²x
f'(x) = -2sinx -2sin²x +2(1-sin²x)
f'(x) = -4sin²x -2sinx +2
f'(x)= (-4sinx+2)(sinx+1)
So f'(x)=0 where sinx = 1/2 or sinx = -1
So x= π/6, 5π/6, and 3π/2 (restricted domain)

To find the nature, just find whether the derivative either side of each of these points is + or -. Eg derivative where x=0, x=π/2, x=π, x=2π

Hope this helps

[Shadowxo]

Hi guys I know this isn't a Methods 3/4 question but I am just asking for some tips.

Are there tips for sketching sine and cosine graphs with horizontal and vertical translations? I often find myself not knowing where to start on the axis because the horizontal scaling is different from the period. Do I just estimate?
 For example, if the equation is y=2sin3(x- (pi)/4)

While the period is 2(pi)/3 and the phase shift is pi/4 to the right, where would I even start on the axis because I can't seem to estimate.

Thank you. 

First, find the y intercept (just plug x=0 into the formula)
Now if you know the amplitude and the vertical translation, draw these in with a dotted line to help you
eg if it's 4cosx +3, draw a dotted line at y=3, and lines at y= -1 (minimum point, 3-4) and y=7 (maximum point, 3+4)
Find a couple x intercepts (these don't help so much with drawing the graph but you often need them anyway)
Find the x values of some max and min points (so if it were 3sinx, find where sinx = ±1 and use this to find x)
You should be able to extrapolate the rest of the graph, using your knowledge of the period, the function and the translations.
There are other ways to do it but I tend to do this. Alternatively, you could draw the basic graph without the translations and then move it left or right (but you'd still need to calculate y intercept, and also sometimes the turning points and x intercepts anyway)

[deStudent]

http://m.imgur.com/ySuf92M
For a and b, is what I wrote acceptable? I'm not actually sure how I can simplify this further to get it in to the literal form they are asking for.

You could make a case that ln(4) and e^2 are constants which fits the form required but I'm not sure if I'm missing a simplification?

Thanks

[peanut]

http://m.imgur.com/ySuf92M
For a and b, is what I wrote acceptable? I'm not actually sure how I can simplify this further to get it in to the literal form they are asking for.

You could make a case that ln(4) and e^2 are constants which fits the form required but I'm not sure if I'm missing a simplification?

Thanks

Yeah, they can be constants. However, just to be safe, I would write it as:
x + (2 + loge(4))
and
4e^2x + 0 (the 0 is important, I think marks were deducted for not including it in a similar question in a VCAA exam)

[ImmaculateJeff]

hey anyone got practice chemistry & methods sacs/tests, if so can you pm me please, thanks!

[lovelyperson]

Noob question but how do I solve trig equations on CAS? I type them in something like this; solve(2sin(2x)=0,x)|0<x<4 but I still get weird solutions with n's within them. Help please 

[clarke54321]

Noob question but how do I solve trig equations on CAS? I type them in something like this; solve(2sin(2x)=0,x)|0<x<4 but I still get weird solutions with n's within them. Help please 

Hey!

I've just tried doing this on my CAS.

I typed in solve(2sin(2x)=0|0<x<4,x

This yielded: x=pi and x=pi/2  (sorry can't find pi symbol)

I think your ,x was in the wrong position. Also, make sure you are in radians!

[Yueni]

Noob question but how do I solve trig equations on CAS? I type them in something like this; solve(2sin(2x)=0,x)|0<x<4 but I still get weird solutions with n's within them. Help please 

On Ti-nspire cx calculators, you can solve it by typing the following

Solve(2\sin(2x)=0,x)|0<x<4, that is fine. The problem arises where you get general solutions when you either

(a) have a domain that is too large

(b) are in a mode that is incorrect to your question ie. degrees instead of radian

[Ahmad_A_1999]

Hey, could someone please explain to me what this question even means 

Question 8 Part E

http://imgur.com/jOCkKmA

And this please, question 13 part D, I don't understand what it means to still be afloat 

http://imgur.com/i1c87Ox

SAC is tomorrow 

Edit: Is it correct to say that changing the base on a logarithm dilates a log graph from the x-axis?

[exit]

Hey, could someone please explain to me what this question even means 

Question 8 Part E

Hey, could someone please explain to me what this question even means 

Question 8 Part E

http://imgur.com/jOCkKmA

And this please, question 13 part D, I don't understand what it means to still be afloat 

http://imgur.com/i1c87Ox

SAC is tomorrow 

Edit: Is it correct to say that changing the base on a logarithm dilates a log graph from the x-axis?

http://imgur.com/jOCkKmA

And this please, question 13 part D, I don't understand what it means to still be afloat 

http://imgur.com/i1c87Ox

SAC is tomorrow 

Edit: Is it correct to say that changing the base on a logarithm dilates a log graph from the x-axis?

For 8E, find the x coordinate of the max point then move two x-values to the right and to the left. (this creates an interval of 4 between these values and due to symmetry, same h value) Then find the height at these two points, they should match and you will get your answer.

13 partD has dodgy wording and I don't think it'll appear, check your textbook answers for clarification

For your log question, remember the change of base rule where log7(34x)=loge(34x)/loge(7)

[Ahmad_A_1999]

http://imgur.com/jOCkKmA

And this please, question 13 part D, I don't understand what it means to still be afloat 

http://imgur.com/i1c87Ox

SAC is tomorrow 

Edit: Is it correct to say that changing the base on a logarithm dilates a log graph from the x-axis?


For 8E, find the x coordinate of the max point then move two x-values to the right and to the left. (this creates an interval of 4 between these values and due to symmetry, same h value) Then find the height at these two points, they should match and you will get your answer.

13 partD has dodgy wording and I don't think it'll appear, check your textbook answers for clarification

For your log question, remember the change of base rule where log7(34x)=loge(34x)/loge(7)

Thanks for the reply Exit!
Question 13 is weird you're right, I won't worry about it.

For 8 e) I get what you are saying and that does give the answer, but I don't know what or why we do that, what does 'h' have anything to do with anything (bare with me I can be extremely stupid)

And for the log question, I remember that rule, but if I were to graph two log functions and change only the base between them, would it be correct to say that the effect of changing that base is a dilation from the x-axis on graph of log whatever it happens to be?

[exit]

Thanks for the reply Exit!
Question 13 is weird you're right, I won't worry about it.

For 8 e) I get what you are saying and that does give the answer, but I don't know what or why we do that, what does 'h' have anything to do with anything (bare with me I can be extremely stupid)

And for the log question, I remember that rule, but if I were to graph two log functions and change only the base between them, would it be correct to say that the effect of changing that base is a dilation from the x-axis on graph of log whatever it happens to be?

It's a time graph. The t on the 'x-axis' represents time. If the depth of water is above the 'h' value, then you match it to the t value on the 'x-axis' and you know at the time 't', the depth is above the depth of the 'h' value. You are finding in this question for how long it is above this h value before it goes back under.

For your log question. I think it is beyond the scope for VCAA to ask that but you never know:

Log3(27x) to log5(27x)  is loge(27x)/loge(3) to loge(27x)/loge(5)

1/loge(3) to 1/loge(5) yes it qualifies as dilation from the x-axis by loge(3)/loge(5) i guess (someone correct if wrong, but weird situation)

[clarke54321]

Could I please have some help with this question!

Thank-you 

[Sine]

Could I please have some help with these questions.

For the general solution one, I don't understand why there aren't 2 solutions.

Thank-you!

The attachments don't work

[clarke54321]

The attachments don't work

Really? On my iPad and computer they seem to work!

I'll try and re-upload them.