Cambridge Prelim MX1 Textbook Marathon/Q&A (3 Viewers)

leehuan · Jul 12, 2015

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

appleibeats said:
Now subbing x = 1, you get y = -3 and so point of contact is (1, -3)

But answer in the book also says at ( -1, 3) the tangent is x + y -2 = 0

Not sure what it means?

Not too sure how that relates to the aforementioned question...

leehuan · Jul 12, 2015

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

$\\ x+2y=4\\ y={ \left( x-1 \right) }^{ 2 }\\ \vdots \\ \left( 4y-9 \right) \left( y-1 \right) =0\\ y=1\quad or\quad y=\frac { 9 }{ 4 } \\ Sub\quad back\quad into\quad x+2y=4\quad for\quad x\quad coordinates:\\ Points\quad are\quad \left( 2,1 \right) \quad and\quad \left( -\frac { 1 }{ 2 } ,\frac { 9 }{ 4 } \right)$

$Now,\quad note\quad that\quad the\quad line\quad has\quad m=-\frac { 1 }{ 2 } \\ y={ \left( x-1 \right) }^{ 2 }\\ \frac { dy }{ dx } =2\left( y-1 \right) \\ We\quad note\quad when\quad x=2,\\ \frac { dy }{ dx } =2\left( 2-1 \right) =2\\ Hence,\quad gradient\quad of\quad tangent\quad at\quad \left( 2,1 \right) \quad is\quad m=2\\ But\quad we\quad note\quad that\quad -\frac { 1 }{ 2 } \times 2=-1\\ Hence\quad { m }_{ 1 }{ m }_{ 2 }=-1\quad so\quad the\quad line\quad is\quad a\quad normal\quad at\quad that\quad point$

faisalabdul16 · Jul 12, 2015

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

appleibeats said:
How do you answer q 19)

Find the points where the line x + 2y = 4 cuts the parabola y = ( x -1)^2, and show that the line is the normal to the curve at one of these points.

I made x = 4 - 2y

and subbed it in to the parabola and got:

4y^2 - 13y + 9 = 0

(4y - 9) (y -1) = 0

Therefore points are ( 2,1) and the second point I am not sure: when I sub y = 9/4 into the parabola I get x = 5/2 but when into the line x = 8/9

Not sure why I am getting different results. And how to show that at one of these points the line is the normal to the curve.

<a href="http://www.codecogs.com/eqnedit.php?latex=x&space;+&space;2y&space;=&space;4&space;\\&space;y&space;=&space;(x-1)^{2}&space;\\&space;\\&space;x&space;+&space;2(x^{2}-2x+1)&space;=&space;4&space;\\&space;x&space;+&space;2x^{2}&space;-4x+-2=0&space;\\&space;2x^{2}-3x-2=0&space;\\(2x+1)(x-2)&space;=&space;0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x&space;+&space;2y&space;=&space;4&space;\\&space;y&space;=&space;(x-1)^{2}&space;\\&space;\\&space;x&space;+&space;2(x^{2}-2x+1)&space;=&space;4&space;\\&space;x&space;+&space;2x^{2}&space;-4x+-2=0&space;\\&space;2x^{2}-3x-2=0&space;\\(2x+1)(x-2)&space;=&space;0" title="x + 2y = 4 \\ y = (x-1)^{2} \\ \\ x + 2(x^{2}-2x+1) = 4 \\ x + 2x^{2} -4x+-2=0 \\ 2x^{2}-3x-2=0 \\(2x+1)(x-2) = 0" /></a>

Points of intersection are (2,1) and (-1/2, 9/4)

Now

<a href="http://www.codecogs.com/eqnedit.php?latex=y&space;=&space;(x-1)^{2}&space;\\&space;y'&space;=&space;2(x-1)&space;\\&space;\text{sub&space;x&space;=&space;2,&space;gradient&space;of&space;tangent&space;to&space;the&space;curve&space;is&space;2.}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y&space;=&space;(x-1)^{2}&space;\\&space;y'&space;=&space;2(x-1)&space;\\&space;\text{sub&space;x&space;=&space;2,&space;gradient&space;of&space;tangent&space;to&space;the&space;curve&space;is&space;2.}" title="y = (x-1)^{2} \\ y' = 2(x-1) \\ \text{sub x = 2, gradient of tangent to the curve is 2.}" /></a>

