VCE Specialist 3/4 Question Thread!

[Lavar Big BBB Balls]

Hi,

For these: http://imgur.com/a/x7Y8L

Q14a) I'm confused in some parts of the book's answer. In their diagram why are they measuring alpha between the line and the y-axis, don't we usually measure between the x-axis? I did the latter which got me the reciprocal of what they had for tan(alpha).

Also, when they write tan(alpha) = at/bt, why do you/are allowed to omit the minus sign? I left the minus sign in there but ended up with an answer with both components being positive which doesn't make sense here.

Q16c) I have x = t^2 -5t -2 and y = 2 which are correct. I was stuck here but the answer concluded that y=2 was the Cartesian equation. How and why did they do this?

[Shadowxo]

Hi,

For these: http://imgur.com/a/x7Y8L

Q14a) I'm confused in some parts of the book's answer. In their diagram why are they measuring alpha between the line and the y-axis, don't we usually measure between the x-axis? I did the latter which got me the reciprocal of what they had for tan(alpha).

Also, when they write tan(alpha) = at/bt, why do you/are allowed to omit the minus sign? I left the minus sign in there but ended up with an answer with both components being positive which doesn't make sense here.

Q16c) I have x = t^2 -5t -2 and y = 2 which are correct. I was stuck here but the answer concluded that y=2 was the Cartesian equation. How and why did they do this?

14:
It says it is in the direction North alphaš West. This means you start at North and measure alphaš West which is why the alpha is located where it is. Then, you know tan(alpha) = 4/3 = opposite/adjacent so they just made it into a triangle with alpha as the angle. If they were to measure from the x-axis you would be using 90-alphaš instead. Bit hard to explain without visuals but let me know if you want a diagram
It's easier to use the diagram for the at/bt part. The x length (the 'opposite' part of the triangle) is a*t ie positive (as they used r(t) = -ai+bj). It must be positive as you cannot have a negative length, and it just uses these positive lengths to find tan(alpha). The y part is just b*t as it's already positive. Then tan(alpha) = at/bt = opp/adj
15:
Think of the path it's travelling. The y value is constant while the x value changes - this is just part of a horizontal straight line. So equation is just y=2 (as the x value can change while y stays the same)

Sorry if this is a bit confusing

[rocksonchan]

Hi,

For these: http://imgur.com/a/x7Y8L

Q14a) I'm confused in some parts of the book's answer. In their diagram why are they measuring alpha between the line and the y-axis, don't we usually measure between the x-axis? I did the latter which got me the reciprocal of what they had for tan(alpha).

Also, when they write tan(alpha) = at/bt, why do you/are allowed to omit the minus sign? I left the minus sign in there but ended up with an answer with both components being positive which doesn't make sense here.

Q16c) I have x = t^2 -5t -2 and y = 2 which are correct. I was stuck here but the answer concluded that y=2 was the Cartesian equation. How and why did they do this?

Hi,

Q14a) In the diagram, they measure alpha between the line and the y-axis because it specifies that alpha is given by Na^oW, which is a compass bearing. How this compass bearings is N tells you to face North, and then turn alpha^o towards West, where North represents the positive direction of the y-axis and West represents the negative direction of the x-axis.

When they write tan(alpha) = at/bt, they omit the minus sign because when working with triangles and trigonometry, the lengths of the sides must be positive for the triangle to exist. To illustrate, when you draw a triangle to represent cos(2pi/3) = -1/2, you still draw the lengths of the sides as 1 and 2, not -1 and 2.

Q16c) The reason the answer concluded y=2 is because no matter what x equals, y will still equal 2. Normally you would be able to express t in terms of x and then substitute but in this case y(t) has no t term to allow such a substitution. That's why the conclusion is y=2. However, it should be noted that y=2 only exists for x greater or equal to -33/4, because the range of x(t) is [-33/4,infinity], which means x only has values for this range.

Hope this helps!!!

[Lavar Big BBB Balls]

Thanks shadowxo and rockson!!

[peanut]

In VCAA exams, if a question asks to sketch and label two graphs for two marks, are endpoints, asymptotes and axes intercepts not required?

[Sine]

In VCAA exams, if a question asks to sketch and label two graphs for two marks, are endpoints, asymptotes and axes intercepts not required?

depends how the marking scheme goes that year

If it's 2 marks you want want at a very minimum at least two things. In my opinion the 2 most important things would be the shape of the graph and the axes intercepts.

As a rule if you aren't sure whether you have enough information on the paper you probably don't and should just keep making your graph as detailed as you like until you are happy and positive you will get full marks for it. 

[geminii]

How on earth am I supposed to graph
f(x) = 3pi/4  -  cos^-1((x-1)/2)
?? Please help, I have no idea where to start!!

[Sine]

How on earth am I supposed to graph
f(x) = 3pi/4  -  cos^-1((x-1)/2)
?? Please help, I have no idea where to start!!

If you know the graph y=cos^-1 (x) just apply those transformations to that graph.

Another trick is to determine the domain of the graph then sub in the end points x values to determine both end points. Now since arccos has rotational symmetry the mid point will be the average of the end points. You still need to take into account reflections to ensure your draw the correct graph

[geminii]

If you know the graph y=cos^-1 (x) just apply those transformations to that graph.

Another trick is to determine the domain of the graph then sub in the end points x values to determine both end points. Now since arccos has rotational symmetry the mid point will be the average of the end points. You still need to take into account reflections to ensure your draw the correct graph

But I don't know how to do the 3pi/4 - (the rest)? like what does minusing the inverse cosine function from 3pi/4 do? i don't understand how to minus the inverse cos function from 3pi/4 and put in on the graph basically. :/

[Quantum44]

But I don't know how to do the 3pi/4 - (the rest)? like what does minusing the inverse cosine function from 3pi/4 do? i don't understand how to minus the inverse cos function from 3pi/4 and put in on the graph basically. :/

That's just a vertical translation and reflection about the x-axis

[Sine]

But I don't know how to do the 3pi/4 - (the rest)? like what does minusing the inverse cosine function from 3pi/4 do? i don't understand how to minus the inverse cos function from 3pi/4 and put in on the graph basically. :/

How on earth am I supposed to graph
f(x) = 3pi/4  -  cos^-1((x-1)/2)
?? Please help, I have no idea where to start!!

[geminii]

Oh my gosh thank you! Yes when it's expressed that way it makes so much more sense! Silly me was for some reason thinking of 3pi/4 as an angle and not an actual translation. Thanks again!!

[Lavar Big BBB Balls]

Hello,

For something like this https://m.imgur.com/a/lwICO

Would VCAA expect us to know that the net force will be pointing in to the third quadrant without using a calculator? it's not obvious if we leave the components in exact form, so I thought the direction was only 25 degrees.

Thanks

[Syndicate]

Hello,

For something like this https://m.imgur.com/a/lwICO

Would VCAA expect us to know that the net force will be pointing in to the third quadrant without using a calculator? it's not obvious if we leave the components in exact form, so I thought the direction was only 25 degrees.

Thanks

Yes, I think you can expect such things from VCAA.
Btw the resultant force is 225 degrees from the positive direction of the x -axis.
(You can work it out by basic trigonometry)

What do you exactly mean by exact form?

[humblepie]

Would really appreciate it if someone could explain how to find the range of the following functions by hand: 

f(x)=(1/3)^2x-x2
f(x)=(1/3)^x2-2x

Thanks in advance!