HSC 2016 MX1 Marathon (archive) (1 Viewer)

[hedgehog_7]

Re: HSC 2016 3U Marathon

This is not the difference between the two functions you gave.

The two position functions were:

x_A(t) = 4te^(-t) and x_B(t) = -4(t^2)e^(-t).

The squared-difference is: s(t) := (x_A(t) – x_B(t))².

Try maximising that. (The distance isn't simply xA – xB, it is |xA – xB|. Since absolute values are less nice to work with than squared-distance, and optimising the squared-distance will be equivalent to optimised the distance, it is easier to work with the squared-distance instead.)

after differentiating xA - xB i ended up with 4(e^-t) (-t^2 + t +1). why does the absolute and the one without the absolute affect the answer? im just wondering

 

[KingOfActing]

Re: HSC 2016 3U Marathon

after differentiating xA - xB i ended up with 4(e^-t) (-t^2 + t +1). why does the absolute and the one without the absolute affect the answer? im just wondering

Distance is a positive measure

The distance between two things is |a-b|

 

[hedgehog_7]

Re: HSC 2016 3U Marathon

Distance is a positive measure

The distance between two things is |a-b|

so whats the difference with just a-b if i was to differentiate it anyways

 

[InteGrand]

Re: HSC 2016 3U Marathon

so whats the difference with just a-b if i was to differentiate it anyways

If you just used xa – xb, in general, maximising this would give us the maximum displacement of A relative to B. So it'd give us the maximum displacement A is ever right of B. So say A at most 1 metre right of B, but at some other time, A was 100 m left of B. Using just the difference rather than the absolute difference, the maximum you'd pick up is the 1m, since this is the max displacement. You'd miss out on the 100 m as the true max. You'd pick it up as a min. though. So basically, if you use xa – xb instead, you just need to find all the extrema candidates and take the one with the biggest absolute value.

(Of course in this specific question, xa – xb is the actual distance as explained in my previous post.)

 

[hedgehog_7]

Re: HSC 2016 3U Marathon

This is not the difference between the two functions you gave.

The two position functions were:

x_A(t) = 4te^(-t) and x_B(t) = -4(t^2)e^(-t).

The squared-difference is: s(t) := (x_A(t) – x_B(t))².

Try maximising that. (The distance isn't simply xA – xB, it is |xA – xB|. Since absolute values are less nice to work with than squared-distance, and optimising the squared-distance will be equivalent to optimised the distance, it is easier to work with the squared-distance instead.)

Actually, in this case, it is easy to see that xA is always greater than or equal to xA is non-negative always, whilst xB is non-positive always. So in this case, the distance actually is xA – xB, and we can just optimise this. What I said above with the squared-distance stuff was for general distance functions, which may change which one is greater than the other over time, in which case the squared-distance trick is handy.

So for this one, we need to optimise the function

f(t) := x_A(t) – x_B(t)

= 4te^(-t) – [-4(t^2)e^(-t)]

= 4te^(-t) [1+t].

I think you posted the distance functions wrong the first time?

agreed but if i was to differentiate that function id end up with 4(e^-t) (-t^2 + t +1) to which i found the maximum value for t is t= (-1+root5)/2 and not what the answer says as t = (1+root 5)/2

 

[hedgehog_7]

Re: HSC 2016 3U Marathon

This is not the difference between the two functions you gave.

The two position functions were:

x_A(t) = 4te^(-t) and x_B(t) = -4(t^2)e^(-t).

The squared-difference is: s(t) := (x_A(t) – x_B(t))².

Try maximising that. (The distance isn't simply xA – xB, it is |xA – xB|. Since absolute values are less nice to work with than squared-distance, and optimising the squared-distance will be equivalent to optimised the distance, it is easier to work with the squared-distance instead.)

Actually, in this case, it is easy to see that xA is always greater than or equal to xA is non-negative always, whilst xB is non-positive always. So in this case, the distance actually is xA – xB, and we can just optimise this. What I said above with the squared-distance stuff was for general distance functions, which may change which one is greater than the other over time, in which case the squared-distance trick is handy.

So for this one, we need to optimise the function

f(t) := x_A(t) – x_B(t)

= 4te^(-t) – [-4(t^2)e^(-t)]

= 4te^(-t) [1+t].

