September 19, 2020, 10:05:52 pm

### AuthorTopic: VCE Specialist 3/4 Question Thread!  (Read 1321141 times) Tweet Share

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#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9675 on: July 13, 2020, 12:21:03 am »
+1
Doesn't seem like they give the initial horizontal velocity. Question is unsolvable with the information given.

EDIT: If hes dropped from the plane that means no initial vert velocity the only thing acting on the person is gravity so use vf^2 = 2ad to find vf after 600 and since the person travels at a constant speed for the rest thats the speed he lands with.

The question states to assume the person falling has the same vertical velocity as the plane, ergo there just be some vertical velocity
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#### withez

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9676 on: July 13, 2020, 10:06:21 am »
0
Doesn't seem like they give the initial horizontal velocity. Question is unsolvable with the information given.

EDIT: If hes dropped from the plane that means no initial vert velocity the only thing acting on the person is gravity so use vf^2 = 2ad to find vf after 600 and since the person travels at a constant speed for the rest thats the speed he lands with.

I reached the conclusion that the question is unsolvable too, but it comes from a vcaa provided sample sac. Which, before you ask, has no solutions.
At the risk of asking a dumb question, what is f?

Thank-you so much
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#### withez

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9677 on: July 13, 2020, 10:09:10 am »
0
The question states to assume the person falling has the same vertical velocity as the plane, ergo there just be some vertical velocity
Yes, I was working off that basis but I still couldn't work it out.
Thank-you so much
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#### S_R_K

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9678 on: July 13, 2020, 10:55:18 am »
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The question states to assume the person falling has the same vertical velocity as the plane, ergo there just be some vertical velocity

It states that the parachutist has the same initial velocity as the plane, and since the plane is "flying horizontally", it seems we should assume that the vertical component of the plane's velocity is zero.

But it seems to me that we need the horizontal component of the plane's velocity, otherwise it's not possible to find the distance of the landing point from O.

Or perhaps when the question states that the parachutist "drops in free fall for 600 m vertically" we are meant to assume that there is no horizontal component to his motion? If you make that assumption, the question is straightforward to solve, although clearly very stupid because it ignores inertia.

#### withez

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9679 on: July 13, 2020, 01:02:12 pm »
0
It states that the parachutist has the same initial velocity as the plane, and since the plane is "flying horizontally", it seems we should assume that the vertical component of the plane's velocity is zero.

But it seems to me that we need the horizontal component of the plane's velocity, otherwise it's not possible to find the distance of the landing point from O.

Or perhaps when the question states that the parachutist "drops in free fall for 600 m vertically" we are meant to assume that there is no horizontal component to his motion? If you make that assumption, the question is straightforward to solve, although clearly very stupid because it ignores inertia.
indeed
issue is, I can't find the horizontal velocity
If I could find that, the whole question is simple.
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#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9680 on: July 13, 2020, 01:50:05 pm »
0
indeed
issue is, I can't find the horizontal velocity
If I could find that, the whole question is simple.

Well then, how about we ignore this sillyness, and instead choose to work under the assumption that the initial velocity is some u with direction to the right. Makes the question more general, and you still get practice - are you able to answer the question this way, or would you still like some help?
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#### withez

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9681 on: July 13, 2020, 05:53:50 pm »
+1
Excellent idea
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#### withez

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9682 on: July 15, 2020, 12:03:32 pm »
0
Hello all
Another question:
L=40/*x+1) + b/(6-x)
Find range of values of b given a=40 such that min value of L is always 16 or higher
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#### Sine

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9683 on: July 15, 2020, 01:40:12 pm »
+2
Hello all
Another question:
L=40/*x+1) + b/(6-x)
Find range of values of b given a=40 such that min value of L is always 16 or higher
what aspect of this question are you finding difficult? What have you tried already?

#### 1729

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9684 on: July 15, 2020, 02:42:50 pm »
+3
Hello all
Another question:
L=40/*x+1) + b/(6-x)
Find range of values of b given a=40 such that min value of L is always 16 or higher
I'm assuming a is what you already replaced with 40 there, and also that you have some knowledge of calculus. Knowing that, we can easily enough find $L'(x) = \frac{-40}{(x+1)^2} + \frac{b}{(6-x)^2}$
We want to find where L has its minimum value, so we want to find points where L' = 0, so we can multiply through by the common denominator of the two fractions to make things convenient.
$L'(x) \cdot (x+1)^2 \cdot (6-x)^2 = (x+1)^2b - 40(6-x)^2$
The left side will only be 0 when L'=0, x=6, or x=-1, so we can set it to 0 as long as we're careful that our value of x satisyfing that isn't 6 or -1, giving us: $0 = (x+1)^2b - 40(6-x)^2$ Which is a quadratic equation that isn't yet in a nicely reduced form.

