Cambridge Prelim MX1 Textbook Marathon/Q&A (2 Viewers)

appleibeats · Apr 27, 2016

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

si2136 said:
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.

Paradoxica · Apr 27, 2016

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

appleibeats said:
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.

Blitz_N7 · May 2, 2016

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

InteGrand · May 2, 2016

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

Blitz_N7 said:
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 c_j 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*}$

Blitz_N7 · May 2, 2016

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

Thank you Integrand!

AfroNerd · May 2, 2016

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

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

Drongoski · May 2, 2016

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)

appleibeats · May 3, 2016

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?

InteGrand · May 3, 2016

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

appleibeats said:
View attachment 33172

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,$

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

appleibeats · May 3, 2016

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 ??

InteGrand · May 3, 2016

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

appleibeats said:
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 $\left(x(T)\right)^2$).$

appleibeats · May 3, 2016

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

InteGrand said:
$\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

Answer says 30 degrees

InteGrand · May 3, 2016

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

appleibeats said:
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

Answer says 30 degrees

$\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).$

appleibeats · May 3, 2016

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

Screen Shot 2016-05-03 at 9.43.39 pm.png

Need help with part b

appleibeats · May 4, 2016

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) ?

davidgoes4wce · May 5, 2016

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

appleibeats said:
View attachment 33174

Need help with part b

$Year 12 Cambridge Ex 3G Q 16 (a) (b)$

$(a) \ For \ Particle \ P_{1}, \ \theta, V$

$\ddot{y}=-g$

$\dot{y}=-gt+V sin \theta$

$y=\frac{-gt^{2}}{2}+Vt \ sin \theta$

$$ At the top of its flight, the vertical component of the ball's velocity is zero \ $ \dot{y}=0$

$0=-gt+V sin \theta$

$gt=V sin \theta$

$t=\frac{V sin \theta}{g} \textcircled{1}$

$$ Substituting $ t=\frac{V sin \theta}{g} $ into y $ =\frac{-gt^{2}}{2}+Vt \ sin \theta$

$y=\frac{-g}{2}(\frac{V \ sin \theta}{g})^{2} +V(\frac{V \ sin \theta}{g}) \ sin \theta$

$y=\frac{-V^{2} sin^{2} \theta}{2g}+ \frac{{V^{2} \ sin^{2} \theta}}{g}=\frac{V^{2} sin^{2} \theta}{2g}$

$(b) (i) $ For Particle 2, $ P_{2}, \frac{\theta}{2}, \frac{3V}{2}$

$\ddot{y}=-g$

$\dot{y}=-gt+\frac{3V}{2} \ sin \frac{\theta}{2}$

$$ At the top of its flight, the vertical component of the ball's velocity is zero \ $ \dot{y}=0$

$t= \frac{3V}{2g} \ sin \frac{\theta}{2} \textcircled{2}$

$$ Equate \textcircled{1} and \textcircled{2}$

$\frac{V sin \theta}{g}=\frac{3V}{2g} \ sin \frac{\theta}{2}$

$$ Simplifying gives $ \ sin \theta= \frac{3}{2} \ sin \frac{\theta}{2} \textcircled{3}$

$$ Knowing the trigonometric identity$ sin 2\theta= 2 \ sin \theta \ cos \theta$

$sin \theta = (2 \ sin(\frac{\theta}{2}) \ cos(\frac{\theta}{2})) \textcircled{4}$

$Replace \ \textcircled{4} \ into \ \textcircled{3}$

$2 \ sin(\frac{\theta}{2}) \ cos(\frac{\theta}{2})= \frac{3}{2} \ sin \frac{\theta}{2}$

