1. ## 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:

2. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
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?

3. ## Re: International Baccalaureate Marathon

Originally Posted by boredofstudiesuser1
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.

4. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
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.

5. ## Re: International Baccalaureate Marathon

Originally Posted by InteGrand
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.

6. ## Re: International Baccalaureate Marathon

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

7. ## Re: International Baccalaureate Marathon

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

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

$Find the Area of the triangle ABC$

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

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

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

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

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

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

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

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

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

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

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

8. ## Re: International Baccalaureate Marathon

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

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

$Find the Area of the triangle ABC$

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

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

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

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

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

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

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

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

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

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

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

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

9. ## 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)?

10. ## Re: International Baccalaureate Marathon

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

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

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

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

11. ## 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.

$Don't enter it in the calculator as$

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

12. ## 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 .

13. ## Re: International Baccalaureate Marathon

Originally Posted by davidgoes4wce
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.

14. ## Re: International Baccalaureate Marathon

OK yep I see it

should be 0.8192 metres and 0.65536 metres

15. ## Re: International Baccalaureate Marathon

Originally Posted by seanieg89
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?

Page 5 of 5 First ... 345

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•