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February 25, 2020, 05:36:52 pm

Author Topic: 4U Maths Question Thread  (Read 311181 times)  Share 

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RuiAce

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Re: 4U Maths Question Thread
« Reply #2445 on: January 07, 2020, 11:02:52 pm »
+1
In this question we need to find the area of the curve bounded by the abscissas. So we need to make x the subject. I have many operations but I can't find a way of making x the subject.
\begin{align*}y &= \frac{x}{\sqrt{x^2+5}}\\ y^2 &= \frac{x^2}{x^2+5}\\ y^2(x^2+5) &= x^2\\ 5y^2 &= x^2 - x^2y^2\\ 5y^2 &= x^2(1-y^2)\\ x^2 &= \frac{5y^2}{1-y^2} \\ x &= \frac{\sqrt{5}y}{\sqrt{1-y^2}}\end{align*}
Note that the positive square root was taken here because we are working in the first quadrant. In the first quadrant, \(x \geq 0\).

(Note also that this question could be done without remaking \(x\) the subject. One way is to consider the area of a rectangle, minus the area of some region by the \(x\)-axis. But both methods would've relied on the same integration technique.)

shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2446 on: January 08, 2020, 12:07:46 pm »
0
\begin{align*}y &= \frac{x}{\sqrt{x^2+5}}\\ y^2 &= \frac{x^2}{x^2+5}\\ y^2(x^2+5) &= x^2\\ 5y^2 &= x^2 - x^2y^2\\ 5y^2 &= x^2(1-y^2)\\ x^2 &= \frac{5y^2}{1-y^2} \\ x &= \frac{\sqrt{5}y}{\sqrt{1-y^2}}\end{align*}
Note that the positive square root was taken here because we are working in the first quadrant. In the first quadrant, \(x \geq 0\).

(Note also that this question could be done without remaking \(x\) the subject. One way is to consider the area of a rectangle, minus the area of some region by the \(x\)-axis. But both methods would've relied on the same integration technique.)

Oh, that makes sense. Thanks.
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shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2447 on: January 25, 2020, 07:33:43 pm »
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Hi, I have a question. Would both these answers be considered valid.

Convert -4i into polar form
4 cis (3pi/2)
4 cis (-pi/2)
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2448 on: January 26, 2020, 11:24:17 am »
+2
Hi, I have a question. Would both these answers be considered valid.

Convert -4i into polar form
4 cis (3pi/2)
4 cis (-pi/2)
They're both correct, but ideally try to use the principle argument where possible. Here, \( \operatorname{Arg}(z) = -\frac\pi2\).

shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2449 on: January 27, 2020, 01:58:39 pm »
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They're both correct, but ideally try to use the principle argument where possible. Here, \( \operatorname{Arg}(z) = -\frac\pi2\).
Ok. Thanks
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milie10

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Re: 4U Maths Question Thread
« Reply #2450 on: January 31, 2020, 11:49:25 pm »
0
Hi!


Is my working out right for this? it says answers may vary, but how did they get the answer below?:


Also, I'm really struggling with these questions:




Thanks!
« Last Edit: February 01, 2020, 01:28:50 am by milie10 »

RuiAce

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Re: 4U Maths Question Thread
« Reply #2451 on: February 01, 2020, 02:55:17 am »
+3
Hi!

(Image removed from quote.)
Is my working out right for this? it says answers may vary, but how did they get the answer below?:
(Image removed from quote.)

Also, I'm really struggling with these questions:
(Image removed from quote.)

(Image removed from quote.)

