IB Maths Marathon (1 Viewer)

davidgoes4wce · Feb 15, 2017

Re: International Baccalaureate Marathon

Was going through worked example from Cambridge tonight and there was a mistake in the working out for part (c), vector BC should be -2-k. Small mistake but its important that I point it out to all IBers in 2017:

boredofstudiesuser1 · Feb 15, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
Was going through worked example from Cambridge tonight and there was a mistake in the working out for part (c), vector BC should be -2-k. Small mistake but its important that I point it out to all IBers in 2017:

Wow, how much maths do you do a day?

davidgoes4wce · Feb 15, 2017

Re: International Baccalaureate Marathon

boredofstudiesuser1 said:
Wow, how much maths do you do a day?

Not that much. I just went to the gym for about an 1.5 hour, also did a bit of driving and I went out to the library for a bit of fresh air.

InteGrand · Feb 15, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
Was going through worked example from Cambridge tonight and there was a mistake in the working out for part (c), vector BC should be -2-k. Small mistake but its important that I point it out to all IBers in 2017:

What do you mean? The vectors b and c have no k, so vector BC would have no k.

davidgoes4wce · Feb 15, 2017

Re: International Baccalaureate Marathon

InteGrand said:
What do you mean? The vectors b and c have no k, so vector BC would have no k.

I meant vector AB, should have been -2-k.

InteGrand · Feb 15, 2017

Re: International Baccalaureate Marathon

$\noindent Oh yeah, they accidentally did $-2-2k$ instead.$

davidgoes4wce · Feb 15, 2017

Re: International Baccalaureate Marathon

$$ Points A,B and C have position vectors A $ \begin{pmatrix} 1\\-19\\5 \end{pmatrix}, B \ \begin{pmatrix} 3.2\\3.6\\2 \end{pmatrix} , C \ \begin{pmatrix} -6\\-15\\7 \end{pmatrix}$

$$ The respective magnitudes: $ |\overrightarrow{AB}|=\sqrt{524.6} , |\overrightarrow{AC}|=\sqrt{69} , |\overrightarrow{BC}|=\sqrt{455.6} , |\overrightarrow{CB}|=\sqrt{455.6}$

$Find the Area of the triangle ABC$

$I preceded to find vectors \overrightarrow{AB}, \overrightarrow{AC} $ and $ \overrightarrow{BC}$

$This is a diagram I drew and then I preceded to work out the individual angles of each of the vertices.$

$\overrightarrow{AB}=\begin{pmatrix} 2.2\\22.6\\-3 \end{pmatrix} \ , \ \overrightarrow{AC}= \begin{pmatrix} -7\\4\\2 \end{pmatrix} \ , \overrightarrow{BC}= \begin{pmatrix} -9.2\\-18.6\\5 \end{pmatrix} \ , \overrightarrow{CB}= \begin{pmatrix} 9.2\\18.6\\-5 \end{pmatrix}$

$cos \ \theta=\frac{\overrightarrow{AB} \ . \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|}=\frac{69}{\sqrt{524.6} \times \sqrt{69}} \implies \theta=68.7^\circ$

$cos \ \beta=\frac{\overrightarrow{BC} \ . \overrightarrow{AC}}{|\overrightarrow{BC}| |\overrightarrow{AC}|}=0 \implies \beta=90^\circ$

$cos \ \alpha=\frac{\overrightarrow{AB} \ . \overrightarrow{CB}}{|\overrightarrow{AB}| |\overrightarrow{CB}|} \implies cos \ \alpha=\frac{455.6}{\sqrt{524.6} \times {\sqrt{455.6}}}=21.3^\circ$

$As a test I then worked out the individual areas of the triangles, use either of the 3 methods and I didn't get the same values. Well one of them was close to the actual answer of 88.7$

$A=\frac{1}{2} |\overrightarrow{BC}| |\overrightarrow{AC}| \ sin \beta=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{69} \ sin \ 90^\circ=88.7 \ units^{2}$

$A=\frac{1}{2} |\overrightarrow{AC}| |\overrightarrow{AB}| \sin \theta=\frac{1}{2} \times \ 13 \ \sqrt{524.6} \ sin \ 68.7^\circ=88.6 \ units ^{2}$

$A=\frac{1}{2} |\overrightarrow{CB}| |\overrightarrow{AB}| \ sin \alpha=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{524.6} \times \ sin \ 21.3^\circ =88.8 units^{2}$

