VCE Specialist 3/4 Question Thread!

[VanillaRice]

Hi,

Could I have some help with the VCAA 2014 E2?

MCQ14) they say the slope field lines that are on the line y = x are undefined. But how can we know this when the axes don't have values on them?

MCQ18) When resolving the force F2, why can't we split the 60 degree angle with F3 in half, so we have F2sin(30) for the vertical component?

Q2c) using the answer's diagram, why can't we use the triangle formed by the points (k, -2), (0, -2) and (0, -2-k)? Why can't we take cos(45) of this triangle where the 45 angle is at the vertice of (k, -2)?

Also, for that question for some reason I'm not convinced the angle made by the dotted line and y-axis is 45 degrees?

14) I would agree that it's probably a bit hard to tell from the unlabelled axes. I would personally just play around with the diagram on my calc.

18) I'm pretty sure you can. Are you referring to resolving the forces which are vertical? (Note: sin(60) = cos(30) by complementary properties)
We can resolve them as
F₁+F₂sin(30)-F₃sin(30) = 0

2c) I believe the problem lies in the horizontal and vertical sides of the triangle you are referring to both having the same length of k. k is an arbitrary constant, and so the angle will be 45 degrees no matter what k is. For example, if k = 1, the triangle will have the two perpendicular sides with length, and the angle is 45 degrees. If k = 2, the angle will still be 45 degrees. The reason that they used the other triangle (as per the examiner's report) is that we know that the hypotenuse is 2 - so we can solve for k. Meanwhile, in the triangle you have indicated, we have no certain side lengths (they are k, k and sqrt(2k²)) - we only know that they are in the same ratio. Hope this makes sense.
Also, regarding the angle - if we know the gradient of the line is 1, it must make an angle of 45 degrees with the positive x-axis. The dotted line joins a point of tangent of the circle with its centre - which must mean the intersection of the dotted line and the solid line is perpendicular. Combining these two facts, you can work out that the angle is 45 degrees.

Hope this helps

[Rieko Ioane]

14) I would agree that it's probably a bit hard to tell from the unlabelled axes. I would personally just play around with the diagram on my calc.

18) I'm pretty sure you can. Are you referring to resolving the forces which are vertical? (Note: sin(60) = cos(30) by complementary properties)
We can resolve them as
F₁+F₂sin(30)-F₃sin(30) = 0

2c) I believe the problem lies in the horizontal and vertical sides of the triangle you are referring to both having the same length of k. k is an arbitrary constant, and so the angle will be 45 degrees no matter what k is. For example, if k = 1, the triangle will have the two perpendicular sides with length, and the angle is 45 degrees. If k = 2, the angle will still be 45 degrees. The reason that they used the other triangle (as per the examiner's report) is that we know that the hypotenuse is 2 - so we can solve for k. Meanwhile, in the triangle you have indicated, we have no certain side lengths (they are k, k and sqrt(2k²)) - we only know that they are in the same ratio. Hope this makes sense.
Also, regarding the angle - if we know the gradient of the line is 1, it must make an angle of 45 degrees with the positive x-axis. The dotted line joins a point of tangent of the circle with its centre - which must mean the intersection of the dotted line and the solid line is perpendicular. Combining these two facts, you can work out that the angle is 45 degrees.

Hope this helps

Thanks a lot!

---
For VCAA SM E1:
Q7b) Would it be wrong to include the y = -x/2 asymptote?

Also, do double check, the end points of arccos graphs have to have an undefined gradients at its end points?

[VanillaRice]

Thanks a lot!

---
For VCAA SM E1:
Q7b) Would it be wrong to include the y = -x/2 asymptote?

Also, do double check, the end points of arccos graphs have to have an undefined gradients at its end points?

Which year are you referring to?

Yes, the gradient (derivative) is undefined at any endpoint (as well as any sharp points).

[Rieko Ioane]

Which year are you referring to?

Yes, the gradient (derivative) is undefined at any endpoint (as well as any sharp points).

Thanks. I forgot to say it was the 2013 exam.

Also, I might've been unclear. I meant the direction of the slope at the actual endpoints of an arccos graph have to be undefined/vertically up (not referring to its derivative function).

[VanillaRice]

Thanks. I forgot to say it was the 2013 exam.

Also, I might've been unclear. I meant the direction of the slope at the actual endpoints of an arccos graph have to be undefined/vertically up (not referring to its derivative function).

2013 Q7b) If you did include that asymptote and I was marking, I would personally not deduct any marks for that. However, I don't think that's necessary to put it in anyway, since you're not drawing any part of the curve which is approaching the asymptote.

RE: arcos graph - in theory, the final part of the curve should be pointing directly up/downwards, although you'd need a microscope to see it actually pointing upwards. As long as it looks like its approaching the up/down direction, you should be fine. I wouldn't recommend trying to put in a flat line pointing up/down at the nds. It's the same as if you were drawing a normal cos(x) graph - you don't really notice that the tangent is 0 at the turning points.

Hope this helps

[Ahmad_A_1999]

Hey guys, could someone please explain MC 10 on the spesh 2016 exam 2, I really don't understand differential fields at all, and given that 65% got this right, something is seriously wrong with me
Thank you!

[VanillaRice]

Hey guys, could someone please explain MC 10 on the spesh 2016 exam 2, I really don't understand differential fields at all, and given that 65% got this right, something is seriously wrong with me
Thank you!

Recall the constant of integration when you evaluate an indefinite integral. In essence, directional fields show you the family of curves which you could have when you solve a differential equation. Each line is a tangent to the curve at that point.