Then all you got to do is, expand (x+1)^2 * b - 40 * (6-x)^2. Giving $(b-40)x^2 + (2b+480)x + (b-1440)$ At which point you could apply the quadratic formula.

However, I am now realizing this is probably the wrong way of approaching the problem because this is ugly anyway you might be able to make something work by following that train of thought, I guess. I am pretty sure there is an easier way, but I can't seem to think of it now

Anyways hope this helps!
« Last Edit: July 15, 2020, 02:48:15 pm by 1729 »

#### chemistrykind

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9685 on: July 16, 2020, 03:27:20 pm »
+1
Hello all
Another question:
L=40/*x+1) + b/(6-x)
Find range of values of b given a=40 such that min value of L is always 16 or higher

This looks like a mandatory CAS question to me (although like 1729 said, the first thing to do is find the derivative and solve for zero). After defining

$L(x)=\frac{40}{(x+1)}+\frac{b}{6-x}$

you should get

$L'(x)=\frac{b}{(x-6)^2}-\frac{40}{(x+1)^2}$

Set this equal to 0, and you obtain the equation

$(b-40)x^2+(2b+480)x+(b-1440)=0$

Before jumping to solutions, note that the discriminant of this quadratic is 7840b. This means that b>0 for x to even exist where the gradient of L(x) is zero (which is a minimum, by the way - you can check this in graphs by substituting in values for b). Now you have your quadratic, solve for x:

$x=\frac{-(b+240)+14\sqrt{10b}}{b-40} , x=\frac{-(b+240)-14\sqrt{10b}}{b-40}$

Substitute this into your old L(x) and get two cool! new! L(b)s:

1. $L1(b)=\frac{(\sqrt{b}+2\sqrt{10})(b-40)}{7(\sqrt{b}-2\sqrt{10})}$

2. $L2(b)=\frac{(\sqrt{b}-2\sqrt{10})(b-40)}{7(\sqrt{b}+2\sqrt{10})}$

Now we need both L1 and L2 to be greater than or equal to 16. You can hit the good ol menu 3 1 for this:

1. $b\geq-8(2\sqrt{70}-19)\approx18.13$

2. $b\geq8(2\sqrt{70}+19)\approx285.866$

Since 2 is greater than 1, in order to make the statement true, we choose 2:

$b\geq8(2\sqrt{70}+19)$

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#### keltingmeith

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9686 on: July 16, 2020, 03:31:58 pm »
+2
However, I am now realizing this is probably the wrong way of approaching the problem because this is ugly

If maths were always pretty, it wouldn't be interesting. Don't assume you're wrong just because what you have looks ugly
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#### chemistrykind

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9687 on: July 16, 2020, 03:47:53 pm »
0
If maths were always pretty, it wouldn't be interesting. Don't assume you're wrong just because what you have looks ugly

if i see a pretty question on a CAS exam i know its fake lmao
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#### AlphaZero

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9688 on: July 16, 2020, 03:49:52 pm »
+1
Hello all
Another question:
L=40/*x+1) + b/(6-x)
Find range of values of b given a=40 such that min value of L is always 16 or higher

...
...

The question asks us to find the set of values of $b$ such that $L(x)\geq 16\ \,\forall\,x$. Both these solutions (incorrectly) argue that the minimum of $L$ must occur at a point where $L'(x)=0$.

In fact, $L(x)$ isn't even bounded below since $L(x)\to -\infty\ \ \text{as}\ \ x\to-1^-\ \ \forall\,b\in\mathbb{R}.$ Thus, the correct answer is just the empty set!
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#### chemistrykind

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##### Re: VCE Specialist 3/4 Question Thread!
« Reply #9689 on: July 16, 2020, 03:55:10 pm »
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The question asks us to find the set of values of $b$ such that $L(x)\geq 16\ \,\forall\,x$. Both these solutions (incorrectly) argue that the minimum of $L$ must occur at a point where $L'(x)=0$.

In fact, $L(x)$ isn't even bounded below since $L(x)\to -\infty\ \ \text{as}\ \ x\to-1^-\ \ \forall\,b\in\mathbb{R}.$ Thus, the correct answer is just the empty set!

I made the assumption that the question meant local minimum but ur so right omg :00
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