VCE Specialist 3/4 Question Thread!

[#J.Procrastinator]

I'm having trouble getting my head around using trig identities for integration. Are there any tips or ways that can help me approach these sort of questions? Like, how would I integrate tan^2(2x) or cosec^2(x-pi/2). Any help is appreciated!

[Thorium]

I'm having trouble getting my head around using trig identities for integration. Are there any tips or ways that can help me approach these sort of questions? Like, how would I integrate tan^2(2x) or cosec^2(x-pi/2). Any help is appreciated!

Here are some general tips:

1) When you have a cos times sin, and they both have same power, eg. 4Sin^2(x/2)cos^2(x/2), then you use the sin compound fornula to change it to: 4[1/2sin(x)]^2=4x1/4sin^2(x)=sin^2(x)
Then you use the cos compound formula to get rid of the power: =1/2(1-cos2x), an antidiff it.

2) Usually when given sin or cos with even power (mostly when squared), use the cos compoud formula to get rid of the power, which makes antidiffing easier. The last step of the previous example is a good example for this.

3) When given sin or cos with odd power (usually cubed), eg cos^3(x), separate one cos so that u have a cos with even power, then use pythagoras so that u have a sin with your cos: cos(1-sin^2(x))
Then let the sin^2=u, and use the substitution method to get rid of cos. Then u will just untidiff the expression in terms of u.

4) When you have cos and sin multiplyed and one has odd and one has even power, then use tip 3 to antidiff it interms of u.

5) when u have a sin and cos with same power but different transformations, eg. Cos(x)sin(2x), then try to make them have same transformation: cos(x)(2sinxcosx)=2sinxcos^2(x). Then use the previous tip.

6) when u have tan^2(x), then use this identity: tan^2(x)+1=sec^2(x) to change it to: sec^2(x)-1, then u can antidiff it as diff of tanx is sec^2(x).

7) When u have only tanx, change it to sin/cos, let cos=u and use substitution

 I cant garantee that these tips will work definitely everywhere, but this is i how usually approach the questions

Hope that helps

[#J.Procrastinator]

Here are some general tips:

1) When you have a cos times sin, and they both have same power, eg. 4Sin^2(x/2)cos^2(x/2), then you use the sin compound fornula to change it to: 4[1/2sin(x)]^2=4x1/4sin^2(x)=sin^2(x)
Then you use the cos compound formula to get rid of the power: =1/2(1-cos2x), an antidiff it.

2) Usually when given sin or cos with even power (mostly when squared), use the cos compoud formula to get rid of the power, which makes antidiffing easier. The last step of the previous example is a good example for this.

3) When given sin or cos with odd power (usually cubed), eg cos^3(x), separate one cos so that u have a cos with even power, then use pythagoras so that u have a sin with your cos: cos(1-sin^2(x))
Then let the sin^2=u, and use the substitution method to get rid of cos. Then u will just untidiff the expression in terms of u.

4) When you have cos and sin multiplyed and one has odd and one has even power, then use tip 3 to antidiff it interms of u.

5) when u have a sin and cos with same power but different transformations, eg. Cos(x)sin(2x), then try to make them have same transformation: cos(x)(2sinxcosx)=2sinxcos^2(x). Then use the previous tip.

6) when u have tan^2(x), then use this identity: tan^2(x)+1=sec^2(x) to change it to: sec^2(x)-1, then u can antidiff it as diff of tanx is sec^2(x).

7) When u have only tanx, change it to sin/cos, let cos=u and use substitution

 I cant garantee that these tips will work definitely everywhere, but this is i how usually approach the questions

Hope that helps

Thank you very much!

[psyxwar]

Quick vectors question:

Dot product of vector a and vector a gives us the magnitude of vector a squared, which is a squared (without the tilde?). Can someone clarify what the a without a tilde means?

[keltingmeith]

The tilde means it's a vector quantity. Without the tilde it's a scalar quantity.

To clarify why we can do this, let a=xi + yj + zk. Then, a^2 = a.a = x^2 + y^2 + z^2, which is no longer a vector quantity, so we can just ditch the tilde and magnitude signs.

[psyxwar]

Oh okay fair enough, thanks. So a sans tilde is equivalent to |a| then?

[kinslayer]

Oh okay fair enough, thanks. So a sans tilde is equivalent to |a| then?

Not in general, but there is nothing stopping you from setting |a| = a if you want to, since they are both scalar quantities (numbers). I'd steer clear of doing that if possible and just stick to using |a|. Less possibility for confusion that way.

[psyxwar]

Not in general, but there is nothing stopping you from setting |a| = a if you want to, since they are both scalar quantities (numbers). I'd steer clear of doing that if possible and just stick to using |a|. Less possibility for confusion that way.

Yeah that's what I thought, but Math Quest uses a randomly in the solutions and I got confused. Thanks.