$2 \ cos (\frac{\theta}{2})=\frac{3}{2}$

$cos \frac{\theta}{2}=\frac{3}{4} \ \textcircled{5}$

$$ Knowing the trigonometric identity $ cos 2 \ \theta = 2 \ cos^{2} {\theta}-1$

$$ Which can be rewritten as $ cos \theta = 2 \ cos^{2} \frac{\theta}{2}-1 \textcircled{6}$

$$ Substitute \textcircled{5} into \textcircled{6}$

$cos \ \theta = 2 (\frac{3}{4})^{2}-1$

$cos \ \theta=2(\frac{9}{16})-1=\frac{18}{16}-1=\frac{2}{16}=\frac{1}{8}$

$cos \ \theta=\frac{1}{8}$

$$ By taking the inverses of both sides $ , \ \theta=cos^{-1} (\frac{1}{8})$

$\therefore \theta=cos^{-1} (\frac{1}{8})$

(ii)

$$ For Particle 1 $ , P_{1}, $ the time it took to reach the maximum height was $ t=\frac{V \ sin \theta}{g}$

$$ For Particle 2 $ , P_{2}, $ the time it took to reach the maximum height was $ t=\frac{3V \ sin (\frac{\theta}{2})}{2g}$

$$ From the above diagram, one can see that $ sin {\theta} = \frac{\sqrt{63}}{8}=\frac{3\sqrt{7}}{8}$

$$ and sin $ \frac{\theta}{2}=\frac{\sqrt{7}}{4}$

$$ For Particle \ 1 $, t=\frac{V}{g} \ sin \theta = \frac{V}{g} \times \frac{3\sqrt{7}}{8}=\frac{3V\sqrt{7}}{8}$

$$ For Particle \ 2 $, t=\frac{3V}{2g} \ sin \frac{\theta}{2} =\frac{3V}{2g} \times \frac{\sqrt{7}}{4}=\frac{3V\sqrt{7}}{8}$

$\therefore $ The times taken to reach the maximum heights are the same time. $$

appleibeats · May 7, 2016

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

Screen Shot 2016-05-07 at 3.28.39 pm.png

Stuck on part ii)

I found T2 when y = 0

t2 = 2Usin(alpha) / g

Now not sure how to get the RHS eqn.

InteGrand · May 7, 2016

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

appleibeats said:
View attachment 33180

Stuck on part ii)

I found T2 when y = 0

t2 = 2Usin(alpha) / g

Now not sure how to get the RHS eqn.

$\noindent Observe that$

$\begin{align*}\frac{T_1}{1-\cot \alpha} &= \frac{\frac{2U}{g}\left(\sin \alpha - \cos \alpha \right)}{1 - \cot \alpha} \\ &=\frac{2U}{g} \times \frac{\sin \alpha - \cos \alpha}{1-\frac{\cos \alpha}{\sin \alpha}} \\ &=\frac{2U}{g} \times \frac{\sin \alpha \left(\sin \alpha - \cos \alpha\right)}{\sin \alpha - \cos \alpha} \quad (\text{multiplying top and bottom by }\sin \alpha) \\ &= \frac{2U}{g}\sin \alpha \\ &= T_2 \quad (\text{using the result you found}). \end{align*}$

InteGrand · May 7, 2016

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

Incidentally, part (iii) was clear without any calculations at all. We are told that the parabolic trajectory passes through the line y = x. This can clearly only happen if the projection angle was higher than 45 degrees to start with. For if it wasn't, the trajectory would pass below the line y = x to start with and would never reach that line, as it is concave down always. (Note that y = x is a tangent to the flight path when the angle of projection is 45 degrees. If the projection angle is α < 45°, then the tangent to the parabola at the origin is a line y = mx, with m < 1. Since the parabola is concave, its tangent lies above it always, so the line y = mx would lie above the parabola always. So the line y = x would too, as the line y = x is always above the line y = mx for x > 0.)

davidgoes4wce · May 10, 2016

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

appleibeats said:
View attachment 33180

Stuck on part ii)

I found T2 when y = 0

t2 = 2Usin(alpha) / g

Now not sure how to get the RHS eqn.

Anyone done Q 17 b (ii) to that question?Cambridge Ex 3G Q 17 b ii

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