STEP Prep Thread 2015

If you didn't write on your UCAS that you're taking STEP/AEA, do you have to update it or something if end up taking it?? or would you just send a copy of your results to your uni on results day???

Yeah it's included in one of my offers


Posted from TSR Mobile

Can anyone give me any hints for the last part of Q6 III 1995, please?

Here is what I've done so far:

Hence to meet the condition
$\frac{cos(\theta)}{r}=-1 \Leftrightarrow r=-cos(\theta)$
So rewrite $w=-cos(\theta) e^{i\theta}$

So, we clearly want points whose modulus are |w|=|-cos(\theta)|, thinking about how this graph looks in polar coordinates I arrive at the following:

So, if we use the locus $|w+0.5| = 0.25$ we get the circle radius 0.5 centred at (-0.5, 0).

Consider any point on the radius of that circle. Join a triangle from it to the origin and the centre of the circle. We get the following triangle (shown in picture)
Hence by the cosine rule $|w|=cos(\theta)$ which agrees with the above.
Noting that for $|w+0.5| = 0.25$ to hold $\pi /2 \leq \theta \leq \pi \ or \ - \pi < \theta \leq - \pi/2$ Hence if we take the modulus of $|\frac{1}{w}| = |\frac{1}{cos(\theta)}|$ and use the argument of $\frac{1}{w} = -\theta$ we see $w=|\frac{1}{cos(\theta)}| (cos(\theta) + isin(\theta)= -1 + isin(\theta)$.

I think a simpler method is just to consider 1/w = -1 + iy, but your method seems fine

I think a simpler method is just to consider 1/w = -1 + iy, but your method seems fine

As soon as he posted it, thats what I thought of, but I got to

w=-1/(1+y^2)-y/(1+y^2)i

how can you show that this is a circle centre (-0.5,0) and radius 0.5?

Let u = real part and let v = imaginary part. Then eliminate y

On the topic of solving parametric equations, something has been bugging me ever since we did them in C4. How come sometimes we can just square equations without losing information?
The most common parametric equation you see is the circle one, x=cost, y=sint, and the way we are taught to eliminate t is by squaring both equations and adding to give x^2+y^2=1, which is a unit circle as we expect. But why does this method work even though we are inventing new solutions in the act of squaring?

Squaring is okay. It's square rooting that causes a problem.

In theory, you should be able to solve equations by squaring, providing that all your variables are positive; otherwise you would need to use modulus signs, which can get messy.

hey guys, are you planning on to tackle STEP stats question this summer??

can someone pls explain STEP 1 1999 Qu 8 the last part, I've looked at the solutions and it looks like a few vague jumps are made

Yes I understand that you can square when variables are all positive like when calculating lengths in a mechanics/geometry problem. But in parametrics x and y take both positive and negative values, as do sines and cosines.

You're not "inventing new solutions" because you're not "solving" anything in the first place!

If x = cos t and y = sin t, then it's a fact of life that x^2 + y^2 = 1 because that follows from an identity.

What you can't do is start from x^2 + y^2 = 1 and write y = (some function of x) because typically for each x value there will be 2 values of y on the circle so you don't have a function that returns a single y value.

You're not "inventing new solutions" because you're not "solving" anything in the first place!

If x = cos t and y = sin t, then it's a fact of life that x^2 + y^2 = 1 because that follows from an identity.

What you can't do is start from x^2 + y^2 = 1 and write y = (some function of x) because typically for each x value there will be 2 values of y on the circle so you don't have a function that returns a single y value.