$Now I spent roughly a couple of hours doing this question tonight, to calculate the area of a triangle in vector form (I'm actually not sure how to word it properly), but what I take from doing a couple of questions today , is you either have to multiply 2 vectors coming in or 2 vectors coming out together to get the required area. So what I mean by this is, at point A in my diagram, both of the vectors are going out. At point C, both of the vectors are coming in. At point B, I had to reverse one of the vectors so that it showed them both coming in.$

InteGrand · Feb 16, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
$$ Points A,B and C have position vectors A $ \begin{pmatrix} 1\\-19\\5 \end{pmatrix}, B \ \begin{pmatrix} 3.2\\3.6\\2 \end{pmatrix} , C \ \begin{pmatrix} -6\\-15\\7 \end{pmatrix}$

$$ The respective magnitudes: $ |\overrightarrow{AB}|=\sqrt{524.6} , |\overrightarrow{AC}|=\sqrt{69} , |\overrightarrow{BC}|=\sqrt{455.6} , |\overrightarrow{CB}|=\sqrt{455.6}$

$Find the Area of the triangle ABC$

$I preceded to find vectors \overrightarrow{AB}, \overrightarrow{AC} $ and $ \overrightarrow{BC}$

$This is a diagram I drew and then I preceded to work out the individual angles of each of the vertices.$

$\overrightarrow{AB}=\begin{pmatrix} 2.2\\22.6\\-3 \end{pmatrix} \ , \ \overrightarrow{AC}= \begin{pmatrix} -7\\4\\2 \end{pmatrix} \ , \overrightarrow{BC}= \begin{pmatrix} -9.2\\-18.6\\5 \end{pmatrix} \ , \overrightarrow{CB}= \begin{pmatrix} 9.2\\18.6\\-5 \end{pmatrix}$

$cos \ \theta=\frac{\overrightarrow{AB} \ . \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|}=\frac{69}{\sqrt{524.6} \times \sqrt{69}} \implies \theta=68.7^\circ$

$cos \ \beta=\frac{\overrightarrow{BC} \ . \overrightarrow{AC}}{|\overrightarrow{BC}| |\overrightarrow{AC}|}=0 \implies \beta=90^\circ$

$cos \ \alpha=\frac{\overrightarrow{AB} \ . \overrightarrow{CB}}{|\overrightarrow{AB}| |\overrightarrow{CB}|} \implies cos \ \alpha=\frac{455.6}{\sqrt{524.6} \times {\sqrt{455.6}}}=21.3^\circ$

$As a test I then worked out the individual areas of the triangles, use either of the 3 methods and I didn't get the same values. Well one of them was close to the actual answer of 88.7$

$A=\frac{1}{2} |\overrightarrow{BC}| |\overrightarrow{AC}| \ sin \beta=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{69} \ sin \ 90^\circ=88.7 \ units^{2}$

$A=\frac{1}{2} |\overrightarrow{AC}| |\overrightarrow{AB}| \sin \theta=\frac{1}{2} \times \ 13 \ \sqrt{524.6} \ sin \ 68.7^\circ=88.6 \ units ^{2}$

$A=\frac{1}{2} |\overrightarrow{CB}| |\overrightarrow{AB}| \ sin \alpha=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{524.6} \times \ sin \ 21.3^\circ =88.8 units^{2}$

$Now I spent roughly a couple of hours doing this question tonight, to calculate the area of a triangle in vector form (I'm actually not sure how to word it properly), but what I take from doing a couple of questions today , is you either have to multiply 2 vectors coming in or 2 vectors coming out together to get the required area. So what I mean by this is, at point A in my diagram, both of the vectors are going out. At point C, both of the vectors are coming in. At point B, I had to reverse one of the vectors so that it showed them both coming in.$

The area values you got are all very close. Haven't checked your calculations but if they're all right, the small discrepancies would just be due to the fact that you used rounded values for the angle.

$\noindent To calculate $\sin \theta$ exactly as a surd here, you can use $\sin \theta =\sqrt{1 -\cos^{2}\theta}$ and use the value you found for $\cos \theta$.$

davidgoes4wce · Feb 18, 2017

Re: International Baccalaureate Marathon

Not sure what the best way to study for sequences and series, whether it be HSC or IB but should we tell the students to memorise the formulae (even though the formuale sheet is provided)?

davidgoes4wce · Feb 18, 2017

Re: International Baccalaureate Marathon

$Anyway I came across this question tonight and it has me a bit stumped. The first term is 1, sum of the first 6 terms is 1.24992. Find the value of the common ratio if:$

$S_n=\frac{u_1(r^n-1)}{r-1}$

$1.24992=\frac{1(r^6-1)}{r-1}$

$I tried using my graphics calculator and it gave me an r value=1$

davidgoes4wce · Feb 18, 2017

Re: International Baccalaureate Marathon

OK problem solved if you guys are familiar with the Casio fx-9860G AU, you should leave the fraction to the right and not cross multiply in Solver.

