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March 29, 2024, 10:54:27 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164658 times)  Share 

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S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9720 on: August 27, 2020, 01:03:14 pm »
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Make sure you read the question. You may be asked to label points with coordinates, in which case your labelling should be (a, b) rather than a + bi. Alternatively, you may be asked to write down points in the form a + bi. If no specific form is asked for, then either is acceptable.

This is of course a hilariously pedantic non-issue (since C is isomorphic to R^2), but don't give VCAA the opportunity to punish you for this.

Note this also applies with vectors - if asked to find the location / position, then check whether your answer should be given as a vector (ie. a linear combination of i, j) or as coordinates.

angrybiscuit

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9721 on: August 28, 2020, 03:28:45 pm »
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How do I verify a point of inflection? Other than having the double derivative equal to zero?
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keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9722 on: August 28, 2020, 03:42:01 pm »
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How do I verify a point of inflection? Other than having the double derivative equal to zero?

Erm, context of the question?

angrybiscuit

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9723 on: August 28, 2020, 04:15:22 pm »
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Erm, context of the question?
We are given the double derivative and have to find a POI, then verify it is a POI. Exam report says having double derivative equal zero is insufficient to verify it.
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keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9724 on: August 28, 2020, 04:32:36 pm »
+3
We are given the double derivative and have to find a POI, then verify it is a POI. Exam report says having double derivative equal zero is insufficient to verify it.

OH okay, so this comes down to the mathematical definition of sufficient. All points of inflection require that the double derivative is 0, but the double derivative being 0 doesn't mean something is a point of inflection. For example, consider f(x)=x^4. f''(x)=12x^2, and f''(0)=0 - but f(0) is a local minimum, not a POI. Because of this, we say that f''(x)=0 is insufficient for something to being a point of inflection - not because you CAN'T use f''(x)=0, but because you need to do another test just to make sure what you have is a point of inflection.

To PROVE that something is a POI, first you need to find a list of candidates - you find these by solving f''(x)=0. Next, you need to look at the gradient either side of the point where f''(x)=0 - if the gradients are the same sign, then what you have is a POI. If you're unsure why, try drawing the slope table and seeing what it looks like.

For example, see f(x)=x^3-3x. f'(x)=3x^2-3, f''(x)=6x. So, we know we need to test when x=0. At x=-0.1, f'(x)=3(-0.1)^2-3=3*0.01-3=-2.97<0. At x=+0.1, f'(x)=3(0.1)^2-3=3*0.01-3=-2.97<0. Since the gradient on either side is the same, what we have is a POI.

p0kem0n21

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9725 on: August 30, 2020, 07:12:11 pm »
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when factorizing polynomials, does vcaa accept synthetic division //// ik it's kind of a methods question too but

keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9726 on: August 30, 2020, 07:14:11 pm »
+4
when factorizing polynomials, does vcaa accept synthetic division //// ik it's kind of a methods question too but

Yes - they will accept any method as long as it is mathematically correct. However, I will urge caution - make sure you understand how synthetic division works for non-linear divisors if you're going to use it, because it gets complicated fast

Chocolatemilkshake

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9727 on: August 30, 2020, 10:27:25 pm »
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Hello,
When calculating the angle of projection or the angle a particle strikes the ground for a vector function, do you use tanθ=dy/dx (when t=0 for projection or t=time particle hits ground)? Why don't you use tanθ=y/x, and is there any time where you do use this? Is this different from calculating the angle of elevation at a given time?
Thanks, I'm just for some reason really confused about this.
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keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9728 on: August 30, 2020, 11:09:09 pm »
+4
Hello,
When calculating the angle of projection or the angle a particle strikes the ground for a vector function, do you use tanθ=dy/dx (when t=0 for projection or t=time particle hits ground)? Why don't you use tanθ=y/x, and is there any time where you do use this? Is this different from calculating the angle of elevation at a given time?
Thanks, I'm just for some reason really confused about this.

I think you may be thinking too much of formulas that you regurgitate, instead of thinking about the physical system. Golden rule for maths - if you can graph it, you should graph it. So, consider the picture I've attached to this post.

This picture shows the parabolic motion for whatever is moving along your typical projectile motion. Maybe the motion isn't parabolic - but let's pretend it is, just for this picture. If you want to figure out the angle a particle is projected at, or the angle it'll land the ground at, then you need to draw a triangle, and use SOH-CAH-TOA to figure out the angle.