Therefore gradient of normal to the curve is <a href="http://www.codecogs.com/eqnedit.php?latex=\frac{-1}{2}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{-1}{2}" title="\frac{-1}{2}" /></a> which is the gradient of the line. Hence line is normal to curve at (2,1)

leehuan · Jul 12, 2015

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

^Done more neatly

Drongoski · Jul 12, 2015

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

Ex 8E - Q20 (for lollipop)

Let radius of semi-circles be 'r'

.: each side of rectangle = (1000 - 2 x pi x r)/2 = 500 - pi x r

.: area of shaded rectangle A(r) = 2r x (500 - pi x r) . . . a quadratic in 'r' = 0 at r = 0, r = 500/pi

The mid-point of this 2 zeroes is r = (0 + 500/pi)/2 = 250/pi

The Max for A(r) occurs at this value of r (concave down parabola, vertex at r = 250/pi).

The corresponding dimensions of the rectangle: 2r x (500 - pi x r) = 2 x 250/pi x (500 - pi x 250/pi) = 500/pi x 250 metres

Drongoski · Jul 12, 2015

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

Ex 8F - Q14 (for lollipop)

Let 'm' be the gradient of the line thru (1,7); its eqn is: y - 7 = m(x - 1) or: y = m(x - 1) + 7

Where this line meets: y = (2 - x)(1 + 3x) = -3x² + 5x + 2

m(x - 1) + 7 = -3x² + 5x + 2

.: 3x² + (m - 5)x + (5 - m) = 0

Since line is tangent to the parabola, this equation has a repeated root, or:

discriminant "b² - 4ac" = (m - 5)² - 4 x 3 x (5 - m) = 0

.: (m - 5)(m + 7) = 0

.: gradient m = -7, 5

appleibeats · Jul 12, 2015

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

For Q 9 in 7E:

a) Find the equation of the tangent to y = 1/ x - 4 at the point L where x = b

b) Hence find the equations of the tangents passing through: i) the origin ii) W (6,0)

I was able to find the equation:

x + y(b - 4)^2 = 2b - 4

Now I though to answer b) you just subbed in the coordinates, but that doesn't get the answer.

Answer for b)

i) x + 4y = 0
ii) x + y = 6

rand_althor · Jul 12, 2015

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

$\begin{align*}x+y(b-4)^2&=2b-4 \\\textup{Tangent through the origin:} \\\textup{Sub in }x&=0\, ,y=0 \\0+0(b-4)^2&=2b-4 \\2b&=4 \\b&=2 \\\textup{When }b=2 &\textup{ the tangent equation is:} \\x+y(2-4)^2&=2(2)-4 \\x+4y&=0 \\\textup{Tangent through the W(6,0):} \\\textup{Sub in }x&=6\, ,y=0\\6+0(b-4)^2&=2b-4\\2b&=10\\b&=5 \\\textup{When }b=5 &\textup{ the tangent equation is:} \\x+y(5-4)^2&=2(5)-4 \\x+y&=6 \\\end{align*}$

appleibeats · Jul 12, 2015

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

For Q22 IN 7D

Find the equation of the tangent to the parabola y = (x - 3)^2 at the point T where x = a, find the coordinates of the x -intercept A and y - intercept B of the tangent, and find the midpoint M of AB. For what value of a does M coincide with T?

I can answer all parts of the question but don't know how to work out and answer the last part. FOR WHAT VALUE OF A DOES M COINCIDE WITH T ?

