Cambridge Prelim MX1 Textbook Marathon/Q&A (8 Viewers)

InteGrand · Oct 13, 2015

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

appleibeats said:
So we know x degree is xpi/180 rad

Hence find:

Derivative of (tan (x degrees + 45 degrees))

Convert everything to radians and then just differentiate with the chain rule.

appleibeats · Oct 13, 2015

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

I have converted everything to radians. Do I then add the fractions??

appleibeats · Oct 13, 2015

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

I got it.

InteGrand · Oct 13, 2015

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

appleibeats said:
I have converted everything to radians. Do I then add the fractions??

No, the 45 degrees is just a constant, don't bother adding it.

Once it's all converted to radians, the argument of the tan function will be a linear function of x. So to differentiate tan of a linear function of x, the answer is sec² of that linear function of x, times the derivative of that linear function of x (which is just the coefficient of the x, i.e. π/180).

appleibeats · Oct 14, 2015

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

Carrotsticks said:
$\\ \frac{1}{2} \left ( \sin(m+n)x + \sin(m-n)x \right )= \sin(mx) \cos(nx) \\\\ \begin{align*} \frac{d}{dx} \sin(mx) \cos(nx) &= \frac{d}{dx} \left ( \frac{1}{2} \left (\sin(m+n)x + \sin(m-n)x \right ) \right ) \\\\ &= \frac{1}{2} \times \frac{d}{dx} \left (\sin(m+n)x + \sin(m-n)x \right ) \\\\ &= \frac{1}{2} \left [(m+n) \cos (m+n)x + (m-n) \cos (m-n)x \right ] \end{align*}$

How do you get from the 2nd last line to the last line??

InteGrand · Oct 14, 2015

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

appleibeats said:
How do you get from the 2nd last line to the last line??

Chain rule.

appleibeats · Oct 14, 2015

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

Do I need to use product rule?? I tried using the chain rule and it is a big mess and getting me nowhere.

InteGrand · Oct 14, 2015

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

appleibeats said:
Do I need to use product rule?? I tried using the chain rule and it is a big mess and getting me nowhere.

$No product rule. Maybe the notation confused you a bit; the functions being differentiated are like: $\sin \left( (m+n)x\right)$, so sine of a constant times $x$. So by the chain rule, this one would have derivative: cosine of the constant times $x$, multiplied by the derivative of that constant times $x$, which is just the constant $m+n$. In other words, $\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin \left( (m+n)x\right)\right) = \cos \left( (m+n)x\right)\times \frac{\mathrm{d}}{\mathrm{d}x} \left( (m+n)x\right)= \cos \left( (m+n)x\right)\times (m+n) = (m+n) \cos \left( (m+n)x\right)$.$

appleibeats · Oct 14, 2015

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

If y = sin x

prove

i) y' = sin(pi/2 + x)

ii) y'' = sin ( pi +x )

iii) y''' = sin (3pi/2 + x )

Hence deduce an expression for y^n

InteGrand · Oct 14, 2015

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

appleibeats said:
If y = sin x

prove

i) y' = sin(pi/2 + x)

ii) y'' = sin ( pi +x )

iii) y''' = sin (3pi/2 + x )

Hence deduce an expression for y^n

$The first three derivatives are, respectively: $\cos x,-\sin x, -\cos x$. Use trig. identities to show that these are equal to the given expressions. The expression for $y^{(n)}$ is $\sin \left(\frac{n\pi}{2}+x \right)$.$

appleibeats · Oct 15, 2015

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

I was able to do the first and second derivative but am struggling with the 3rd.

InteGrand · Oct 15, 2015

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

appleibeats said:
I was able to do the first and second derivative but am struggling with the 3rd.