Thanks!
Your working makes perfect sense to me. But just take note that \( \frac{\lambda}{\frac12 \lambda - \frac32} \) reads like a fraction: "lambda over (half lambda - 3 halves)". Try to keep to vector notation: \( \begin{pmatrix} \lambda\\ \frac12 \lambda - \frac32\end{pmatrix} \).
\[ \text{Their working out looks as though they parametrised }y\text{ instead.}\\ \text{That is, they set }y=\lambda.\quad (\lambda \in \mathbb{R}) \]
\[ \text{Then }x=3+2y\text{, so }x=3+2\lambda.\\ \text{Continue with the same approach to get to their answer.} \]
In general, it doesn't matter which variable you choose to parametrise.
___________________________________________________________________________

You're expected to know and understand that in the vector equation \( \vec{u} = \begin{pmatrix}a\\b\end{pmatrix} + \lambda \begin{pmatrix}x\\y\end{pmatrix} \) that \( \begin{pmatrix}a\\b\end{pmatrix} \) is the position vector of a point on the line. With that in mind, for both questions 3 and 4 you should be able to figure out the \( \begin{pmatrix}a\\b\end{pmatrix} \) vector.

As for the direction vector \( \begin{pmatrix}x\\y\end{pmatrix} \):

- For Q3, you've already begun to use \(m=\tan\theta\). In general, if \(m\) is the gradient of the line, then a direction vector for the line is \( \begin{pmatrix}1\\m\end{pmatrix}\). Think about why! (Or if you can't think about why intuitively, prove it by finding the vector equation of the line itself.)
 
- For Q4, note that the angle is not made with respect to the positive \(x\)-axis this time. Rather, it's perpendicular to some random line, in particular the line \(x+3y+1=0\). Using only vector techniques, you would probably want to convert this line into its vector equation form.

Once you've done this, apply the scalar product formula \( \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta \), where \(\theta = 45^\circ\), to find the possible direction vectors for the lines you seek. Set \(\vec{a}\) to be the direction vector of \(x+3y+1=0\), and in the same spirit as Q3, set \(\vec{b} = \begin{pmatrix}1\\m\end{pmatrix} \).

(The syllabus is still new, so I'm not aware of any concrete rules of thumb in doing these problems. I'm relying on what makes sense based off what I know, and what the syllabus allows for.)
___________________________________________________________________________

The last problem is the shortest distance between a point in a line problem, but necessitating vector methods. When vectors are required for shortest distances, always think projections.

Suppose that you want the shortest distance from a point \( P \) to a line. Let \( \vec{p}\) be the position vector of the point. Furthermore, suppose that the equation of the line is \( \vec{u}=\vec{a}+\lambda\vec{b}\).

1. Then the vector \( \vec{a} - \vec{p}\) is a vector that joins the point \(P\), to some random point on the line.
2. Now, \( \operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) \) represents the projection of the vector you found above, onto this line.
3. Therefore, \( \operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) - \vec{p}\) is another vector joining \(P\), to a point on the line. However, this vector is special, in that it is perpendicular to the line.
4. The shortest distance between a point and a line is always the perpendicular distance between them. Hence, the answer you seek can be found by taking the magnitude of this vector, i.e. \(|\operatorname{proj}_{\vec{b}} (\vec{a}-\vec{p}) - \vec{p}|\).

Note: All of the above is hard to visualis in your head. You should use a diagram when going through all of that.

In your case, \( \vec{p}= \begin{pmatrix}2\\3\end{pmatrix}\). You have to convert your line into its vector equation to write down your \(\vec{a}\) and \(\vec{b}\).

shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2452 on: February 12, 2020, 06:19:59 pm »
0
Hi. I am not quite sure why we integrate with the following limits.
Thanks.
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fun_jirachi

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Re: 4U Maths Question Thread
« Reply #2453 on: February 12, 2020, 09:59:17 pm »
+2
Hi. I am not quite sure why we integrate with the following limits.
Thanks.

Hey there!

One of the ideas that was stressed to me as a student was 'look at what you're working towards'. It's become a really handy hint - note that in the result, the logarithm does not use 'n', rather, it uses '1/n'. Since we know the integral of a linear function is a logarithm, and we can directly see the connection between the given result and the result we're working towards (particularly in the middle term), it makes sense to try out 1/n as an upper limit. As a result of this, this means the right hand side will become 1/n, meaning everything will be multiplied by n, leaving a power on the logarithm. This kind of foresight often reduces the amount of work that needs to be done trialling different limits amongst other things - attempt to use it wherever you can.