$Don't enter it in the calculator as$

$1.24992(r-1)=1(r^6-1) $ it won't solve it for an unknown reason. $$

davidgoes4wce · Feb 21, 2017

Re: International Baccalaureate Marathon

This is a Band 7 IB question (equivalent to a HSC Band 6 aiming for 95%+ on paper)

With regards to (b) , would it be OK to not write the answer involving sequences? i.e doing it manually in order to get the marks

$The following was the solution:$

$t_n=2+2 \sum_{k=1}^n h_k$

$t_8=2+16(1-0.8^8)=15.3 \ metres$

My way I did it was just using the 'common sense method':

$$ Total distance ball has traveled before 9th bounce : $=2+2(1.6)+2(1.28)+2(1.024)+2(0.892)+2(0.655536)+2(0.524288)+2(0.4194304)+2(0.33554432)=15.46 \ metres$

Notice both answers do give slightly different decimal values .

seanieg89 · Feb 21, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
This is a Band 7 IB question (equivalent to a HSC Band 6 aiming for 95%+ on paper)

With regards to (b) , would it be OK to not write the answer involving sequences? i.e doing it manually in order to get the marks

$The following was the solution:$

$t_n=2+2 \sum_{k=1}^n h_k$

$t_8=-2+16(1-0.8^8)=15.3 \ metres$

My way I did it was just using the 'common sense method':

$$ Total distance ball has traveled before 9th bounce : $=2+2(1.6)+2(1.28)+2(1.024)+2(0.892)+2(0.655536)+2(0.524288)+2(0.4194304)+2(0.33554432)=15.46 \ metres$

Notice both answers do give slightly different decimal values .

0.892 and 0.655536 are wrong. By doing it manually like this you are greatly increasing the number of points at which you can mess up the calculation. And of course this won't be feasible if 9 is replaced by 1000.

The correct answer is the expression t8 (although the 2 is +2, not -2), which in decimal form is exactly 15.31564544.

Ps. alarm bells should be ringing in your head at "slightly different decimal values". The only step in which any rounding is involved is at the very end, so these answers should coincide exactly prior to rounding, which they clearly don't.

davidgoes4wce · Feb 21, 2017

Re: International Baccalaureate Marathon

OK yep I see it

should be 0.8192 metres and 0.65536 metres

davidgoes4wce · Feb 21, 2017

Re: International Baccalaureate Marathon

seanieg89 said:
0.892 and 0.655536 are wrong. By doing it manually like this you are greatly increasing the number of points at which you can mess up the calculation. And of course this won't be feasible if 9 is replaced by 1000.

The correct answer is the expression t8 (although the 2 is +2, not -2), which in decimal form is exactly 15.31564544.

Ps. alarm bells should be ringing in your head at "slightly different decimal values". The only step in which any rounding is involved is at the very end, so these answers should coincide exactly prior to rounding, which they clearly don't.

I'm aware that there was a slight error with my working out but in terms of 'method' of getting to the answer, there are 2 methods in order to get to the right answer.

Would my method be suitable in order to answer a question like this, whether it be HSC and/or IB?

seanieg89 · Feb 22, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
I'm aware that there was a slight error with my working out but in terms of 'method' of getting to the answer, there are 2 methods in order to get to the right answer.

Would my method be suitable in order to answer a question like this, whether it be HSC and/or IB?

Depends what you mean by suitable.

a) Will the answer be the same?
Yes, obviously. Summing a geometric series manually will give you the same result as applying the formula for the sum of a geometric series.

b) Will markers treat the solution as being equally valid?
I don't know the marking criteria, but I can't imagine any reason why not.

c) Do I recommend manually summing geometric/arithmetic series?
Definitely not, unless the series is only a couple of terms long. As I said in my previous post, you are greatly increasing the number of places at which you can make mistakes. The method is also slower (assuming proficiency with both), and not feasible for series with many terms. You need to know how to sum general arithmetic and geometric series for the course anyway, so why not use this knowledge?

