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March 28, 2024, 10:38:00 pm

Author Topic: 4U Maths Question Thread  (Read 659804 times)  Share 

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RuiAce

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Re: 4U Maths Question Thread
« Reply #735 on: December 23, 2016, 12:02:58 pm »
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Find a and b.
2+i=(1+31)/(a+bi)
Is that 1+31 or 1+3i there on the top?

Also, this is just elementary complex numbers division. Have you been taught how to do this

Jakeybaby

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Re: 4U Maths Question Thread
« Reply #736 on: December 23, 2016, 12:04:11 pm »
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Is that 1+31 or 1+3i there on the top?

Also, this is just elementary complex numbers division. Have you been taught how to do this
I'd assume that it's 1+3i, yet again, too quick for me
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bluecookie

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Re: 4U Maths Question Thread
« Reply #737 on: December 23, 2016, 12:22:55 pm »
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I'd assume that it's 1+3i, yet again, too quick for me


Yes its 1+3i.

Is that 1+31 or 1+3i there on the top?

Also, this is just elementary complex numbers division. Have you been taught how to do this

Yep. Aha, I just figured it out then! Thanks :)

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Re: 4U Maths Question Thread
« Reply #738 on: December 23, 2016, 05:57:37 pm »
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Prove by induction.

a) Conjugate(z1+z2+...zn)=conjugate(z1)+conjugate(z2)+...conjugate(zn)
b) Conjugate(z1*z2*...zn)=conjugate(z1)*conjugate(z2)*...conjugate(zn)

RuiAce

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Re: 4U Maths Question Thread
« Reply #739 on: December 23, 2016, 10:08:13 pm »
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Prove by induction.

a) Conjugate(z1+z2+...zn)=conjugate(z1)+conjugate(z2)+...conjugate(zn)
b) Conjugate(z1*z2*...zn)=conjugate(z1)*conjugate(z2)*...conjugate(zn)
When typing in words, feel free to use conj(z1) for the conjugate




The addition question is very similar to this. I will leave it as your exercise to do.
« Last Edit: December 23, 2016, 10:09:52 pm by RuiAce »

bluecookie

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Re: 4U Maths Question Thread
« Reply #740 on: December 24, 2016, 10:24:33 am »
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Thank you :)

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Re: 4U Maths Question Thread
« Reply #741 on: January 15, 2017, 05:07:05 pm »
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7. A chord AB of a circle makes an angle theta with the diameter passing through A. If the area of the minor segment is one-quarter the area of the circle, show that sin2theta=pi/2 - 2theta. Sovle this equation graphically.
8. A chord AB of a circle subtends an angle theta at the centre of the circl. If the perimeter of the minor segment is one-half the circumference of the circle, show that 2*sin(theta/2)=pi-theta. Solve this equation graphically.

jakesilove

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Re: 4U Maths Question Thread
« Reply #742 on: January 15, 2017, 07:23:16 pm »
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7. A chord AB of a circle makes an angle theta with the diameter passing through A. If the area of the minor segment is one-quarter the area of the circle, show that sin2theta=pi/2 - 2theta. Sovle this equation graphically.
8. A chord AB of a circle subtends an angle theta at the centre of the circl. If the perimeter of the minor segment is one-half the circumference of the circle, show that 2*sin(theta/2)=pi-theta. Solve this equation graphically.

First, make sure to draw out the relevant diagram, and label anything that's easy to figure out. Labeling the origin as 'O', and joining a line between O and B, we find that



As they are both the radius of the circle. As such,



As the base angles in an isosceles triangles are equal. Therefore, since the angle sum of a triangle is pi radians,



Using the area of a minor segment formula, we know that (for subtending angle alpha)



Subbing in the angle we found above, the area of the minor segment is going to be



As the area of the segment is a quarter of the area of the circle, we note that





Play around with this last line, and you'll quickly get the desired result. Then, plot the graph of sin(2theta) and pi/2 - 2theta. Find out where they intersect; it should be around 0.42.

I'm working on the second question now.

The second question is probably much easier; we know that the length of an arc will be



As we need the perimeter of the minor segment, so we also need the length of the chord AB. We can imagine that half the length is like a right angle triangle, with angle theta/2 and hypotenuse r. Therefore, the total length of the chord AB will be



The total perimeter is therefore



We know that the perimeter is half the circumference of the circle. Therefore,



Again, if you play around with this line, you'll get the desired result. Draw it graphically, you should get a number around 1.7
« Last Edit: January 15, 2017, 07:33:16 pm by jakesilove »
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kiwiberry

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Re: 4U Maths Question Thread
« Reply #743 on: January 21, 2017, 01:48:16 am »
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Hey, could I please get some help with this question :)

1) P(3secθ,2tanθ) and Q(3secθ,-2tanθ) are two points on the hyperbola x2/9 - y2/4 = 1. The normal at P meets the line OQ at R. Show that the locus of R is a concentric hyperbola

Thank you!!!
« Last Edit: January 21, 2017, 01:55:44 am by kiwiberry »
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jakesilove

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Re: 4U Maths Question Thread
« Reply #744 on: January 21, 2017, 09:16:11 am »
+1
Hey, could I please get some help with this question :)

1) P(3secθ,2tanθ) and Q(3secθ,-2tanθ) are two points on the hyperbola x2/9 - y2/4 = 1. The normal at P meets the line OQ at R. Show that the locus of R is a concentric hyperbola

Thank you!!!

Hey! Let's start by finding the equation for the normal. I'm going to use implicit differentiation here.





We want to find the normal, so



At P,



The equation of the normal will be



At the point P,







Okay, now let's find the line OQ.



So the equation is



Using simultaneous equations, we can find the point of intercept.



Jesus, this is getting complicated. I assume I made a mistake; I'll leave this here, and explain how the rest would work. Find the point of intersection (both x and y value), then figure out the locus (by subbing one into the other). I have to head out or I'd redo this; good luck!
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kiwiberry

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Re: 4U Maths Question Thread
« Reply #745 on: January 21, 2017, 02:27:36 pm »
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Hey! Let's start by finding the equation for the normal. I'm going to use implicit differentiation here.





We want to find the normal, so



At P,



The equation of the normal will be



At the point P,







Okay, now let's find the line OQ.



So the equation is



Using simultaneous equations, we can find the point of intercept.



Jesus, this is getting complicated. I assume I made a mistake; I'll leave this here, and explain how the rest would work. Find the point of intersection (both x and y value), then figure out the locus (by subbing one into the other). I have to head out or I'd redo this; good luck!

Thanks Jake I worked it out, you got the gradient for OQ the wrong way round haha :)
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jakesilove

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Re: 4U Maths Question Thread
« Reply #746 on: January 22, 2017, 06:11:18 pm »
+1
Thanks Jake I worked it out, you got the gradient for OQ the wrong way round haha :)

Glad that you got there! This is why you can't do 4U maths in a rush :D
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RuiAce

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Re: 4U Maths Question Thread
« Reply #747 on: January 24, 2017, 12:45:22 pm »
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I know I've been missing but can I just say how happy I am seeing Jake answer all the 4U maths questions for once (without me beating him all the time) :P

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Re: 4U Maths Question Thread
« Reply #748 on: January 26, 2017, 11:04:45 pm »
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Hi Could i please get some help on this question
Much appreciated
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RuiAce

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Re: 4U Maths Question Thread
« Reply #749 on: January 26, 2017, 11:07:19 pm »
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Hi Could i please get some help on this question
Much appreciated
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