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

DatAtarLyfe · Jul 9, 2015

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

Drongoski said:
I take it to mean: x^2 = (28/5)y

In that case simply express in the form: x² = 4ay

For this question: x^2 = 4* (7/5)*y

so, the focal length "a" is 7/5.

Lol, i was right but deleted my post cause i was like "wait wtf, it's 28/5y"

appleibeats · Jul 9, 2015

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

i though the focus length was the distance from the vertex and the focus.

I got the focus was (0,7/5)

and the vertex ( 14/5, 7/5)

so I got the focus length = 14/5

Where did i go wrong?

appleibeats · Jul 9, 2015

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

DatAtarLyfe said:
Lol, i was right but deleted my post cause i was like "wait wtf, it's 28/5y"

sorry about not making it clear

VBN2470 · Jul 9, 2015

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

For $x^2=\dfrac{28}{5}y$ , vertex is $(0,0)$ , focus is $(0, \dfrac{7}{5})$ , hence focal length is $\dfrac{7}{5}$ .

DatAtarLyfe · Jul 9, 2015

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

appleibeats said:
i though the focus length was the distance from the vertex and the focus.

I got the focus was (0,7/5)

and the vertex ( 14/5, 7/5)

so I got the focus length = 14/5

Where did i go wrong?

Your vertex is incorrect. If you look at your original equation, x^2=28/5y, your vertex is actually (0,0)
To determine the vertex from your equation, you use the standard equation (x-h)^2=4a(y-k), where (h,k) is your vertex. So in your equation, h=0, k=0 and 4a=28/5
Hence your vertex is (0,0), your focal length is 7/5 and thus your focus is (0,7/5)

Drongoski · Jul 9, 2015

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

appleibeats said:
i though the focus length was the distance from the vertex and the focus.

I got the focus was (0,7/5)

and the vertex ( 14/5, 7/5)

so I got the focus length = 14/5

Where did i go wrong?

The focal length is indeed the distance from the vertex to the focus.

DatAtarLyfe · Jul 9, 2015

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

appleibeats said:
i though the focus length was the distance from the vertex and the focus.

I got the focus was (0,7/5)

and the vertex ( 14/5, 7/5)

so I got the focus length = 14/5

Where did i go wrong?

Also, when you have an equation where it is x^2, then the focus is above/below (depending on concavity) the vertex in the Y-DIRECTION. Therefore, the x-oordinate of the vertex and focus will be the same, only the y-oordinates will differ by the magnitude of the focal length.

appleibeats · Jul 9, 2015

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

For y = x^2/7

I know it is a cusp. But why does the derivative have a discontinuity at x=0 instead of a zero

y' = 2/7 x^-5/7

doesn't the derivative equal zero when x = 0

InteGrand · Jul 9, 2015

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

appleibeats said:
y' = 2/7 x^-5/7

doesn't the derivative equal zero when x = 0

$$No, because $x^{-\frac{5}{7}}=\frac{1}{x^{\frac{5}{7}}}$, which is undefined at $x$=0 $(\frac{1}{x^{\frac{5}{7}}}\to \pm \infty$ as $x \to 0^{\pm}$).$

Speed6 · Jul 9, 2015

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

appleibeats said:
For y = x^2/7

I know it is a cusp. But why does the derivative have a discontinuity at x=0 instead of a zero

y' = 2/7 x^-5/7

doesn't the derivative equal zero when x = 0

That is a silly mistake that can be avoidable.

appleibeats · Jul 9, 2015

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

For y = (1 - x ) ^1/4 - 2

the domain is x less than or equal to 1

but y' = 1 / 4(1 - x) ^3/4

so at x = 1

the derivative is undefined. What is the significance of this and do you include the point in the graph?

VBN2470 · Jul 9, 2015

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

For $y=(1-x)^{\frac{1}{4}}-2$ , y(1) = -2 so you certainly include the point (1,-2) when sketching y = f(x). Derivative is undefined at x = 1 (hence not differentiable at that point), so the graph of y = f(x) at x = 1 will contain some sort of 'cusp' or 'sharp' point (i.e. the slope of the graph of f is undefined). You can check this graphically.