The answer is a = 1

Is there a method to getting this answer or is it guess and check??

rand_althor · Jul 12, 2015

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

$\begin{align*}&\textup{T is the point}(a,0)\textup{, A is the point }(\frac{a+3}{2},0)\textup{ and B is the point }(0,9-a^2)\\&\textup{The midpoint of AB, is M}(\frac{a+3}{4},\frac{9-a^2}{2})\\&\textup{For M to coincide with T, their coordinates must be equivalent: }\\x_M&=x_T \\\frac{a+3}{4}&=a \\a+3&=4a \\3a&=3 \\a&=1 \\\end{align*}$

appleibeats · Jul 13, 2015

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

3a from 7F

Find the tangents and normals to these curves a the indicated points:

y = x( 1 - x)^6 at the origin

I found the derivative:

y' = ( 1 - x)^5 (1 - 7x)

Now don't you just sub in (0,0)?

VBN2470 · Jul 13, 2015

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

So your tangent will be of the form $y=mx$ (since it passes through the origin). Find the derivative of $f:=f(x)=x(1-x)^{6}$ which will be $f':=f'(x)=(1-x)^6-6x(1-x)^5=(1-x)^5(1-7x)$ (what you found) and sub. in the point x = 0 to get f'(0) = 1. Hence your tangent to the graph of f at (0, 0) will be y = x. Normal at the origin will then be y = -x.

appleibeats · Jul 13, 2015

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

I am not sure about how to differentiate 8a from 7F

Differentiate y = a(x-a)(x-b) using the product rule

Is this right:

u = a( x-a) u' = not sure whether it is a or -a (do you need to do chain rule?)
v = (x - b) v' = not sure whether 1 or -1

I seem to get the correct answer without the negatives but am unsure why it is not negative. Don't you have to use chain rule in the derivate??

VBN2470 · Jul 13, 2015

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

Factor out the 'a' term and differentiate via product rule as usual $y'(x)=a[(x-b)+(x-a)]=a[2x-(a+b)]$

rand_althor · Jul 13, 2015

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

appleibeats said:
Is this right:

u = a( x-a) u' = not sure whether it is a or -a (do you need to do chain rule?)
v = (x - b) v' = not sure whether 1 or -1

As VBN2470 said you should factor out the 'a'. Hence you would have: u=(x-a), v=(x-b). Both u' and v' would be 1, as the derivative of x is 1. Both 'a', and 'b' are constants, hence when you derive them with respect to x, they become 0.
If you did u=a(x-a), you would not have to do the chain rule, as the 'a' term is independent of x - it is a constant.

DatAtarLyfe · Jul 13, 2015

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

Every time i go to answer a question, someone else says it before me, FML

appleibeats · Jul 14, 2015

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

What do you mean factor out the a.

Also does it matter that the question has alpa and beta in the brackets?

y = a( x - alpha)(x - beta)

rand_althor · Jul 14, 2015

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

It doesn't matter if it is alpha and beta instead of a and b.

$\begin{align*}y&=a(x-\alpha)(x-\beta) \\u&=x-\alpha\rightarrow u'=x \\v&=x-\beta\rightarrow v'=x \\\frac{\mathrm{d}y}{\mathrm{d}x}&=a\times(v'u+u'v) \\&=a(x-\alpha+x-\beta) \\&=a(2x-\alpha-\beta)\end{align*}$

DatAtarLyfe · Jul 14, 2015

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

appleibeats said:
What do you mean factor out the a.

Also does it matter that the question has alpa and beta in the brackets?

y = a( x - alpha)(x - beta)

rand_althor said:
As VBN2470 said you should factor out the 'a'. Hence you would have: u=(x-a), v=(x-b). Both u' and v' would be 1, as the derivative of x is 1. Both 'a', and 'b' are constants, hence when you derive them with respect to x, they become 0.
If you did u=a(x-a), you would not have to do the chain rule, as the 'a' term is independent of x - it is a constant.

^^

appleibeats · Jul 14, 2015

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

For that same question part b)

Show that the tangents at the x- intercepts have opposite gradients and meet at a point M whose x -coordinates is the average of the x -intercepts.

I can show that the tangents at the two x - intercepts have opposite gradients:

at x = a

gradient = a ( a -b )

at x = b
gradient = -a (a - b)

But am struggling on the second part of the question. I though it has something to do with simultaneous equations??

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