[Cort]

You know, I thought I was starting to get pretty okay in physics. Doing this however, made me very sad. Basically every question to this one Ive had to use the worked solutions. I just have no idea how to approach these question - it seems all foreign to me despite my experience with motion in physics.

Especially with b.
When it asks how long it takes to get back to the bike; does that mean we already assume that the police car already caught the bugger?

Thanks,
cort.

[brightsky]

the key to these kinds of questions is to draw a velocity-time graph. however, you need to make sure that your units are consistent. the convention is to use SI units. so velocity (the vertical axis) is in m/s and time (the horizontal axis) is in s. define t = 0 to be the time at which the car passes the policeman. the car is traveling at a constant velocity of 90 km/h (or 25 m/s) forward. hence, the velocity-time graph for the car is simply a horizontal line starting at (0,25) and extending to the right. now, let us consider the velocity-time graph for the policeman. it is given that the policeman is starts off in pursuit 5 s after the car passes him. he moves for 250 m with constant acceleration until he reaches a speed of 120 km/h (or 33 1/3 m/s). hence, the velocity-time graph for the policeman is a slanted line starting at (5,0) and ending at (20,33 1/3). the graph then extends horizontally to the right from (20, 33 1/3).

for part a), we want to find the time it takes for the distance travelled from t = 0 by the car and the policeman to be the same. let this time be t. then we require the area of the rectangle with corners (0,0), (0,25), (t,25) and (t,0) to be equal to the area of the trapezium with corners (5,0), (20,33 1/3), (t,33 1/3), (t,0):

25*t = 1/2*[(t-20) + (t-5)]*33 1/3]
25*t = 100*t/3 - 1250/3
t = 50

hence, the policeman catches up to the car 50 s after the car passes him.

to solve part b), we need to use the extra info given. as described above, the velocity-time graph for the car is simply a horizontal line extending to the right from (0,25) and the velocity-time graph for the policeman is a slanted line starting from (5,0) and ending at (20, 33 1/3) and then a horizontal line extending to the right from (20,33 1/3). at t = 50, we know that the policeman catches up to the car. from that point on, the velocity-time graphs for both the car and the policeman make a nosedive, both ultimately ending up at (53,0) (it should be clear from the velocity-time graphs that the car needs to decelerate more quickly than the policeman but it is assumed that the deceleration for both is constant, so that the two velocity-time graphs from t = 50 onwards should both look like slanted lines with negative gradient. now that we have the complete velocity-time graphs, all we need to do is calculate the areas under each graph and find the difference between the two. the area under the velocity-time graph for the car is 25*50 + 1/2*25*3 = 2575/2 = 1287.5. the area under the velocity-time graph for the policeman is 1/2*15*(33 1/3) + 30*(33 1/3) + 1/2*3*(33 1/3) = 1300. the difference between the two is 1300 - 1287.5 = 12.5. hence the policeman travels 12.5 m too far and so has to walk back 12.5 m from his cycle to the stationary car.

hope this hand-wavy explanation makes sense.

[keltingmeith]

Don't worry - mechanics in physics and mechanics in spec are two completely different things (y'know, even though it's exactly the same concept). Don't be disheartened because you're good in physics and bad here - in fact, I got full marks for mechanics in specialist last year, and have been struggling all year with mechanics in physics (even though I'm meant to be doing a year 12 equivalent unit).

For this question, what you want to do is draw a velocity-time graph. Once you've done that, you need to figure out when the two vehicles meet - obviously, this would be when their displacement is the same. Displacement is given by the area under a velocity-time graph, so we get:



(note: I converted acceleration to m/s to make things easier. Also, I found acceleration and time using constant acceleration formula, and . You could also use this with area formula instead of integrating, but I thought I'd show this method anyway as it can be easier to follow and will work for any function. You'll also note that I cheated by letting t=0 at a bunch of places - this is fine, as long as you remember to fix this up [which hopefully I did])

In regards to your actual question - yes, we do assume that the car has caught the guy, as it says that he has. In fact, beautifully enough, we don't need the previous answer to answer this question.

Using , we get the following equations:



Gone a bit overboard, but I hope I've both answered your question and put your mind at ease. Good effort in getting into mechanics so early!

EDIT: So, brightsky beat me to it. Still, I'll keep this all up simply because I've gone and done it the sucker's way instead of using area formula.

[psyxwar]

So maths quest had a worked example that was like:

u=cos(ti)+sin(tj)

They found the cartesian equations by letting cos(t)=x and sin(t)=y then using the pythagorean identity.

Is there some sort of trig rule I'm forgetting or is this meant to be cos(t)i+sin(t)j?

[pi]

It's a typo, Maths Quest is famous for typos in every edition.

[psyxwar]

It's a typo, Maths Quest is famous for typos in every edition.

Thanks. Yeah i thought so but then two worked examples in a row had that same mistake... Kinda threw me off lol

[Mieow]

Is there are way to find all the posts that you've made in this thread? I need help with a question but I'm pretty sure I already asked about it a while back