Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

DatAtarLyfe · Jul 17, 2015

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

One day i'll learn Latex....

calebwhitaker199 · Jul 18, 2015

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

rand_althor said:
22)

$\begin{align*}&\textup{Using the equation of a parabola with vertex }(h,k): \\(x-h)^2&=4a(y-k) \\x^2&=4a(y-8) \\&\textup{When }x=3,\, y=0: \\3^2&=4a(0-8)\\4a&=\frac{9}{-8} \\a&=\frac{9}{-32} \\\therefore x^2&=4(\frac{-9}{32})(y-8) \\8x^2&=-9(y-8) \\ 8x^2&+9y-72=0 \end{align*}$

$$Focus has coordinates $(h,k+a)\to(0,8+(-\frac{9}{32}))\to(0,\frac{247}{32}) \\$Directrix is the equation $y=k-a\to y=8-(-\frac{9}{32})\to y=\frac{265}{32} \\$

24)

$\begin{align*}&\textup{24. Using the vertex form of a parbola with vertex }(h,k): \\y&=a(x-h)^2+k \\y&=a(x+2)^2+3 \\&\textup{When }x=2,\, y=1 \\1&=a(2+2)^2+3 \\-2&=16a \\a&=-\frac{1}{8} \\\therefore y&=-\frac{1}{8}(x+2)^2+3 \\-8y&=x^2+4x+4-24 \\x^2&+4x+8y-20=0\end{align*}$

Thank you so much for that !

calebwhitaker199 · Jul 18, 2015

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

DatAtarLyfe said:
One day i'll learn Latex....

Hahah thanks anyway !

appleibeats · Jul 18, 2015

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

A parabola has parametric equations:

x = 2t
y = t^2

a) find dx /dt and dy /dt

dx/dt = 2

dy / dt = 2t

b) Hence, find dy/dx in terms of t

dy / dx = 2t x 1/2 = t

c) Find the equations of the tangent at the point t = -3

So I not sure how to get the coordinate of x and y to sub into the y - y1 = m(x - x1)

i know m = -3

but what do y1 and x1 equal to??

rand_althor · Jul 18, 2015

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

$$When $t=-3$, $ x=2(-3)=-6$ and $y=(-3)^2=9$. So you're finding the tangent at the point $(-6,9).$

DatAtarLyfe · Jul 18, 2015

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

rand_althor said:
$$When $t=-3$, $ x=2(-3)=-6$ and $y=(-3)^2=9$. So you're finding the tangent at the point $(-6,9).$

Soo m=-3, (-6,9)
using point gradient:
y-9 = -3 (x+6)
y-9 = -3x - 18
3x + y + 9 = 0
Q.E.D

lol, did all that just to make myself feel better about the fact that rand beat me to it....again

appleibeats · Jul 18, 2015

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

Thanks for the help.

Consider the parabola x^2 = 4y

a) write down the general equation of the normal at a point with parameter t

b) Hence find the points on the parabola at which the normal at that point passes through N (0,3)

Does the general equation has something to do with:

x + py = 2ap + ap^3 ??

rand_althor · Jul 18, 2015

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

$$The general equation for the normal of the parabola $x^2=4ay$ is $x+ty=2at+at^3$. $$If $a=1, $ the equation is $x+ty=2t+t^3.$

$$If the normal passes through N$(0,3):$

$\begin{align*}0+3t&=2t+t^3 \\t^3-t&=0 \\t(t^2-1)&=0 \\t(t-1)(t+1)&=0 \\t&=-1,0,1\end{align*}$

$\begin{align*}&\textup{Using the parametric form of the parabola } (2t,t^2): \\t&=-1 \to (-2,1) \\t&=0 \to (0,0) \\t&=1 \to (2,1)\end{align*}$

appleibeats · Jul 18, 2015

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

The steel frame of a rectangular prism, has length 3x, width x, and heigh h. It is three times as long as it is wide. The prism has a volume of 4374 m^3. Find the dimensions of the frame so that the minimum amount of steel is used.

So firstly am i meant to find the derivative of 3hx^2. But don't you have to get rid of the h. So finding what h is in terms of x?? But that doesn't work.

integral95 · Jul 18, 2015

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

appleibeats said:
The steel frame of a rectangular prism, has length 3x, width x, and heigh h. It is three times as long as it is wide. The prism has a volume of 4374 m^3. Find the dimensions of the frame so that the minimum amount of steel is used.

So firstly am i meant to find the derivative of 3hx^2. But don't you have to get rid of the h. So finding what h is in terms of x?? But that doesn't work.

No you have to consider the sum of all of the sides, that is what the steel frame is.

So the sum of all sides is

$P = (3x+x)4 +4h = 16x+ 4h \\ \\ $Now we know that$ \ \ 3hx^2 = 4374 \\ \\ \\ \Rightarrow h = \frac{4374}{3x^2} \ \ \ Find \ \ \frac{dP}{dx}$

appleibeats · Jul 18, 2015

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

A cardboard box is to have square ends and a volume of 768cm^3. It is to be be sealed using two pieces of tape, one passing entirely around the length and width of the box and the other passing entirely around the height and width of the box. Find the dimensions of the box so that the least amount of tape is used.

Answer: 12 x 8 x 8

Perimeter = 8x + 4y

yx^2 = 768
y = 768/x^2

P = 8x + 3072 / x^2

dp /dx = 8 - 6144 / x^3 = 8x^3 - 6144 / x^3

Dp/Dx = 0 when x = ???