DatAtarLyfe · Jul 9, 2015

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

VBN2470 said:
For $y=(1-x)^{\frac{1}{4}}-2$ , y(1) = 0 so you certainly include the point (1,0) when sketching y = f(x). Derivative is undefined at x = 1 (hence not differentiable at that point), so the graph of y = f(x) at x = 1 will contain some sort of 'cusp' or 'sharp' point (i.e. the slope of the graph of f is undefined). You can check this graphically.

This is the critical point, right?

VBN2470 · Jul 9, 2015

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

DatAtarLyfe said:
This is the critical point, right?

Yes.

appleibeats · Jul 9, 2015

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

Using dy/dx = dy/dt / dx/dt , find any stationary points and sketch the graphs of the functions:

a) x = 6t , y = 3t^2

dy/dx = 6t /6 = t

so dy/dx = 0 when t = 0

Is that right and if so how do I go about graphing the function now?

VBN2470 · Jul 9, 2015

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

So you found that $\dfrac{\text{ d}y}{\text{ d}x}=0$ when t = 0 , which means the derivative is 0 at the point $(x,y) = (6(0), 3(0)^{2}) = (0,0)$ . To sketch the graph of the curve defined by those parametric equations, we eliminate the parameter t to get:
$t = \dfrac{x}{6} \Rightarrow y = 3(\dfrac{x}{6})^{2} \Rightarrow x^2=12y$ Or you could square x first and it will make it bit easier maybe. You will end up having to sketch the standard parabola $x^2=12y$ .

appleibeats · Jul 10, 2015

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

A parabola has Cartesian equation x^2 = - 8/3 y

I have found that the focal length = -2/3

How doe I now find the parametric equations of the parabola?

And hence find the equation of the chord PQ given that the point P has parameter 2 and the point Q has parameter -1. ( Assuming the equation of a chord)

I assume using: y = ( p + q / 2)x - apq

VBN2470 · Jul 10, 2015

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

Remember, focal length is always positive, so it should be $a=\dfrac{2}{3}$ , since your parabola is of the form $x^2=-4ay$ (concave down). Parametric equation of the parabola will given by $x=2at, y=-at^2$ , where a is a positive constant and t is your parameter. So your parametric equations will be $x=\dfrac{4}{3}t, y=-\dfrac{2}{3}t^2$ . The endpoints of the chord will be given by $P(\frac{8}{3}, -\frac{8}{3}), Q(-\frac{4}{3}, -\frac{2}{3})$ (substituting t = 2 and t = -1 respectively). Then, you can find the Cartesian equation of the chord using standard coordinate geometry. Equation of the chord should be 3x + 6y + 8 = 0.

appleibeats · Jul 10, 2015

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

A parabola has parametric equations x = -8t - 7 , y = -4t^2 + 2

i) state the focal length
ii) Without finding the Cartesian equation, write down the coordinates of the vertex, and the equation of the directrix

I began by eliminating the parameter t :

t = x + 7 / -8

y = -4( x + 7 / -8 )^2 + 2

But isn't that find the Cartesian equation? Is there an easier way?

InteGrand · Jul 10, 2015

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

appleibeats said:
A parabola has parametric equations x = -8t - 7 , y = -4t^2 + 2

i) state the focal length
ii) Without finding the Cartesian equation, write down the coordinates of the vertex, and the equation of the directrix

I began by eliminating the parameter t :

t = x + 7 / -8

y = -4( x + 7 / -8 )^2 + 2

But isn't that find the Cartesian equation? Is there an easier way?

$Rearrange the parametric equations into the form $x-h=2at$, $y-k=at^2$ (this is the standard parametrisation of a parabola with a vertical axis of symmetry, focal length $|a|$ and vertex $(h,k)$.)$

$Doing this, we have $x-(-7)=-8t$, $y-2=-4t^2$. So we have $|a|=4$, so the focal length is 4 units. Also, we have $h=-7,k=2$ the vertex is $(-7,2)$. The directrix is $|a|=4$ units above the vertex (since the parabola here is downwards opening as $a=-4<0$), so the directrix is the line $y=6$.$

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