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

drsabz101 · Feb 20, 2016

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

Thanks heaps, I get it now

davidgoes4wce · Feb 20, 2016

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

No worries

drsabz101 · Feb 20, 2016

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

Can u help solve this, like outline the steps taken? Thankyou!

authenticity · Feb 20, 2016

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

A) Differentiate it. Sub in P to find a f(x) equation. Sub in P's x value and find the y value (gradient). Using point-gradient formula, figure out the equation, and then the x and y intercepts
B) Use point line distance formula to find the distance, which should be 13 (same as radius) - circle geo - the perpendicular of a tangent at the point of contact is the radius
C) Knowing the x & y intercepts, use the area formula for triangles (Area = x * y * 1/2)
D) Use distance formula to find the hypotenuse. Then it would be hypotenuse + x + y value = perimeter.

davidgoes4wce · Feb 20, 2016

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

drsabz101 said:
Can u help solve this, like outline the steps taken? Thankyou! View attachment 32930

Q13 (a)
$y=(169-x^2)^{\frac{1}{2}}$

$\frac{dy}{dx}=\frac{1}{2} (169-x^2)^\frac{-1}{2} (-2x)$

$= \frac{-x}{(169-x^2)^\frac{1}{2}}$

$At x =12 , \frac{dy}{dx} = \frac{-12}{(169-12^2)^\frac{1}{2}} =\frac{-12}{5}$

$$ Point of Gradient Formula used. $ y-y_{1}=m(x-x_{1})$

$y-y_{1}=\frac{-12}{5} (x-12)$

$$ x-intercept . Let y=0, x-intercept $ (\frac{169}{12},0).$ Let x=0, y-intercept $ (0,\frac{169}{5}). $ The standard form of the equation is $ 12x+5y-169=0$

$$ Part (b) Perpendicular Distance $ =\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^2+b^2}}$

$The origin is the centre point (0,0) which will be substituted into the formula Perpendicular Distance formula, given we also know the Standard Form of the equation.$

$d=\frac{|12(0+5(0)-169|}{\sqrt{12^2+5^2}}= \frac{169}{\sqrt{169}}=13. $ Hence, 13 is the radius of this circle. $$

$$ Part (c) Given we know the x-intercept is $ \frac{169}{12} , $ the y-intercept is $ \frac{169}{5} $ we can use the formula of the Triangle to solve for the Area of a Triangle, which in this case is $ A=\frac{1}{2} b h$

$A=\frac{1}{2} \times \frac{169}{5} \times \frac{169}{12} = \frac{169^2}{120}$

$$ Part (d) The perimeter of the triangle can be found by adding the distance from the Origin to the x-intercept $ = \frac{169}{12}, $ the distance from the Origin to the y-intercept $ = \frac{169}{5} $ and the diagonal distance by using Pythagoras' thereoem $ \sqrt{\frac{169}{5})^2+(\frac{169}{12})^2}=\frac{2197}{60}. $ By adding up all 3 values we are able to obtain our perimeter=$ \frac{169}{12} +\frac{169}{5}+\frac{2197}{60}= 84.5 or \frac{169}{2}$

drsabz101 · Feb 20, 2016

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

Is it okay if you can outline the steps for q11, like authenticity did previously. just the steps only (thanks for the help again)

davidgoes4wce · Feb 20, 2016

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

drsabz101 said:
Is it okay if you can outline the steps for q11, like authenticity did previously. just the steps only (thanks for the help again)

I'll try my best, I do include comments in my working out as well and formulas used to get to the answer.

drsabz101 · Feb 20, 2016

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

Thankyou, I appreciate it

drsabz101 · Feb 21, 2016

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

Can someone show how to differentiate this using product rule. I am getting the last half wrong , don't know why though.

