Year 13 Maths Help Thread

No problem, cheers for looking anyways. Reckon this will be alright? link

I tried that aforementioned STEP question (it's STEP I 2014 Q4) and my 'solution' to the first part currently contains an unjustified statement.

An accurate clock has an hour hand of length $a$ and a minute hand of length $b$ (where $b>a$), both measured from the pivot at the center of the clock face. Let $x$ be the distance between the ends of the hands when the angle between the hands is $\theta$, where $0\leq\theta<\pi$.

Show that the rate of increase of $x$ is greatest when $x=(b-a)^{\frac{1}{2}}$.

The locus of points touched by the hour hand is the circle $C_1$ with radius $a$ and the locus of points touched by the minute hand is the circle $C_2$ with radius $b$. The center of the two circles is $O$

Let us hold the minute hand fixed at a certain point $B$ and vary $\theta$ from ${0}$ to $\pi$.

Let $A$ be the point the hour hand is currently at. A diagram can show that the maximum rate of increase of $x$ occurs when the line joining A to B is a tangent to $C_1$ so that AOB is a right angled triangle, and by Pythagoras' theorem, $x=(b^2-a^2)^{\frac{1}{2}}$.

I think I'm just restating the question in a different form right now, but don't leave me any hints at the moment as I'm trying to work out a breakthrough in the coming days. I will leave this up here in case I'm still stuck after a while.

Hint below if you come back later and need one

I tried the cosine rule and then used implicit differentiation countless times before and it just leads to dead ends (or I get something which resembles a second order differential equation but 1) second order ODEs aren't in the C1-C4 syllabus 2) it's not actually a second order ODE)...

$x^2=a^2+b^2-2ab\cos\theta$.

I'll try another approach after the many failed implicit differentiation attempts: just substituting
$x=(b-a)^{\frac{1}{2}}$ and see what happens.
If $x=(b-a)^{\frac{1}{2}}$ then of course $b^2-a^2= a^2+b^2-2ab\cos\theta$
so $-2a^2=-2ab\cos\theta$ so $a=b\cos\theta$ which doesn't particularly help either aside from the fact that the triangle formed is a right angled triangle and the line connecting the minute and hour hands is a tangent to the circle.

But I have yet to prove why the rate of increase of $x$ is greatest when this happens which is something I have to figure out...

I can gladly type up the second (much easier) half of the question for marking.

P.S. I know this should go in the STEP prep thread but I am a bit of a wimp. I hope that I'll improve at STEP over the coming months after practice.

I tried the cosine rule....

Im stuck on a coordinate geometry Q! !

"The diagram shows a rectangle (all sides are 90 degree) ABCD. The point A is (2,14), B is (-2,8) and C lies on the x-axis. Find (i) The equation of BC (ii) the coordinates of C and D"

I have done (i) and its correct the answer is y=-2/3+20/3 and ive found the coordinates of C for (ii) which are C(10,0) but ive got no clue for coordinates of D.

x

Are you Zacken's sibling ?

Starting with the cosine rule:
$x^2=a^2+b^2-2ab\cos\theta$
Implicitly differentiating both sides:

$2x\frac{dx}{d\theta}=2ab\sin \theta$

Implicitly differentiating again (the greatest rate of increase happens when $\frac{d^2x}{d{\theta}^2}=0$:

[tex\displaystyle]2x\frac{d^2x}{d{\theta}^2}

What resources can i use to revise and selfteach s3

I now know why I kept messing up my implicit differentiation-it was a stupid mistake which I somehow repeated each and every time at the same stage. Usually $x$ is the letter used for the independent variable but here it is the letter used for the dependent variable so when I differentiated $2x\frac{dx}{d\theta}$, I ended up with a wrong expression.
The universities I'm applying to don't require STEP but apparently STEP is a good preparation for a maths degree._{As I see $x^2$ I'm guessing the next stage involves the use of our original expression involving $x^2$ from the cosine rule or we can exploit the derived equation from the cosine rule $\cos \theta=\frac{x^2-a^2-b^2}{2ab}$, though the latter seems too messy. Will therefore try the former:Starting off from before:}

$(\frac{ab(1-{\cos}^2\theta)}{x^2})=\cos\theta$
_Rearranging:$(\frac{ab(1-{\cos}^2\theta)}{\cos\theta})=x^2$
Exploiting the cosine rule at the start: $x^2=a^2+b^2-2ab\cos\theta$

Let's see if we can get a quadratic in terms of $\cos\theta$:
We're getting close!
$(\frac{ab(1-{\cos}^2\theta)}{\cos\theta})=a^2+b^2-2ab\cos\theta$

Multiplying by $\cos\theta$:
$ab-ab{\cos}^2\theta=a^2\cos\theta+b^2\cos\theta-2ab{\cos}^2\theta$

We're getting closer...

Rearranging gives us:$ab{\cos}^2\theta-(a^2+b^2)\cos\theta+ab=0$

And by means of the quadratic formula:$=\frac{a^2+b^2 \pm (a^2-b^2)}{2ab}$. Our values in terms of $\theta$ are $\frac{b^2}{ab}$ and $\frac{a^2}{ab}$ and when these values are inserted into the cosine rule, we get $x^2=a^2+b^2-2a^2=b^2-a^2$ and $x^2=a^2+b^2-2b^2$. As $b>a$ and $x^2>0$, only the former equation makes sense. So $x=(b^2-a^2)^{\frac{1}{2}}$.J'ai fini.

Great!
That's an interesting way to do that final bit ^.^
Slightly quicker to solve the quadratic for cos(y) (abcos^2(y) + x^2cos(y) - ab = 0) then set it equal to what the cosine rules gives, but your way does work!

I find this question rather long by the standards of the STEP I questions I've done. It's still going to be a while until I am brave enough to join the STEP Prep thread though. I might try STEP I Q5 or Q6 2014 and then move on to some 1990s STEP I questions.

I've not done school level maths for many years but do have query with regard to the notation.

Suppose f is a function of x, then the first differential is written f'. The second diff is written f'' or f(ii).

So shouldn't the sixth diff be written f(vi)?

*

I've always seen f^6(x) - never seen your roman numeral version

I find this question rather long by the standards of the STEP I questions I've done. It's still going to be a while until I am brave enough to join the STEP Prep thread though. I might try STEP I Q5 or Q6 2014 and then move on to some 1990s STEP I questions.