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

appleibeats · Jul 21, 2015

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

Water is leaking out through a hole in the bottom of a bucket at a rate dv/dt =-8(25 - t), where v cm^3 is the volume of liquid in the bucket at time T seconds
Initially there was 2 1/2 litres of water in the bucket. Find v in terms of t, and hence find the time taken for the bucket to be emptied.

InteGrand · Jul 21, 2015

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

appleibeats said:
Water is leaking out through a hole in the bottom of a bucket at a rate dv/dt =-8(25 - t), where v cm^3 is the volume of liquid in the bucket at time T seconds
Initially there was 2 1/2 litres of water in the bucket. Find v in terms of t, and hence find the time taken for the bucket to be emptied.

$We shall use $V$ to denote volume as $v$ is usually used for velocity.$

$Since $\frac{\mathrm{d}V}{\mathrm{d}t}=-8(25-t)$, integrating gives us $V=\int -8(25-t) \text{ d}t$$

$\Rightarrow V = -8 \times \frac{(25-t)^2}{-2}+C$, for some constant $C$

$$\Rightarrow V(t) = 4(25-t)^2+C$.$

$Since $V(0) = \frac{5}{2}$, we have $\frac{5}{2}=4\times (25-0)^2 + C\Rightarrow C = \frac{5}{2}-4\times 25^2 = -\frac{4995}{2}$. Hence $V(t) = 4(25-t)^2-\frac{4995}{2}$.$

$The bucket is empty when $V=0$, i.e. when$

$4(25-t)^2-\frac{4995}{2}=0$

$\Rightarrow 4(25-t)^2=\frac{4995}{2}$

$\Rightarrow (25-t)^2=\frac{4995}{8}$

$\Rightarrow 25-t=\pm \sqrt{\frac{4995}{8}}$

$$\Rightarrow t=25 \mp \sqrt{\frac{4995}{8}}$.$

$Both solutions turn out to be positive, so we take the earlier answer, which is $t=25-\sqrt{\frac{4995}{8}}$, which is roughly 0.0125 seconds for the bucket to empty.$

appleibeats · Jul 22, 2015

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

y = ax(x - a) is the general form of a quadratic whose graph passes through the origin. Hence find the equation of such a quadratic:

d) with no other x -intercepts, and having gradient 1 at x = -2

e) with another x intercept at x = 5, and having gradient 2 at the origin

for d) i get the answer y = 1/4 x^2

however the answer in the book has it as negative my answer. Not sure why.

For e) i know that to have a x intercept at 5

it becomes y = x(x -5)

but not sure how to work out the gradient bit.

rand_althor · Jul 22, 2015

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

$$d) As the only x-intercept is at the origin, the equation of the parabola is $y=ax^2$. If $y=ax^2$, then $\frac{\mathrm{d}y}{\mathrm{d}x}=2ax$. When $x=-2$, $\frac{\mathrm{d}y}{\mathrm{d}x}=1$:$ \\1=2a(-2)\to a=-\frac{1}{4}. \therefore y=-\frac{1}{4}x^2.$

$$e) As there is an x-intercept at the origin and $x=5$, the equation of the parabola is $y=ax(x-5)=ax^2-5ax$. The derivative of this function is $\frac{\mathrm{d}y}{\mathrm{d}x}=2ax-5a$. If $\frac{\mathrm{d}y}{\mathrm{d}x}=2$ when $x=0$, then: $ 2=-5a\to a=-\frac{2}{5}$. $\therefore y=-\frac{2}{5}x(x-5).$

appleibeats · Jul 24, 2015

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

If y = 1 - 2x is a tangent to x^2 = 4ay , find a and the point of contact

I though you first find the gradient of the parabola; y' = x / 2a

Then that equals the gradients of the line; so

x / 2a = - 2

x = -4a

Now not sure where to go from here.

dan964 · Jul 24, 2015

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

Sub the x coordinate into the two equations
one you will get y=8a+1 and the other y=4a
Solve simultaenously to get a and hence y and x.

rand_althor · Jul 24, 2015

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

$\begin{align*}y&=1-2x \to y'=-2 \\y&=\frac{x^2}{4a} \to y'=\frac{x}{2a} \\\textup{Equating the gradients gives you }x&=-4a\textup{, which is the x-coordinate of the point of contact.} \\\textup{Sub }x&=-4a \to y=\frac{x^2}{4a}: \\y&=\frac{(-4a)^2}{4a}=\frac{16a^2}{4a}=4a \\\end{align*}$

$Therefore both the tangent and the parabola pass through the point $(-4a,4a)$. Subbing this point into the tangent equation: \\$4a=1+8a \to a=-\frac{1}{4}$ \\Thus the point is $(-4\times-\frac{1}{4},4\times-\frac{1}{4})\to (1,-1)$

appleibeats · Jul 25, 2015

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

Consider the parabola x^2 = 12y. Line ax + by = 1 is tangent to this parabola.