I don't know where I have gone wrong. X comes out to be a horrible number when it should not be.

rand_althor · Jul 18, 2015

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

$\begin{align*}T&=\textup{Tape required to around length and width }+ \textup{Tape to go around height and width} \\&=4x+2x+2y \\&=6x+2y \\\textup{Volume}&=x^2y\\x^2y&=768 \\y&=\frac{768}{x^2} \\\therefore T&=6x+2\times\frac{768}{x^2} \\&=6x+\frac{1536}{x^2} \\\frac{\mathrm{d}T}{\mathrm{d}x}&=6-\frac{3072}{x^3} \\ \textup{Let }\frac{\mathrm{d}T}{\mathrm{d}x}&=0 \textup{ to find stationary points.}\\6-\frac{3072}{x^3}&=0 \\x^3&=\frac{3072}{6} \\x&=8 \\\end{align*}$

$You can confirm this is a minimum by using the second derivative test.$ \\$When $x=8,\,y=\frac{768}{8^2}=12$. Therefore the dimensions of the box will be $ 8$cm$\times 8$cm$\times 12$cm.$

appleibeats · Jul 19, 2015

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

a) Sketch a graph of the function y = (x - 1)(x + 2) /x(x+4)

b) Prove that the roots of the equation kx(x+ 4) = (x -1)(x +2) are always real.

c) How can you establish this result from the graph you have drawn?

I can do part a and b.

Dont understand answer to c.

" Whatever the value of k, every horizontal line y = k intersects the graph. "

rand_althor · Jul 19, 2015

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

$kx(x+4)=(x-1)(x+2) \to k=\frac{(x-1)(x+2)}{x(x+4)}$

You are essentially equating the line y=k and the original function, to find points of intersections. If you drew your graph correctly, you'll notice that it's range is all real y. Thus, regardless of the value of k, y=k and the original function will always have a point of intersection - i.e. the equation $kx(x+4)=(x-1)(x+2)$ will always have a real root.

appleibeats · Jul 19, 2015

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

What do you mean y is all real. What about asymptotes?

rand_althor · Jul 19, 2015

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

This is the graph:

There are vertical asymptotes at x=-4 and x=0, and a horizontal asymptote at y=1.
However, the range is still all real y.

appleibeats · Jul 20, 2015

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

State in terms of b and c the condition for the roots of x^2 +2bx + 3c = 0 to be :

d) opposite in sign
e)distinct and positive
f) distinct and negative

not sure what to make the discriminant:

4b^2 -12 equal ?

InteGrand · Jul 20, 2015

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

appleibeats said:
State in terms of b and c the condition for the roots of x^2 +2bx + 3c = 0 to be :

d) opposite in sign
e)distinct and positive
f) distinct and negative

not sure what to make the discriminant:

4b^2 -12 equal ?

$(The discriminant is $\Delta =4b^2 - 12c$, where $b$ and $c$ are real numbers for our purposes.)$

$(d) Two real numbers are opposite in sign if and only if their product is negative. So we require that $\Delta > 0 \Longleftrightarrow 4b^2 - 12c > 0$ (in order to have two distinct real roots), and $3c<0\Longleftrightarrow c<0$ (since the product of roots is $3c$, and we need this to be negative). The condition $4b^2 - 12c > 0$ implies that $b^2 > 3c$. Since $b^2 \geq 0 > 3c$ for any real $b$ (since a square of any real number is non-negative), it follows that $b$ can be any real number. So the conditions on $b$ and $c$ are that $c<0$ and $b$ is any real number.$

$(e) For the roots to be distinct and real, we again require $\Delta > 0 \Longleftrightarrow 4b^2 - 12c > 0$. Two real numbers are both positive if and only if their sum is positive \emph{and} their product is positive. So we also require $-2b>0$ (sum of roots) and $3c>0$ (product of roots). The condition $-2b>0$ is equivalent to $b<0$, and the condition $3c>0$ is equivalent to $c>0$. Since $4b^2-12c>0 \Longleftrightarrow b^2 > 3c(>0)$, it follows that we require $b<-\sqrt{3c},c>0$ (since $b$ must be negative). This is the required condition.$

$(f) As the roots are distinct and real, we again require $b^2>3c$ (from the discriminant, as before). Two real numbers are both negative if and only if their sum is negative and their product is positive. So we require $-2b<0\Longleftrightarrow b>0$ (from sum of roots), and $3c>0\Longleftrightarrow c>0$ (from product of roots). So we require $b>\sqrt{3c},c>0$.$

appleibeats · Jul 20, 2015

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

Engineers have determined that the strength s of a rectangular beam varies as the product of the width w and the square of the depth d of the beam, that is s = kwd^2 for some constant k. Find the dimensions of the strongest rectangular beam that can be cut from a cylindrical og with diameter 48 cm.

Not sure how to begin this question. Has to do with maximisation and minimisation.

rand_althor · Jul 20, 2015

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

$\begin{align*}&\textup{The strength is maximised when the width and depth are greatest.} \\&\textup{Using the diagram in the textbook, apply Pythagoras' Theorem:} \\d^2+w^2&=48^2 \to d^2=48^2-w^2 \\s&=kwd^2 \\&=kw(48^2-w^2) \\&=2304kw-kw^3 \\\frac{\mathrm{d}s}{\mathrm{d}w}&=2304k-3kw^2 \\&\textup{Let the derivative equal zero to find the maximum:} \\2304k-3kw^2&=0 \\w^2&=\frac{2304}{3} \\w&=\sqrt{768} \\&=16\sqrt{3} \\&\textup{When }w=16\sqrt{3}: \\d&=\sqrt{48^2-(16\sqrt{3})^2}=16\sqrt{6} \\&\textup{Therefore the strength will be greatest when }w=16\sqrt{3}\textup{ cm, }d=16\sqrt{6}\textup{ cm}\end{align*}$

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