# STEP maths I, II, III 1991 solutions Watch

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#181

(Original post by

I think the problem is, the head could melt slower or faster than the body, which means you

**insparato**)I think the problem is, the head could melt slower or faster than the body, which means you

*could*reach half the initial height without the head or body necessarily being exactly half the respective radius.
When i attempted this question, this is what plagued me... I don't think you can assume that the two spheres are going to amount to half the initial height and be half the initial radii.

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#182

**insparato**)

I think the problem is, the head could melt slower or faster than the body, which means you

*could*reach half the initial height without the head or body necessarily being exactly half the respective radius.

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#183

and

Separating variables

Separating variables on the other one

What do I do with the constants now? They're bugging me.

is the diameter (height) of the upper sphere and is the diameter of the lower sphere. So the height is

So 0.5=h(t) but don't see how this helps. I'm giving up, I have never been able to do rates of change:/

Separating variables

Separating variables on the other one

What do I do with the constants now? They're bugging me.

is the diameter (height) of the upper sphere and is the diameter of the lower sphere. So the height is

So 0.5=h(t) but don't see how this helps. I'm giving up, I have never been able to do rates of change:/

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#184

I'd do something like:

Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.

So

And

So , so for some constant k.

Subbing in for the initial values, we get .

Now find kt when R_u+R_l = 5R/2, etc...

Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.

So

And

So , so for some constant k.

Subbing in for the initial values, we get .

Now find kt when R_u+R_l = 5R/2, etc...

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#188

On the last part of question 15 of STEP II I got instead of . Did anybody else get this? What are the chances they made a typo? I may type up my solution in a bit, but I just typed up 16 and I'm latexed out.

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#189

In my personal opinion, both solutions to STEP III Q1 part b on here fail to find a general term (although this second one is quite nice with two fairly simple general terms), so I am going to post the way I did it, which is quite similar to the above method except that I actually put a general term, rather than a set of general terms. I don't know whether they expect you to do this, but before looking at these I presumed you had to. Anyway, here goes:

It's just that this question would be too easy without this, I think that what they're testing is your ability to think of a way to merge the x

It's just that this question would be too easy without this, I think that what they're testing is your ability to think of a way to merge the x

^{n}and x^{2n}terms, and finding a fairly simple function that returns 1 for even n and 0 for odd n.
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#190

(Original post by

I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...

Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors.

I updated the attachment.

Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.

**Rabite**)I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...

Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors.

I updated the attachment.

Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.

2. No minimum at (1, -4) of any sort.

3. But there is a minimum at (2, -5), which is also necessary for the graph.

4. In the last part, you are defining f

^{-1}(x) so the x we are talking about is a different x to before if that makes sense, ie it is what was on the y-axis before not the x-axis. So your table thing doesn't really make sense.

5. If we ignore point 4, the right-most box is wrong anyway, because the positive root is only valid to the right of 2.

So in the last part you need to consider . The first is undefined, all the others have two possibilities (one of which is the same for all four). It's seriously boring stuff. This question is not particularly challenging, but loads of cases to consider, and loads of room for little errors.

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#191

**STEP III Q16**

Woooooooo! This has taken me almost all day and I've finally done it! I'm almost certain the answer is right as it works for a=2 and a=3, and everything follows nicely.

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#193

**Rabite**)

I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...

Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors.

I updated the attachment.

Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.

For instance for -1<x<0

is undefined

Also, in the interval x<-1

is undefined once x<-5 so the interval should be -5<x<-1

and the inverse functions are undefined for the intervals -1<x<0 and x<-5

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