Well, you can't really integrate that without ending up doing what Brian did.
Let's write v(x) for v, to emphasise that v is a function of x.
Then you have
21v(0)2−21v(d)2=∫0dv(x)ω(v(x)2+V2)dx
So, here's the problem - at this point you have absolutely no idea what v(x) is, which is going to make finding the integral pretty tricky.
So, let's relabel some of the variables: I'm going to write 'x' instead of d, and 't' instead of 'x'.
21v(0)2−21v(x)2=∫0dv(t)ω(v(t)2+V2)dt
Now diff both sides w.r.t. x:
v(x)dxdv=v(x)ω(v(x)2+V2)
Which is Brian's first line. So from here you can solve for v as a function of x, which will give you the desired result.
Obviously it's quicker to just use the v dv/dx form of acceleration to start with.
Thanks, I sort of get what you mean. My thinking was that v is the derivative of x so if we integrated something with v in it we would get something in terms of x.
STEP 1 question 8 A solution has allready been posted using the suggested way but the question let you solve it any other way (apart from newton rapheson et al). This is the way you could do it if , like me, you haven´t done De moivre´s theorem in FP2. (I am using a spanish keyboard which makes typing in latex a nightmare so, for my own ease, I have let theta=x) LHS=cos(4x)=cos(2x+2x)=cos2(2x)−sin2(2x) using the identities: sin2(2x)=1−cos2(2x)[br]cos2x=2cos2x−1 We get: 2(2cos2x−1)2−1=8cos4−8cos2+1 as required.
Next part: cos6x=cos(3x+3x) By standard results: cos3x=4cos3x−3cosx So, we do the same as in the previous part: 2cos23x−1=2(16cos6x−24cos4x+9cos2x)−1[br]=32cos6x−48cos4x+18cos2x−1 as required.
Next part: 16x6−28x4+13x2−1=0 Whenever they give you questions like this, it is worth checking for easy roots. This one has two, x=1 and, noticing that all the powers are even, x=-1. We can then factorise it to get: 16x6−28x4+13x2−1=(x+1)(x−1)(16x4−12x2+1)=0 (16x4−12x2+1)=0 is a quadratic in terms of x^2 so all you have to do is use the quadratic formula and you are done.
Just did the question, and got different answers to you... I diagree with you about the equations of the parabola. For example, the equation for your AHG: y=x(2r−x) does not necessary represent our curve. Although the roots are correct, the equation can be multiplied by a constant a while still holding the two roots to be true. Then we can find the actual value of a using the co-ordinates of the maximum or minimum of the parabola. Your equations just assumes that a=1, right? (correct me if I am wrong...I have just started looking at step)
The final answer I got was, Ap=32(2r−2h)2. This fitted with my answer for the first part, as (2r−2h), is the length HD.
Slighty of topic, but do you think that we would not lose marks for just stating the equations of the parabolas/ having similar "jumps" in our working for STEP? Is the marking really strict, and do we get our papers back?
Sorry but I don't understand a line in your solution to STEP III Q3. In it, you say: "if P=M then Xk=MkX0+Q.O=MkX0⇒Xm=X0 But Mk−1+Mk−2+...+M2+M+I=O when k=m not for any value of k which is what you seem to have assumed. Am I reading it wrongly?
Sorry but I don't understand a line in your solution to STEP III Q3. In it, you say: "if P=M then Xk=MkX0+Q.O=MkX0⇒Xm=X0 But Mk−1+Mk−2+...+M2+M+I=O when k=m not for any value of k which is what you seem to have assumed. Am I reading it wrongly?
I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want.
Is there a slight error in your result for y, I make y=(12t2−24t−29)e−2t−9sinh2t
Here's my solution for Q10, STEP I 1989. Sorry for dodgy presentation, I am practicing using my new graphics tablet I realise that LaTeX would be easier to read
If I have understood this question, I don't like it at all... not a good STEP question. On the other hand I may be wrong in some important way, if so please correct!
Here is my solution for Q11 STEP II 1989. (sorry I called it Q10 in the pic!) I did it because I thought it hadn't been done (because I thought it was Q10, doh!), but anyway you might find my slightly simpler method interesting.
It's a very easy question I think, although it helps a lot to consider relative acceleration.
One last comment: the second part (why the lift comes to rest) I described as "obvious" .. do you see why?
Here's my answer for STEP III 1989 Q12. Note to Brian Everit: I was pleased to see after I did it that we modeled it almost exactly the same way. The only difference is in the end; my reading of the question suggested they didn't want an explicit solution for ϕ but just to deduce from the form of dtdϕ.
But however you look at it, I think you did a great job (That's assuming neither of missed something..)
After comparing with Brian Everit's solution on post 65, I see that we have exactly the same answer. Only thing different is I have a quicker way of doing the first part.
Not the best question ever - asking students to "tell me how you would prove it, but don't prove it" and then asking them to draw 4 different quadratic/quadratic graphs is pretty evil...
After comparing with Brian Everit's solution on post 65, I see that we have exactly the same answer. Only thing different is I have a quicker way of doing the first part.
Not the best question ever - asking students to "tell me how you would prove it, but don't prove it" and then asking them to draw 4 different quadratic/quadratic graphs is pretty evil...
Well done finding that method for the first part. I wish I had thought of it myself.
Well done finding that method for the first part. I wish I had thought of it myself.
Yes, it is clever isn't it, except for the tiny detail that I've just realised it doesn't work! I can't take ln(f(x)) if f(x) is negative! Well, I guess I can if I'm allowing complex answers.
Yes, it is clever isn't it, except for the tiny detail that I've just realised it doesn't work! I can't take ln(f(x)) if f(x) is negative! Well, I guess I can if I'm allowing complex answers.
I don't have the question in front of me, but if you're only interested in the behaviour when |x| is large then f(x) won't be negative then.
I don't have the question in front of me, but if you're only interested in the behaviour when |x| is large then f(x) won't be negative then.
Thanks for the suggestion, it would be OK if it were so, but unfortunately the first part of this monstrously long question asks to prove that f(x)xf′(x)=…, given only that a,b,c,d are real and x not equal to c and d. Only then does it say "... and deduce that if |x| is large..."