InteGrand · Feb 21, 2016

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

drsabz101 said:
Can someone show how to differentiate this using product rule. I am getting the last half wrong , don't know why though.
View attachment 32931

$\noindent The answer is $y^\prime = 3x^2 (x+1)^4 + x^3 \cdot 4\left(x+1\right)^3$.$

drsabz101 · Feb 21, 2016

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

I got that though

drsabz101 · Feb 21, 2016

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

this is the textbook answer I even factorised x^2(x-3)^3 out :

InteGrand · Feb 21, 2016

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

InteGrand said:
$\noindent The answer is $y^\prime = 3x^2 (x+1)^4 + x^3 \cdot 4\left(x+1\right)^3$.$

drsabz101 said:
this is the textbook answer I even factorised x^2(x-3)^3 out : View attachment 32932

$\noindent If we want to factorise it, we have$

$\begin{align*} y^\prime &= x^2 (x+1)^3 \left(3(x+1) + 4x\right) \\ &= x^2 (x+1)^3 (7x+1). \end{align*}$

drsabz101 · Feb 21, 2016

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

Oh , true ! thankyou

drsabz101 · Feb 21, 2016

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

can someone show how to solve for part b

davidgoes4wce · Feb 21, 2016

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

drsabz101 said:
can someone show how to solve for part b View attachment 32936

With this question, its best to visualise it on a graphing facility. It's a shame that the HSC hasn't taken the same route as VCE, WACE and IB Maths in allowing students to use such a tool as I believe the learning is enhanced immensely.

Below is a screenshot of an example I did with the parabola on Desmos.

What you can see is if the gradients are opposite in magnitude, in this case, gradients of 5 and -5, the tangent curves intersect at an external point, (2,-10.25) . You can also see that the normals intersect at (2,2.75). Forget about the coordinates in this case, what you should be doing is focusing on the shape of that quadrilateral. What shape does it look like? The shape is in fact a kite.

$$ The formula for a kite you may remember from junior maths (years 8-10 level) is of the form $ A=\frac{pq}{2}$

Given that you know the x-coordinates from part (a) , the tricky part here is to use algebra in order to find the y-coordinates. You then want to work out the distances of the vertical height.

davidgoes4wce · Feb 21, 2016

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

$$ We found out from part (a) that $ x=\frac{h+m}{2} $ for the x-coordinate at P. Substituting $ x=\frac{h+m}{2} $ into the original equation $ y=(x-h)^2+k, $ gives a y-coordinate of $ \frac{m^2}{4}+k . $ Hence, the coordinate $ P(\frac{h+m}{2}, \frac{m^2}{4}+k)$

$$ Doing the same thing for Q by instead we substitute $ x=h-\frac{m}{2}. $ The coordinate for Q turns out to be $ Q (h-\frac{m}{2}, \frac{m^2}{4}+k )$

$What we should then do is equate the Normal Equations Together. Given the gradient of a tangent are m and -m respectively. The reciprocals of these would give us the gradients of the normals, which in this case would be $ \frac{1}{m} and \frac{-1}{m}. $ And given we know the coordinates of P and Q, by using the Point Gradient formula $ y-y_{1}=m(x-x_{1}) $ we can find the equations of the Normals and the Tangents.$

$The equation of the Normals at P and Q$

$$ At P $, y-(\frac{m^2}{4}+k)=\frac{-1}{m}[x-(h+\frac{m}{2})] \textcircled{1}$

$$ At Q $, y-(\frac{m^2}{4}+k)=\frac{1}{m}[x-(h-\frac{m}{2})] \textcircled{2}$

$Let equations $ \textcircled{1} and \textcircled{2} $ be equal to each other. Remember what we want is from that previous diagram, is the y-coordinate, So we have to think of a way to eliminate 'x' , so what I did was equated x terms equal to each other by rearranging them to get them as the subject.$

$$ Rearranging,$ \textcircled{1} , x=-m[y-(\frac{m^2}{4}+k)+(h+\frac{m}{2})]$

$$ Rearranging,$ \textcircled{2}, x=m[y-(\frac{m^2}{4}+k)+(h-\frac{m}{2})]$

$-m[y-(\frac{m^2}{4}+k)+(h+\frac{m}{2})=m[y-(\frac{m^2}{4}+k)+(h-\frac{m}{2})]$

$$ The y-coordinate at the Normals intersecting is $ y=\frac{1}{2}+(\frac{m^2}{4}+k)$

$Now we have to do the same thing and equate the equations of the Tangents in order to find the y-coordinate of the intersection. This time we use the gradient of m for Point P and gradient of -m for Point Q$

$$ The tangent equation at Point P $, y-(\frac{m^2}{4}+k)=m[x-(h+\frac{m}{2})] \textcircled{3}$

$$ The tangent equation at Point Q $, y-(\frac{m^2}{4}+k)=-m[x-(h-\frac{m}{2})] \textcircled{4}$