$We just need to show that $\sin \left(\frac{3\pi}{2}+x \right)=-\cos x$. Proof:$

$$LHS = \sin \left(\frac{3\pi}{2}+x -2\pi \right)$ (periodicity of sine)$

$$= \sin \left(x -\frac{\pi}{2} \right)$ (simplifying)$

$$= -\cos x$ (either notice this by shifting the graph of $y=\sin x$ right by $\frac{\pi}{2}$ to get the graph of $y= \sin \left(x -\frac{\pi}{2} \right)$ and noticing it coincides with the graph of $y=-\cos x$, or note that $\sin \left(x -\frac{\pi}{2} \right) = - \sin \left(\frac{\pi}{2}-x \right)=-\cos x$, by the odd property of the sine function and the complementary angles trig. identity ($\sin \left(\frac{\pi}{2}-\alpha \right)=\cos \alpha$))$

$=RHS$. $\blacksquare$

appleibeats · Oct 18, 2015

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

PQ is a diameter on the given circle and S is a point on the circumference. T is the point on PQ such that PS = PT . Let Angle SPT = alpha

a) Show that the are A of triangle SPT is A = 1/2 d^2 cos^2a sina, where d is the diameter of the circle.

b) Hence show that the maximum are of triangle SPT as S varies on the circle is 1/9 d^2 root3 units^2

I can do part a), unsure how to approach b) ?

kawaiipotato · Oct 18, 2015

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

appleibeats said:
PQ is a diameter on the given circle and S is a point on the circumference. T is the point on PQ such that PS = PT . Let Angle SPT = alpha

a) Show that the are A of triangle SPT is A = 1/2 d^2 cos^2a sina, where d is the diameter of the circle.

b) Hence show that the maximum are of triangle SPT as S varies on the circle is 1/9 d^2 root3 units^2

I can do part a), unsure how to approach b) ?

$$ A $ = \frac{1}{2}d^{2}\cos ^{2}\alpha \sin \alpha$

$To find maximum or minimum, differentiate and let that equal zero$

$$ But first, use the fact that $ \cos ^{2}\alpha = 1 - \sin ^{2} \alpha ($Didn't want to use product rule$)$

$$A$ = \frac{1}{2}d^{2}(\sin \alpha - \sin ^{3}\alpha)$

$Note that d is a constant, and as S varies, $ \alpha $ varies, which causes A to vary$

$$ Differentiating A with respect to $ \alpha ,$

$\frac{dA}{d \alpha} = \frac{1}{2}d^{2}(\cos \alpha - 3 \cos \alpha \sin^{2} \alpha )$

$$ let $ \frac{dA}{d \alpha} = 0$

$\cos \alpha = 0 ($reject, since $ \alpha \neq \frac{\pi}{2}), \sin ^{2}\alpha = \frac{1}{3}$

$Sub. $ \sin ^{2}\alpha = \frac{1}{3} $into A to find maximum area$

$\therefore A_{max} = \frac{1}{2}d^{2}(1 - \frac{1}{3})(\frac{1}{\sqrt3})$

$note: You have to prove that it's a maximum with the second derivative$

appleibeats · Oct 18, 2015

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

A straight line passes through the point 2,1 and has positive x and y int at P and Q respectivley. Let Angle ORQ = a where O is the origin.

So the A of the Triange = (2tan a + 1 )^2 / 2tan a

Now how do I show that this area is maximised when tan a = 1/2

kawaiipotato · Oct 18, 2015

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

appleibeats said:
A straight line passes through the point 2,1 and has positive x and y int at P and Q respectivley. Let Angle ORQ = a where O is the origin.

So the A of the Triange = (2tan a + 1 )^2 / 2tan a

Now how do I show that this area is maximised when tan a = 1/2

$An obvious way would be to let $ \frac{dA}{d \alpha} = 0 $ and then test with the second derivative.$