Show that 3a^2 + b = 0

found the gradient for each function. Thus the x coordinate. Sub it into one of the functions and get the y coordinate. Then subbed it into the line and simplified. However still didn't get the right answer.

Is this the correct method of answer this question?

rand_althor · Jul 25, 2015

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

Your method is correct.

$$Find the x-coordinate by equating the gradients, and then find the y-coordinate by subbing the x-coordinate into the equation of the parabola. You should find the coordinates of the point of contact between the tangent and the parabola to be $ (-\frac{6a}{b}, \frac{3a^2}{b}).$ Sub this point into the equation of the tangent: \\$-\frac{6a}{b}\times a +\frac{3a^2}{b^2}\times b =1 $ \\Simplify this equation to obtain the required result: $3a^2+b=0$.$

appleibeats · Jul 25, 2015

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

For the same question,

Hence, find the tangents to the parabola with y intercept -27

Do I sub in (0, -27) into the equation of the line: ax + by = 1

Then find b = - 1/27

Then sub this into 3a^2 + b = 0

So, 3a^2 = 1/27

a^2 = 1/81

a = + 1/9 and - 1/9

So, the equations of the tangents are:

1/9 x - 1/27 y = 1 ; 3x - y - 27 = 0

and

-1/9 x - 1/27 y = 1 ; 3x + y + 27 = 0

Is this correct? Not really sure if my method is right.

rand_althor · Jul 25, 2015

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

Yes that is correct.

wiliaamnguuyen · Jul 25, 2015

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

How do you graph the regions of: (a) |y| > |x|
(b) 1/x > 1/y

InteGrand · Jul 25, 2015

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

wiliaamnguuyen said:
How do you graph the regions of: (a) |y| > |x|
(b) 1/x > 1/y

$(a) $|y|>|x|\Longleftrightarrow y>|x|$ or $y<-|x|$. So sketch the graphs of of $y=|x|$ and $y=-|x|$ in dotted lines, then shade in the region above the graph of $y=|x|$ and below that of $y=-|x|$, i.e. the region shaded here:$

http://www.wolframalpha.com/input/?i=|y|>|x|

$(b) $\frac{1}{x}>\frac{1}{y},x,y\neq 0$$

$For $x>0$, this means that either $y<0$ or $y>x$. So for the right half of the plane ($x>0$), sketch in the region where $y<0$ (i.e. the fourth quadrant) and the region above the line $y=x,x>0$ (sketch the line dotted).$

$For $x<0$, the inequality means firstly that $y$ cannot be positive, since that would make the R.H.S. positive, whereas the L.H.S. is negative. So $y<0$. This means that $y>x$. We can see this either by thinking about it, or via algebra: $\frac{1}{x}>\frac{1}{y},x,y<0\Rightarrow y>x$ (multiplying through by $xy$, and noting that this means that the inequality sign does not change, since $xy>0$, since $x$ and $y$ are both negative). So for the left half of the plane, shade in the region above the line $y=x$, but below the $x$-axis (as $y$ must be negative).$

$The overall region is shaded in the link below:$

http://www.wolframalpha.com/input/?i=1/x+>+1/y

appleibeats · Jul 25, 2015

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

AB is the chord contact of the parabola x^2 = 4ay from P (x0, y0). The line AB meets the directrix of the parabola at D. Prove that PD subtends a right angle at the focus.

InteGrand · Jul 25, 2015

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

appleibeats said:
AB is the chord contact of the parabola x^2 = 4ay from P (x0, y0). The line AB meets the directrix of the parabola at D. Prove that PD subtends a right angle at the focus.

$We are asked to prove that $\angle PSD = 90^\circ \Longleftrightarrow DS \perp PS$, where $S=(0,a)$ is the focus.$

$The equation of the chord of contact $AB$ is $xx_0 = 2a(y+y_0)\Longleftrightarrow x = \frac{2a}{x_0}(y+y_0)$ (you would need to derive this in an exam first; check the textbook for its proof)$

$As $D$ lies on this line, and $y_D = -a$ (directrix), we have $x_D=\frac{2a}{x_0}(-a+y_0)= \frac{2a}{x_0}(y_0-a)$. So $D=\left(\frac{2a}{x_0}(y_0-a),-a \right)$.$

$$\therefore m_{DS}=\frac{y_D - y_S}{x_D - x_S}=\frac{-a-a}{ \frac{2a}{x_0}(y_0-a)-0}=\frac{-2a}{ \frac{2a}{x_0}(y_0-a)}=\frac{x_0}{-(y_0 -a)}$.$