Hope this helps :)
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milie10

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Re: 4U Maths Question Thread
« Reply #2454 on: February 15, 2020, 01:11:59 am »
0
Hi!

Another vector question I'm struggling with:
"Find the shortest distance between the line through the points (1, 3, 1) and (1, 5, -1) and the line through the points (0, 2, 1) and (1, 2, -3)"

how do you find the equations using vector methods? what would the subsequent steps be?

Thanks!!!
« Last Edit: February 15, 2020, 01:18:52 am by milie10 »

shekhar.patel

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Re: 4U Maths Question Thread
« Reply #2455 on: February 15, 2020, 11:29:31 am »
0
Hey there!

One of the ideas that was stressed to me as a student was 'look at what you're working towards'. It's become a really handy hint - note that in the result, the logarithm does not use 'n', rather, it uses '1/n'. Since we know the integral of a linear function is a logarithm, and we can directly see the connection between the given result and the result we're working towards (particularly in the middle term), it makes sense to try out 1/n as an upper limit. As a result of this, this means the right hand side will become 1/n, meaning everything will be multiplied by n, leaving a power on the logarithm. This kind of foresight often reduces the amount of work that needs to be done trialling different limits amongst other things - attempt to use it wherever you can.

Hope this helps :)

Ok thanks.
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RuiAce

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Re: 4U Maths Question Thread
« Reply #2456 on: February 16, 2020, 05:49:00 pm »
0
Hi!

Another vector question I'm struggling with:
"Find the shortest distance between the line through the points (1, 3, 1) and (1, 5, -1) and the line through the points (0, 2, 1) and (1, 2, -3)"

how do you find the equations using vector methods? what would the subsequent steps be?

Thanks!!!
This is actually an extremely demanding question given the scope of MX2. The shortest distance between two lines usually requires knowledge of planes in 3D space or the cross product, both of which arenít in the syllabus. What is the source of this question?

milie10

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Re: 4U Maths Question Thread
« Reply #2457 on: February 17, 2020, 08:10:00 pm »
0
This is actually an extremely demanding question given the scope of MX2. The shortest distance between two lines usually requires knowledge of planes in 3D space or the cross product, both of which arenít in the syllabus. What is the source of this question?

I think my coaching taught us some out of syllabus content for vectors- they said that for completeness of this topic, learning the cross product is necessary. I'd still love to know how it would be done though, but I'm not too fussed about it since it won't be tested :)

Also, could someone explain this proofs question: Q10d from the cambridge textbook? I've forgotten how roots work- am I meant to use the sum and product of roots?


thanks so much :D
« Last Edit: February 17, 2020, 08:15:51 pm by milie10 »

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Re: 4U Maths Question Thread
« Reply #2458 on: February 17, 2020, 09:26:26 pm »
+1
I'll leave the vector question for a more competent user while I continue getting accustomed to them myself, but I'll answer your second question :)

There are a few ways to prove to yourself this is true:
- You can prove to yourself that they are one and the same by doing the sum of roots and product of roots of both those quadratics
- You can also consider the roots as per the quadratic equation for each quadratic, inverting them and rationalising the denominator
- Also, consider what happens when you have \(f(\alpha)\) in the first quadratic, and factor out \(\alpha ^2\), and vice versa

Hopefully this will start to make sense! :)
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milie10

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Re: 4U Maths Question Thread
« Reply #2459 on: February 17, 2020, 09:32:00 pm »
0
I'll leave the vector question for a more competent user while I continue getting accustomed to them myself, but I'll answer your second question :)

There are a few ways to prove to yourself this is true:
- You can prove to yourself that they are one and the same by doing the sum of roots and product of roots of both those quadratics
- You can also consider the roots as per the quadratic equation for each quadratic, inverting them and rationalising the denominator
- Also, consider what happens when you have \(f(\alpha)\) in the first quadratic, and factor out \(\alpha ^2\), and vice versa

Hopefully this will start to make sense! :)

Thank you so much!!