So, can we use the displacement curve? Well, no - because we're interested in the angle at a point, and this triangle has straight lines that only touch the curve once. In fact, that one time is tangent to the curve - so maybe we can use the derivative? In fact, that triangle's two side lengths are going to be dy/dt and dx/dt, because that's the derivative - and so, the the triangle as a whole should have:



Which is the equation we wanted! Important to note - we didn't use the fact that our motion was "parabolic" in any way, so this "proof" will still work, REGARDLESS of the type of motion. However, by pretending it was and drawing it out, we were able to come to a better understanding of what was going on, and come up with a formula to solve for the setting we were in. Do you think, being able to see this diagram, you can think of any time we might use a formula tan(theta)=y/x? If you can't think of any time we might see this, why do you think we might not? This is just a question to test your understanding, so just give it a go, you have nothing to lose if you get it wrong

Chocolatemilkshake

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9729 on: August 31, 2020, 06:26:58 am »
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Wow thank you so much this makes so much more sense I wish my teacher had it explained it like that.

Do you think, being able to see this diagram, you can think of any time we might use a formula tan(theta)=y/x? If you can't think of any time we might see this, why do you think we might not? This is just a question to test your understanding, so just give it a go, you have nothing to lose if you get it wrong
Okay I'll give it ago...Would you use tanθ=y/x when the particle is hovering at a certain point on the curve and you want to calculate the angle of elevation? You wouldn't be able to use the derivative because you want the angle that the object makes with the origin (and therefore you would use tan(θ)=y/x (the j and i components of the vectors). However, if you wanted the angle at which the particle hits the ground you would use the derivatives again because you want the exact angle it makes when it hits the ground?
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keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9730 on: August 31, 2020, 12:32:28 pm »
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Wow thank you so much this makes so much more sense I wish my teacher had it explained it like that.
Okay I'll give it ago...Would you use tanθ=y/x when the particle is hovering at a certain point on the curve and you want to calculate the angle of elevation? You wouldn't be able to use the derivative because you want the angle that the object makes with the origin (and therefore you would use tan(θ)=y/x (the j and i components of the vectors). However, if you wanted the angle at which the particle hits the ground you would use the derivatives again because you want the exact angle it makes when it hits the ground?

Nope - you would still use tan(theta)=dy/dx even then, because if you think about it, the angle it makes with the ground is the same as it makes with any line parrallel to the x-axis - see attached. In truth, I don't think there is a time you'd use tan(theta)=y/x - unless you wanted to find an approximate angle (just like with approximate rates of change in methods). It's good to see you thinking about the situation, though :)

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9731 on: August 31, 2020, 12:55:02 pm »
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KM your first post is correct but your second is not. In some cases it is necessary to use the position vector, rather than the velocity vector, to find an angle.

Compare the following two questions:

1. Find the angle, relative to the horizontal, of the particle's path / trajectory / direction of motion at a time t.

2. Find the angle of elevation of the particle's position at a time t.

The first question requires use of the derivative, since the angle of the path / trajectory / direction of motion is given by the gradient of the tangent to the path.

The second question requires use of the position vector function, since it simply asks what is the angle between the vector from the origin to the particle's position, and the projection of that vector in the horizontal plane.

Compare the following two questions from past VCAA exams: 2007 Exam 2 4c; 2015 Exam 2 4aii.

Also, while the tan(theta) = dy/dx method is useful for finding the angle of a path with the horizontal axis, students should also know how to find the angle between a path and a vector in any direction. See 2016 Exam 4c. This can be done using tan(theta) = dy/dx and a compound angle formula, or by using the dot product of the two derivative vectors.

keltingmeith

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9732 on: August 31, 2020, 01:40:35 pm »
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KM your first post is correct but your second is not. In some cases it is necessary to use the position vector, rather than the velocity vector, to find an angle.

Compare the following two questions:

1. Find the angle, relative to the horizontal, of the particle's path / trajectory / direction of motion at a time t.

2. Find the angle of elevation of the particle's position at a time t.

The first question requires use of the derivative, since the angle of the path / trajectory / direction of motion is given by the gradient of the tangent to the path.

The second question requires use of the position vector function, since it simply asks what is the angle between the vector from the origin to the particle's position, and the projection of that vector in the horizontal plane.

Compare the following two questions from past VCAA exams: 2007 Exam 2 4c; 2015 Exam 2 4aii.

Also, while the tan(theta) = dy/dx method is useful for finding the angle of a path with the horizontal axis, students should also know how to find the angle between a path and a vector in any direction. See 2016 Exam 4c. This can be done using tan(theta) = dy/dx and a compound angle formula, or by using the dot product of the two derivative vectors.

Oops, your right, I was misreading the question. That's what I get for answering things while rushing off to meetings

p0kem0n21

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9733 on: September 03, 2020, 10:08:11 am »
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Can someone explain to me what the scalar resolute shows in relation to the vector resolute/what is it used for? (general question because I don't understand the concept)

S_R_K

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Re: VCE Specialist 3/4 Question Thread!
« Reply #9734 on: September 03, 2020, 11:16:16 am »
+2
Can someone explain to me what the scalar resolute shows in relation to the vector resolute/what is it used for? (general question because I don't understand the concept)

The absolute value of the scalar resolute is the magnitude of the vector resolute.

So, thinking of vectors as arrows, the absolute value of the scalar resolute of a in the direction of b gives the length of the component of a that is parallel to b.