$Also, $m_{PS}=\frac{y_P - y_S}{x_P - x_S}=\frac{y_0 - a}{x_0 - 0}=\frac{y_0 - a}{x_0}$.$

$$\therefore m_{DS}\times m_{PS}=\frac{x_0}{-(y_0 -a)}\times \frac{y_0 - a}{x_0} = -1\Rightarrow DS \perp PS.$ \blacksquare$

appleibeats · Jul 28, 2015

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

Indefinite Integral of ( (ax)^1/2 + 1/(ax)^1/2 ) dx

Question is to find each of the indefinite integrals:

I get the answer 2(ax)^3/2 / 3 + 2(ax)^1/2 + C

The answer in the book gives instead of my 2(ax)^ 1/2 as 2(ax)^1/2 all over a

Not sure where the a comes from?? Have i done the integral wrong???

appleibeats · Jul 28, 2015

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

Indefinite Integral of ( (ax)^1/2 + 1/(ax)^1/2 ) dx

Question is to find each of the indefinite integrals:

I get the answer 2(ax)^3/2 / 3 + 2(ax)^1/2 + C

The answer in the book gives instead of my 2(ax)^ 1/2 as 2(ax)^1/2 all over a

Not sure where the a comes from?? Have i done the integral wrong???

InteGrand · Jul 28, 2015

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

appleibeats said:
Indefinite Integral of ( (ax)^1/2 + 1/(ax)^1/2 ) dx

Question is to find each of the indefinite integrals:

I get the answer 2(ax)^3/2 / 3 + 2(ax)^1/2 + C

The answer in the book gives instead of my 2(ax)^ 1/2 as 2(ax)^1/2 all over a

Not sure where the a comes from?? Have i done the integral wrong???

$$We need to divide by the coefficient of $x$, remembering that $\int (Ax+B)^{n}\text{ d}x = \frac{(Ax+B)^{n+1}}{A(n+1)}+c$$.$

$So $\int \left((ax)^{\frac{1}{2}}+\frac{1}{(ax)^{\frac{1}{2}}} \right)\text{ d}x=\int \left((ax)^{\frac{1}{2}}+(ax)^{-\frac{1}{2}} \right)\text{ d}x$$

$=\frac{(ax)^{\frac{1}{2}+1}}{a\left(\frac{1}{2}+1 \right)}+\frac{(ax)^{-\frac{1}{2}+1}}{a\left(-\frac{1}{2}+1 \right)}+c$

$=\frac{(ax)^{\frac{3}{2}}}{a\left(\frac{3}{2} \right)}+\frac{(ax)^{\frac{1}{2}}}{a\left(\frac{1}{2} \right)}+c$

$=\frac{2(ax)^{\frac{3}{2}}}{3a}+\frac{2(ax)^{ \frac{1}{2}}}{a}+c$

$$=\frac{\frac{2}{3}(ax)^{\frac{3}{2}}+2(ax)^{ \frac{1}{2}}}{a}+c$.$

appleibeats · Jul 28, 2015

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

InteGrand said:
$We shall use $V$ to denote volume as $v$ is usually used for velocity.$

$Since $\frac{\mathrm{d}V}{\mathrm{d}t}=-8(25-t)$, integrating gives us $V=\int -8(25-t) \text{ d}t$$

$\Rightarrow V = -8 \times \frac{(25-t)^2}{-2}+C$, for some constant $C$

$$\Rightarrow V(t) = 4(25-t)^2+C$.$

$Since $V(0) = \frac{5}{2}$, we have $\frac{5}{2}=4\times (25-0)^2 + C\Rightarrow C = \frac{5}{2}-4\times 25^2 = -\frac{4995}{2}$. Hence $V(t) = 4(25-t)^2-\frac{4995}{2}$.$

$The bucket is empty when $V=0$, i.e. when$

$4(25-t)^2-\frac{4995}{2}=0$

$\Rightarrow 4(25-t)^2=\frac{4995}{2}$

$\Rightarrow (25-t)^2=\frac{4995}{8}$

$\Rightarrow 25-t=\pm \sqrt{\frac{4995}{8}}$

$$\Rightarrow t=25 \mp \sqrt{\frac{4995}{8}}$.$

$Both solutions turn out to be positive, so we take the earlier answer, which is $t=25-\sqrt{\frac{4995}{8}}$, which is roughly 0.0125 seconds for the bucket to empty.$

The answer in the book says 25 seconds. I understand your working, but why does neither of your two answer equal 25 seconds??

InteGrand · Jul 28, 2015

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

appleibeats said:
The answer in the book says 25 seconds. I understand your working, but why does neither of your two answer equal 25 seconds??

I realised I made a silly mistake, I should have converted the initial volume V(0) to cubic cm, since the dV/dt is given in cubic cm per second.

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