omegadot · Feb 22, 2017

Re: International Baccalaureate Marathon

davidgoes4wce said:
$$ Points A,B and C have position vectors A $ \begin{pmatrix} 1\\-19\\5 \end{pmatrix}, B \ \begin{pmatrix} 3.2\\3.6\\2 \end{pmatrix} , C \ \begin{pmatrix} -6\\-15\\7 \end{pmatrix}$

$$ The respective magnitudes: $ |\overrightarrow{AB}|=\sqrt{524.6} , |\overrightarrow{AC}|=\sqrt{69} , |\overrightarrow{BC}|=\sqrt{455.6} , |\overrightarrow{CB}|=\sqrt{455.6}$

$Find the Area of the triangle ABC$

$I preceded to find vectors \overrightarrow{AB}, \overrightarrow{AC} $ and $ \overrightarrow{BC}$

$This is a diagram I drew and then I preceded to work out the individual angles of each of the vertices.$

$\overrightarrow{AB}=\begin{pmatrix} 2.2\\22.6\\-3 \end{pmatrix} \ , \ \overrightarrow{AC}= \begin{pmatrix} -7\\4\\2 \end{pmatrix} \ , \overrightarrow{BC}= \begin{pmatrix} -9.2\\-18.6\\5 \end{pmatrix} \ , \overrightarrow{CB}= \begin{pmatrix} 9.2\\18.6\\-5 \end{pmatrix}$

$cos \ \theta=\frac{\overrightarrow{AB} \ . \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|}=\frac{69}{\sqrt{524.6} \times \sqrt{69}} \implies \theta=68.7^\circ$

$cos \ \beta=\frac{\overrightarrow{BC} \ . \overrightarrow{AC}}{|\overrightarrow{BC}| |\overrightarrow{AC}|}=0 \implies \beta=90^\circ$

$cos \ \alpha=\frac{\overrightarrow{AB} \ . \overrightarrow{CB}}{|\overrightarrow{AB}| |\overrightarrow{CB}|} \implies cos \ \alpha=\frac{455.6}{\sqrt{524.6} \times {\sqrt{455.6}}}=21.3^\circ$

$As a test I then worked out the individual areas of the triangles, use either of the 3 methods and I didn't get the same values. Well one of them was close to the actual answer of 88.7$

$A=\frac{1}{2} |\overrightarrow{BC}| |\overrightarrow{AC}| \ sin \beta=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{69} \ sin \ 90^\circ=88.7 \ units^{2}$

$A=\frac{1}{2} |\overrightarrow{AC}| |\overrightarrow{AB}| \sin \theta=\frac{1}{2} \times \ 13 \ \sqrt{524.6} \ sin \ 68.7^\circ=88.6 \ units ^{2}$

$A=\frac{1}{2} |\overrightarrow{CB}| |\overrightarrow{AB}| \ sin \alpha=\frac{1}{2} \times \sqrt{455.6} \times \sqrt{524.6} \times \ sin \ 21.3^\circ =88.8 units^{2}$

$Now I spent roughly a couple of hours doing this question tonight, to calculate the area of a triangle in vector form (I'm actually not sure how to word it properly), but what I take from doing a couple of questions today , is you either have to multiply 2 vectors coming in or 2 vectors coming out together to get the required area. So what I mean by this is, at point A in my diagram, both of the vectors are going out. At point C, both of the vectors are coming in. At point B, I had to reverse one of the vectors so that it showed them both coming in.$

$Not sure if you know about the \textit{vector cross product}, but if you do one of the geometric interpretations for the magnitude of the cross product between two vectors corresponds to the area of a triangle.$

$If two sides of a triangle are formed by the vectors$ \overrightarrow{u} $ and $\overrightarrow{v}$ the area $\mathcal{A}$ of the triangle is given by$

$\mathcal{A} = \frac{1}{2} \|\overrightarrow{u} \times \overrightarrow{v}\|$

$$So if the two sides of a triangle are formed by the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ the area $\mathcal{A}$ of the triangle with vertices $A$, $B$, and $C$ will be given by$

$\mathcal{A} = \frac{1}{2} \|\overrightarrow{AB} \times \overrightarrow{AC}\|.$

$Now, for your example, we have $ \overrightarrow{AB} = \langle 3.2 - 1,3.6 + 19, 2-5 \rangle = \langle 2.2,22.6,-37 \rangle$ and $\overrightarrow{AC} = \langle -6 - 1, -15 + 19, 7 - 5 \rangle = \langle -7, 4, 2 \rangle.$ Note here for brevity I am using triangluar brackets to denote a vector.$

