Thread: Cambridge Prelim MX1 Textbook Marathon/Q&A

1. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
If v^2 = 24 - 6x -3x^2, find the acceleration of the particle at the particles greatest displacement from the origin.

Greatest displacement at amplitude where v = 0

When v = 0

( x - 2) (x + 4) = 0

so x = 2 or x = -4

Now acceleration = -3 -3x

when x = -4 ( greatest displacement )

acceleration = 9 m /s^2

But the answers says Max displacement occurs when x = 2 so acceleration is -9m/s^2

Why is it when x = 2 and not x = -4. Isn't the displacement from the origin greatest when x = -4 ??
The distance from the origin is greatest at x = -4. The displacement includes the sign, so is greatest at x = +2.

2. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by Nailgun
what do you mean lel?
When your identity is numerical, i.e there are no variables, you should be aiming to prove it without shifting anything.

When your identity has variables, you can shift things around, maybe prove it using calculus.

Or you could be me and do it by inspection.

3. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by InteGrand
The distance from the origin is greatest at x = -4. The displacement includes the sign, so is greatest at x = +2.
So displacement can be either positive or negative?
Whereas distance is positive or 0 only.

4. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
So displacement can be either positive or negative?
Whereas distance is positive or 0 only.
Distance is a scalar (magnitude only) whereas displacement is a vector (magnitude and direction)

Therefore yes

5. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
If v^2 = 24 - 6x -3x^2, find the acceleration of the particle at the particles greatest displacement from the origin.

Greatest displacement at amplitude where v = 0

When v = 0

( x - 2) (x + 4) = 0

so x = 2 or x = -4

Now acceleration = -3 -3x

when x = -4 ( greatest displacement )

acceleration = 9 m /s^2

But the answers says Max displacement occurs when x = 2 so acceleration is -9m/s^2

Why is it when x = 2 and not x = -4. Isn't the displacement from the origin greatest when x = -4 ??
Do you mind if I ask where you got that question from? Can't seem to find that in the Cambridge text book.

6. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by davidgoes4wce
Do you mind if I ask where you got that question from? Can't seem to find that in the Cambridge text book.
It's from the Grammar Trial 2015

7. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
It's from the Grammar Trial 2015
Uhh

Sorry for nitpicking but can you please leave this thread for the textbook questions

.

10. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Have you learnt Implicit Differentiation?

11. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by si2136
Have you learnt Implicit Differentiation?
Yes, but the questions from the chapter about sums and products. So I think they want us to use that.

12. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Yes, but the questions from the chapter about sums and products. So I think they want us to use that.
equate both sides, you have a quartic equation

α is a tangent, thus, a double root.

The four roots are α,α,β,1 where β is some other root.

Use the product and sum of roots formulae to determine the values of the roots.

13. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Is there a way to do this question without "equating coefficients"?
http://pasteboard.co/C1Wjawh.png

14. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by Blitz_N7
Is there a way to do this question without "equating coefficients"?
http://pasteboard.co/C1Wjawh.png
Yes, there is also a combinatorial proof that can be given. (Note that cj is just n choose j, from the Binomial Theorem.)

$\noindent Imagine a group of n males and n females (total 2n people) from which we wish to select n+2 of them to form a committee. On the one hand, this is done in \binom{2n}{n+2} =\frac{\left(2n\right)!}{\left(n+2\right)!\left(n-2\right)!} ways. On the other hand, we can count this no. of ways by considering how many we choose from the n males, and how many we choose from the n females. The cases are we can pick k males and then \left(n+2-k\right) females, where k goes from 2 up to n (because we must pick at least 2 males to get a committee of n+2).$

$\noindent The no. of ways to pick exactly k males and then n+2-k females is just \binom{n}{k}\binom{n}{n+2-k}. Then summing up from k=2 to n gives us the total no. of ways (because we're adding up all the cases). So equating our two ways of counting the same thing, we get$

\begin{align*}RHS = \frac{\left(2n\right)!}{\left(n+2\right)!\left(n-2\right)!} &= \sum_{k=2}^{n} \binom{n}{k}\binom{n}{n+2-k} \\ &= \sum _{k=2}^{n} c_k c_{n+2-k} \\ &= \sum _{k=2}^{n} c_k c_{k-2} \quad (\text{as }c_j = c_{n-j} \text{ from binomial coefficient properties}) \\ &= c_2 c_0 + c_3 c_1 + c_4 c_2 +\cdots + c_{n}c_{n-2} \\ &= LHS. \end{align*}

15. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Thank you Integrand!

16. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

polynomial.png
induction.png

Hey guys, need help with these questions on polynomials and induction

17. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Say roots are a and a2

.: sum of roots: a + a2 = -m ==> m = -a(a+1)

product of roots: a x a2 = a3 = n

.: m3 = a3(-a3 - 3a2 - 3a -1)

= n(-n +3m -1)

= n(3m - n -1)

18. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Screen Shot 2016-05-03 at 7.58.06 pm.png

Not sure how to answer part b

Do i get the cartisan form?

19. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Screen Shot 2016-05-03 at 7.58.06 pm.png

Not sure how to answer part b

Do i get the cartisan form?
$\noindent At time T, we have y(T) = -40, since the stone is at the lake, which is 40 m below the cliff. So from the equation for y(t), we find -5T^2+20T \sin \theta=-40\Rightarrow 20T\sin \theta = 5T^2 -40. So 400T^2 \sin^2 \theta = \left(5T^2 -40\right)^2 (squaring both sides of the previous equation). Therefore,$

\begin{align*}\left(x(T)\right)^2 &= \left(20T\cos \theta\right)^2 \\ &= 400T^2 \cos^2 \theta \\ &= 400T^2 \left(1-\sin^2 \theta \right) \\ &= 400T^2 - 400T^2 \sin^2 \theta \\ &= 400T^2 -\left(5T^2 -40\right)^2. \end{align*}

20. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

For part c) Do i differentiate this equation in part b) in respects to T ? If so how do you find the derivate of x(T) squared ??

21. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
how do you find the derivate of x(T) squared ??
$\noindent Differentiate this with respect to T: 400T^2 -\left(5T^2 -40\right)^2 (since that's \left(x(T)\right)^2).$

22. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by InteGrand
$\noindent Differentiate this with respect to T: 400T^2 -\left(5T^2 -40\right)^2 (since that's \left(x(T)\right)^2).$

Found T to be 4, now how do I find max theta. Do I sub t = 4 into displacements of x and y and differentiate parametrically. I tried and got -cos(x)/sin(x) x for theta

23. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Originally Posted by appleibeats
Found T to be 4, now how do I find max theta. Do I sub t = 4 into displacements of x and y and differentiate parametrically. I tried and got -cos(x)/sin(x) x for theta

$\noindent Once you've found the T (it's 4 you say), we just need to find the corresponding \theta (recall that T and \theta are \emph{not} independent; the time to hit the bottom clearly depends on \theta, or if we choose the time to hit the bottom, then \theta depends on this T). We have y(4) = -40\Longleftrightarrow -5 \times 16 + 20 \times 4 \sin \theta = -40. Solve this for \theta and you'll find that \theta = 30^{\circ} (since \theta is acute).$

25. Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Is it right for part b) to plug The new velocity and angle into part a max height and then equate it with part a) ?

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