$But you can also do it via inequalities$

$$ A $ = \frac{4 \tan ^{2} \alpha + 4 \tan \alpha + 1}{2 \tan \alpha}$

$= 2 \tan \alpha + 2 + \frac{1}{2 \tan \alpha} = 2 \tan \alpha + \frac{1}{2 \tan \alpha} + 2$

$$ Consider $ (a-b)^{2} \geq 0$

$a^2 + b^{2} \geq 2ab$

$\frac{a}{b} + \frac{b}{a} \geq 2$

$a \longrightarrow 2 \tan \alpha , b \longrightarrow 1$

$$ So $ 2 \tan \alpha + \frac{1}{2 \tan \alpha} \geq 2$

$2 \tan \alpha + \frac{1}{2 \tan \alpha} + 2 \geq 4$

$\therefore $A$ \geq 4$

$$ A $ = \frac{4 \tan ^{2} \alpha + 4 \tan \alpha + 1}{2 \tan \alpha} \geq 4$

$$ So minimum A at A = 4$

$4 \tan ^{2} \alpha + 4 \tan \alpha + 1 = 8 \tan \alpha ($since$ \tan \alpha \geq 0)$

$4 \tan ^{2} \alpha - 4 \tan \alpha + 1 = 0$

$\tan \alpha = \frac{1}{2} $ for minimum area $$

Paradoxica · Oct 18, 2015

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

appleibeats said:
A straight line passes through the point 2,1 and has positive x and y int at P and Q respectivley. Let Angle ORQ = a where O is the origin.

So the A of the Triange = (2tan a + 1 )^2 / 2tan a

Now how do I show that this area is maximised when tan a = 1/2

make the substitution x = tan a
all relevant quantities are strictly non-negative
2x + 1 >= 2 sqrt(2x) (easy to prove am-gm for n=2)
declare equality and hence solve for x.

beaten by kawaiipotato

appleibeats · Oct 20, 2015

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

Use mathematical induction to prove that for all positive integers n:

a + (a+ d) + (a + 2d) + ... + ( a + ( n - 1 )d) = 1/2 . n (2a + (n - 1)d)

appleibeats · Oct 27, 2015

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

SOlve integral of (16 - x^2)^1/2 dx using the substitution x = 4sinE

InteGrand · Oct 27, 2015

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

appleibeats said:
SOlve integral of (16 - x^2)^1/2 dx using the substitution x = 4sinE

$Let $\mathcal{I} = \int \sqrt{16 - x^2} \text{ d}x$.$

$Use the substitution $x=4\sin \theta$, $\theta \in \left[-\frac{\pi}{2},\frac{\pi}{2} \right]$ \Big{(}choosing this domain so that our substitution is one-to-one with $\theta = \sin^{-1} \left( \frac{x}{4}\right)$ and so $\cos \theta \geq 0$, so that $\cos \theta = +\sqrt{1-\sin^2 \theta}$\Big{)}. Now, $\mathrm{d}x = 4\cos \theta \text{ d}\theta$ and $16 - x^2 = 16 - 16\sin^2 \theta = 16 \left(1-\sin^2 \theta \right)\Rightarrow \sqrt{16-x^2} = 4 \cos \theta$. Therefore, we have$

$\begin{align*}\mathcal{I} &= \int 4 \cos \theta \cdot 4 \cos \theta \text{ d}\theta \\ &= \int 16 \cos^2 \theta \text{ d}\theta \\ &= 16 \int \frac{1}{2} \left(1-\cos 2\theta \right) \text{ d}\theta \\ &= 8 \left(\theta - \frac{\sin 2 \theta}{2} \right)+C \\ &= 8 \left[\sin^{-1} \left( \frac{x}{4}\right) - \frac{2\sin \theta \cos \theta}{2} \right]+C \\ &=8 \left[\sin^{-1} \left( \frac{x}{4}\right) - \frac{x}{4}\cdot \sqrt{1-\sin^2 \theta} \right]+C \\ &= 8\sin^{-1} \left( \frac{x}{4}\right) - 2x \sqrt{1 - \left(\frac{x}{4} \right)^2} +C \\ &= 8\sin^{-1} \left( \frac{x}{4}\right) - \frac{1}{2}x \sqrt{16-x^2} +C . \end{align*}$

Cambridge Prelim MX1 Textbook Marathon/Q&A (8 Viewers)

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