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STEP 1 clock geometry / calculus problem

STEP 1 2004 Question 4:

https://pmt.physicsandmathstutor.com/download/Maths/STEP/Papers/2014%20STEP%201.pdf

I'm struggling to see how the first and second derivatives relate to the rate of increase of x here.

My first idea was to try to consider the numerator of the first derivative, dx/d(theta), which is absin(theta), and since ab is constant and positive, deduce that the rate of increase of x must be greatest when sin(theta) is greatest at pi/2. This doesn't lead to the expression sought, but is it otherwise technically correct?

Thanks.
Original post by jameshyland29
STEP 1 2004 Question 4:

https://pmt.physicsandmathstutor.com/download/Maths/STEP/Papers/2014%20STEP%201.pdf

I'm struggling to see how the first and second derivatives relate to the rate of increase of x here.

My first idea was to try to consider the numerator of the first derivative, dx/d(theta), which is absin(theta), and since ab is constant and positive, deduce that the rate of increase of x must be greatest when sin(theta) is greatest at pi/2. This doesn't lead to the expression sought, but is it otherwise technically correct?

Thanks.


What expression do you obtain for dxdθ\dfrac{dx}{d\theta} as a whole?
Original post by jameshyland29
STEP 1 2004 Question 4:

https://pmt.physicsandmathstutor.com/download/Maths/STEP/Papers/2014%20STEP%201.pdf

I'm struggling to see how the first and second derivatives relate to the rate of increase of x here.

My first idea was to try to consider the numerator of the first derivative, dx/d(theta), which is absin(theta), and since ab is constant and positive, deduce that the rate of increase of x must be greatest when sin(theta) is greatest at pi/2. This doesn't lead to the expression sought, but is it otherwise technically correct?

Thanks.

No, it's incorrect. You have to consider the whole fraction, not just the numerator (assuming your denominator isn't constant w.r.t. theta).
Original post by DFranklin
No, it's incorrect. You have to consider the whole fraction, not just the numerator (assuming your denominator isn't constant w.r.t. theta).


Ahh I see (I think!), so I guess that's because of the almost alternating nature of cos and sin -- I mean, if the term in the denominator was also sin(theta) rather than cos(theta), then might it be possible to see at a glance that dx/d(theta) would be a maximum when theta = pi/2?

Thanks.
Original post by jameshyland29
Ahh I see (I think!), so I guess that's because of the almost alternating nature of cos and sin -- I mean, if the term in the denominator was also sin(theta) rather than cos(theta), then might it be possible to see at a glance that dx/d(theta) would be a maximum when theta = pi/2?

Thanks.

No, it's got nothing to do with sin and cos. You can't possibly ignore the denominator when looking for the minimum/maximum value. To see why this can't work, note that, for example, every function f(x)/g(x) can be written as 1g(x)/f(x)\dfrac{1}{g(x) / f(x)}, at which point looking at the denominator (i.e. 1) doesn't tell you anything useful.
Original post by DFranklin
No, it's got nothing to do with sin and cos. You can't possibly ignore the denominator when looking for the minimum/maximum value. To see why this can't work, note that, for example, every function f(x)/g(x) can be written as 1g(x)/f(x)\dfrac{1}{g(x) / f(x)}, at which point looking at the denominator (i.e. 1) doesn't tell you anything useful.


Sorry I wasn't very clear before, I meant could you essentially "ignore" the denominator if the only variable it contains is the same variable as in the numerator, so that both numerator and denominator increase together, for instance:

g(x) = f(x) / (3 + f(x))

so that g(x) is always a maximum when f(x) is a maximum. Would there be any reason to bother taking the second derivative to find the maximum in such a case?

Thanks.
Original post by jameshyland29
Sorry I wasn't very clear before, I meant could you essentially "ignore" the denominator if the only variable it contains is the same variable as in the numerator, so that both numerator and denominator increase together, for instance:

g(x) = f(x) / (3 + f(x))

so that g(x) is always a maximum when f(x) is a maximum. Would there be any reason to bother taking the second derivative to find the maximum in such a case?

Thanks.

*Still* no. E.g. consider f(x)/(1+f(x)^2). Numerator and denominator are both maximized when f(x) is, but the function as a whole is maximized when f(x)=1.

There are scenarios where it does work, but you would need to justify why a particular expression was one of them.
Reply 7
Obviously agree with the comments about having to consider the whole fraction, in general.
Have you managed to solve the question by doing this? Just been through it and its not too bad once you get started.
Original post by DFranklin
*Still* no. E.g. consider f(x)/(1+f(x)^2). Numerator and denominator are both maximized when f(x) is, but the function as a whole is maximized when f(x)=1.

There are scenarios where it does work, but you would need to justify why a particular expression was one of them.


Got it. Thanks.

Original post by mqb2766
Obviously agree with the comments about having to consider the whole fraction, in general.
Have you managed to solve the question by doing this? Just been through it and its not too bad once you get started.


Thanks for the reply. Do you mean have I solved it by considering the whole fraction and determining its maximum value by inspection? If so I just tried it and it seems rather hard but I'm clear about why it's rarely the best idea to attempt such a thing now, which is the main thing! :smile:
Reply 9
Original post by jameshyland29
Got it. Thanks.



Thanks for the reply. Do you mean have I solved it by considering the whole fraction and determining its maximum value by inspection? If so I just tried it and it seems rather hard but I'm clear about why it's rarely the best idea to attempt such a thing now, which is the main thing! :smile:

Its honestly not too bad and there are some nice simplifications in there. If you want a few pointers, just let me know.

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