$Finding the cross product between these two vectors.$

$\begin{align*}\overrightarrow{AB} \times \overrightarrow{AC} &=\begin{array}{|ccc|}\hat{e}_x & \hat{e}_y & \hat{e}_z\\2.2 & 22.6 & -3\\-7 & 4 & 2\end{array}\\&= \hat{e}_x (22.6 \cdot 2 + 3 \cdot 4) - \hat{e}_y (2.2 \cdot 2 - (-3) \cdot (-7)) + \hat{e}_z (2.2 \cdot 4 + 22.6 \cdot 7)\\&= \langle 57.2, -16.6,167 \rangle \end{align*}$

$Now finding the magnitude of this vector we have$

$\|\overrightarrow{AB} \times \overrightarrow{AC}\| = \sqrt{57.2^2 + (-16.6)^2 + 167^2} = \sqrt{31436.4}.$

$So for the area we have$

$\mathcal{A} = \frac{1}{2} \sqrt{31436.4} \doteq 88.65$ units$^2$

davidgoes4wce · Feb 23, 2017

Re: International Baccalaureate Marathon

omegadot said:
$Not sure if you know about the \textit{vector cross product}, but if you do one of the geometric interpretations for the magnitude of the cross product between two vectors corresponds to the area of a triangle.$

$If two sides of a triangle are formed by the vectors$ \overrightarrow{u} $ and $\overrightarrow{v}$ the area $\mathcal{A}$ of the triangle is given by$

$\mathcal{A} = \frac{1}{2} \|\overrightarrow{u} \times \overrightarrow{v}\|$

$$So if the two sides of a triangle are formed by the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ the area $\mathcal{A}$ of the triangle with vertices $A$, $B$, and $C$ will be given by$

$\mathcal{A} = \frac{1}{2} \|\overrightarrow{AB} \times \overrightarrow{AC}\|.$

$Now, for your example, we have $ \overrightarrow{AB} = \langle 3.2 - 1,3.6 + 19, 2-5 \rangle = \langle 2.2,22.6,-37 \rangle$ and $\overrightarrow{AC} = \langle -6 - 1, -15 + 19, 7 - 5 \rangle = \langle -7, 4, 2 \rangle.$ Note here for brevity I am using triangluar brackets to denote a vector.$

$Finding the cross product between these two vectors.$

$\begin{align*}\overrightarrow{AB} \times \overrightarrow{AC} &=\begin{array}{|ccc|}\hat{e}_x & \hat{e}_y & \hat{e}_z\\2.2 & 22.6 & -3\\-7 & 4 & 2\end{array}\\&= \hat{e}_x (22.6 \cdot 2 + 3 \cdot 4) - \hat{e}_y (2.2 \cdot 2 - (-3) \cdot (-7)) + \hat{e}_z (2.2 \cdot 4 + 22.6 \cdot 7)\\&= \langle 57.2, -16.6,167 \rangle \end{align*}$

$Now finding the magnitude of this vector we have$

$\|\overrightarrow{AB} \times \overrightarrow{AC}\| = \sqrt{57.2^2 + (-16.6)^2 + 167^2} = \sqrt{31436.4}.$

$So for the area we have$

$\mathcal{A} = \frac{1}{2} \sqrt{31436.4} \doteq 88.65$ units$^2$

Yep thanks for your response.

I did some vector cross product before , a large chapter was devoted to that in Calculus Stewart book. I remember when I did 2nd year Engineering, I spent a lot of time doing those questions. It's a very efficient way of calculating an area.

davidgoes4wce · Mar 5, 2017

Re: International Baccalaureate Marathon

$Cambridge question$

$Q1. Kenny is offered a choice of two investment plans, each requiring an initial investment of \$10,000.$

$Plan A offers a fixed return of \$800 per year.$

$Plan B offers a return of 5 \% each year, reinvested in the plan$

$Over what period of time is plan A better than plan B ?$

$$ I graphed the two functions respectively , $ A_n=10000+800n , B_n=10000(1.05)^n$

The answer in the book states that ' ..for the first 19 years $A_n > B_n$

Want to check if I could say something along the lines of '18 years and 11 months $A_n > B_n$ or if we could leave the answer as a decimal (with 1 d.p), 18.4 years?

$P.S I do know that 0.838 of a year is roughly 10.056 months$

He-Mann · Mar 5, 2017

Re: International Baccalaureate Maths Marathon

If you were pedantic, then expressing your answer in years only is more appropriate since our question was with respect to years. i.e. maintain consistency of units

IB Maths Marathon (1 Viewer)

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

Vexed?

Users Who Are Viewing This Thread (Users: 0, Guests: 1)