In terms of doing this question, what calculator do you have? Hopefully you have the CAS TI-nspire, because that's what I'm using to show you how to do this question
1) Open a new calculator page, and go Menu > 3 > 7
2) Enter in your function. Here, you need to pretend dy/dx is on the left, and everything else is on the right. So enter -x-y1 Make sure you include the 1 after y (y1), it's important (don't worry about why)!
3) In the two boxes underneath where you enter your function, put in the coordinates (0,-1). These are your initial conditions.
4) Hit enter. You will see the directional field, plus a graph. This is the solution which passes through the points (0,-1).
5) Go Menu > 5:Trace
6) Hit right/left arrows to find the coordinates which best matches the answer. In this case, you should get (3.5, -2.55), which best matches A.


Alternatively, you could try to trace a graph using the tangents, but this is quite error prone (but still works for some).
Alternatively again, you could simply solve the differential equation in your calculator using the initial condition (0,-1), and see which of the options best fits your solution.

Hope this helps

[Sine]

Hey guys, could someone please explain MC 10 on the spesh 2016 exam 2, I really don't understand differential fields at all, and given that 65% got this right, something is seriously wrong with me
Thank you!

beaten by VanillaRice but Alternative Solution

locate the point (0,-1) in the differential field then try to determine which "lines" of the differential field that point is associated with.
In this case there is a line exactly on the point (0,-1) so try to follow it onto nearby lines (of the same solution)
Draw the curve in to make it easier for you.
Now "check" each of the answers and see whether they fall on the curve you have drawn in.

[Ahmad_A_1999]

Recall the constant of integration when you evaluate an indefinite integral. In essence, directional fields show you the family of curves which you could have when you solve a differential equation. Each line is a tangent to the curve at that point.

In terms of doing this question, what calculator do you have? Hopefully you have the CAS TI-nspire, because that's what I'm using to show you how to do this question
1) Open a new calculator page, and go Menu > 3 > 7
2) Enter in your function. Here, you need to pretend dy/dx is on the left, and everything else is on the right. So enter -x-y1 Make sure you include the 1 after y (y1), it's important (don't worry about why)!
3) In the two boxes underneath where you enter your function, put in the coordinates (0,-1). These are your initial conditions.
4) Hit enter. You will see the directional field, plus a graph. This is the solution which passes through the points (0,-1).
5) Go Menu > 5:Trace
6) Hit right/left arrows to find the coordinates which best matches the answer. In this case, you should get (3.5, -2.55), which best matches A.


Alternatively, you could try to trace a graph using the tangents, but this is quite error prone (but still works for some).
Alternatively again, you could simply solve the differential equation in your calculator using the initial condition (0,-1), and see which of the options best fits your solution.

Hope this helps

Thanks for the reply! I've got the CAS classpad, so I can't do that with this but with how you said you could solve the differential equation using the calc with the initial condition, how would I go about doing that? there's a y in the term and not sure how to go about.

beaten by VanillaRice but Alternative Solution

locate the point (0,-1) in the differential field then try to determine which "lines" of the differential field that point is associated with.
In this case there is a line exactly on the point (0,-1) so try to follow it onto nearby lines (of the same solution)
Draw the curve in to make it easier for you.
Now "check" each of the answers and see whether they fall on the curve you have drawn in.

Thanks Sine! I get the answer doing it that way

[Ahmad_A_1999]

Ahhh could I please get help with spesh 2016 MC 14 

[Syndicate]

Ahhh could I please get help with spesh 2016 MC 14 

This triangle is a right angled triangle. The angle between T2 and the horizontal is arcos(3/5), whilst the angle between T1 and the horizontal is arcos(4/5)

Acos(arccos(3/5)) = Bcos(arccos(4/5)) (for the sake of simplicity, A represents T2 and B represents T1)

3A/5 = 4B/5

We want the ratio T1/T2 (B/A)

Therefore B/A = 3/4 (Option B)

[VanillaRice]

Thanks for the reply! I've got the CAS classpad, so I can't do that with this but with how you said you could solve the differential equation using the calc with the initial condition, how would I go about doing that? there's a y in the term and not sure how to go about.

Not sure about the classpad, but on the ti-nspire, there's a function that allows you to input a differential equation and solve.

Sorry, can't be of more help on this unfortunately 

[Ahmad_A_1999]

This triangle is a right angled triangle. The angle between T2 and the horizontal is arcos(3/5), whilst the angle between T1 and the horizontal is arcos(4/5)

Acos(arccos(3/5)) = Bcos(arccos(4/5)) (for the sake of simplicity, A represents T2 and B represents T1)

3A/5 = 4B/5

We want the ratio T1/T2 (B/A)

Therefore B/A = 3/4 (Option B)

THANK YOU!!!

Not sure about the classpad, but on the ti-nspire, there's a function that allows you to input a differential equation and solve.

Sorry, can't be of more help on this unfortunately 

All good!! I'll wing it

[Rieko Ioane]

Hi,

Would I lose marks for drawing the y = -x asymptote? https://imgur.com/a/EH1An

I didn't draw the quadrant 4 part of the hyperbola but I did draw the asymptote that it would go towards.

Does the restriction on t affect the asymptotes? If so how?

[keltingmeith]

Hi,

Would I lose marks for drawing the y = -x asymptote? https://imgur.com/a/EH1An

I didn't draw the quadrant 4 part of the hyperbola but I did draw the asymptote that it would go towards.

Does the restriction on t affect the asymptotes? If so how?

Whether you would lose marks or not, I don't know - but, asymptotes only occur as x and y go to infinity. The restriction on t isn't important, but how it affects y and x is.