$Like the previous part for the normals what we want to do is try to rearrange those equations to get x as the subject.$

$$ Rearranging $ \textcircled{3} \frac{1}{m} [y-(\frac{m^2}{4}+k)+(h+\frac{m}{2})]=x$

$$ Rearranging $ \textcircled{4} \frac{-1}{m}[y-(\frac{m^2}{4}+k)+(h-\frac{m}{2})]=x$

$$ Equating equations $ \textcircled{3} and \textcircled{4}, \frac{1}{m} [y-(\frac{m^2}{4}+k)+(h+\frac{m}{2})=\frac{-1}{m}[y-(\frac{m^2}{4}+k)+(h-\frac{m}{2})]$

$$ The y-coordinate at the tangents intersecting is $ y=k-\frac{m^2}{4}$

$$ What you can then do is work out the differences in distance between the y-coordinate of the Tangent and the y-coordinate of the Normal which in this case is $ (\frac{1}{2}+\frac{m^2}{4}+k)-(k-\frac{m^2}{4})=\frac{1}{2} \times (m^2+1)$

$$From the previous part$ (a) , $ one can work out the horizontal distance between the 2 coordinates simply by subtracting the 2 x-coordinate values. Given at $ P, x={h+\frac{m}{2}} $ and at Q $, x=h-\frac{m}{2}$

$$ The difference in the horizontal distance is simply $ = (h+\frac{m}{2})-(h-\frac{m}{2})=m .$

$$ Now that we have the horizontal distance=m and vertical distance $ =\frac{1}{2} \times (m^2+1) $ we can substitute them into the formula $, A=\frac{pq}{2}$

$A=[\frac{1}{2} \times (m^2+1)] m \frac{1}{2} = \frac{1}{4}m (m^2+1)$

davidgoes4wce · Feb 21, 2016

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

I also want to say that every Development or Extension questions in the exercises, that you can knock off in Cambridge, you feel a greater appreciation for the mathematics language. Think any student in Year 11 or 12 should practice as many questions from here as they are exam standard pretty much. (especially the Development questions). It's also great in the fact that key concepts are highlighted in blue. (if you know those main formuale, equations and theory and go all over it again and again. You won't go wrong). Last year I said Fitzpatrick was a better book, I was wrong. Its Cambridge>Fitzpatrick>Lee>Grove but if your a keen maths student you would look at all resources possible.

As well as having a formulae book (which I purchased) for this subject, I make the effort in making my own theory notes. Just notes and formuale that I can go over again and again when I'm trying to refresh my memory on a certain question. I think all up I have around 100-120 pages worth of formulae, notes that I rigorously go all over and over, till its stored. (From time to time you do forget something )

Shuuya · Feb 22, 2016

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

I'm having a mind blank

Could someone remind me how to do these please? I know that it says what to do in the question but I'm getting stuck

InteGrand · Feb 22, 2016

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

Shuuya said:
I'm having a mind blank

Could someone remind me how to do these please? I know that it says what to do in the question but I'm getting stuck

$\noindent We can do each of these (except the last one) using the fact that $\frac{\mathrm{d}}{\mathrm{d}x} \left(a^{mx+b}\right) = a^{mx+b}\cdot m \ln a$.$

$\noindent To prove this fact, we can let $y = a^{mx+b}$, and then we have $y=\mathrm{e}^{\left(mx+b\right) \ln a}$. Let $u=(mx+b)\ln a = m\ln a\cdot x + b\ln a$. Then $\frac{\mathrm{d}u}{\mathrm{d}x} = m\ln a$. Now by the chain rule, noting $y=\mathrm{e}^{u} = a^{mx+b}$, we have$

$\begin{align*} y^\prime &= \frac{\mathrm{d}y}{\mathrm{d}u} \frac{\mathrm{d}u}{\mathrm{d}x} \\ &= \mathrm{e}^{u} \cdot \left(m\ln a\right) \\ &= a^{mx+b}\cdot \left(m\ln a\right), \end{align*}$

$\noindent as claimed.$

$\noindent The main rule is that $\frac{\mathrm{d}}{\mathrm{d}u} \left(a^u \right) = \ln a \cdot a^u$. This, along with the chain rule, allows us to do all the questions above.$

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