I think you posted the distance functions wrong the first time?

agreed but if i was to differentiate that function id end up with 4(e^-t) (-t^2 + t +1) to which i found the maximum value for t is t= (-1+root5)/2 and not what the answer says as t = (1+root 5)/2

 

[InteGrand]

Re: HSC 2016 3U Marathon

agreed but if i was to differentiate that function id end up with 4(e^-t) (-t^2 + t +1) to which i found the maximum value for t is t= (-1+root5)/2 and not what the answer says as t = (1+root 5)/2

Oh. I asked what was the function being optimised rather than the derivative you obtained (I didn't bother differentiating it). But assuming you differentiated correctly, the derivative is 0 iff -t^2 + t +1 = 0, i.e. t^2 - t - 1 = 0. This is the famous defining quadratic of the golden ratio, so the optimum occurs at t = phi, where phi = (1+sqrt(5))/2 is the golden ratio. (This is obtained just by using the quadratic formula or completing the square etc.; there's nothing important for this Q. about it being the golden ratio.)

So I think you just made a silly mistake when solving the quadratic.

 

[Paradoxica]

Re: HSC 2016 3U Marathon

agreed but if i was to differentiate that function id end up with 4(e^-t) (-t^2 + t +1) to which i found the maximum value for t is t= (-1+root5)/2 and not what the answer says as t = (1+root 5)/2

Hmm... double check your derivative.

 

[hedgehog_7]

Re: HSC 2016 3U Marathon

Oh. I asked what was the function being optimised rather than the derivative you obtained (I didn't bother differentiating it). But assuming you differentiated correctly, the derivative is 0 iff -t^2 + t +1 = 0, i.e. t^2 - t - 1 = 0. This is the famous defining quadratic of the golden ratio, so the optimum occurs at t = phi, where phi = (1+sqrt(5))/2 is the golden ratio. (This is obtained just by using the quadratic formula or completing the square etc.; there's nothing important for this Q. about it being the golden ratio.)

So I think you just made a silly mistake when solving the quadratic.

the problem is that when i use quadratic for -t^2 + t +1 = 0 i get t = (-1 + sqrt (5))/2) and not t = (1+ sqrt (5))/2)

 

[hedgehog_7]

Re: HSC 2016 3U Marathon

FUCK I SEE IT .............................................................. alright thanks a lot xddd

 

[KINGOM 885]

Re: HSC 2016 3U Marathon

I have a question.
A body of 200 degrees is put inside a room with temperature 5 degrees. 25 minutes later, the body is 50 degrees. The cooling stops at 37 minutes. If the cooling follows Newton's Law of Cooling, what is the temperature at 38 minutes

 

[InteGrand]

Re: HSC 2016 3U Marathon

I have a question.
A body of 200 degrees is put inside a room with temperature 5 degrees. 25 minutes later, the body is 50 degrees. The cooling stops at 37 minutes. If the cooling follows Newton's Law of Cooling, what is the temperature at 38 minutes

Spoiler

 

[davidgoes4wce]

Re: HSC 2016 3U Marathon

I have a question.
A body of 200 degrees is put inside a room with temperature 5 degrees. 25 minutes later, the body is 50 degrees. The cooling stops at 37 minutes. If the cooling follows Newton's Law of Cooling, what is the temperature at 38 minutes

There is a horizontal asymptote as T=5 , as the temperature never goes below 5 degrees. I have shown it in a diagram above.

Would be good if someone could confirm.

 

[KINGOM 885]

Re: HSC 2016 3U Marathon

There is a horizontal asymptote as T=5 , as the temperature never goes below 5 degrees. I have shown it in a diagram above.

Would be good if someone could confirm.

But the cooling stops at 37 minutes. Wouldn't you have to calculate something extra instead of just substituting for t?

 

[davidgoes4wce]

Re: HSC 2016 3U Marathon

But the cooling stops at 37 minutes. Wouldn't you have to calculate something extra instead of just substituting for t?

Do you happen to have an answer for it? My thinking is you have to work the temperature at t=37 minutes then? I must admit I've done a few of these heating and cooling questions but never seen a question worded like that. (I missed the stop cooling at 37 minutes part) Maybe there was a typo in the question?

 

[davidgoes4wce]

Re: HSC 2016 3U Marathon

What happens between he 37th and 38th minute will affect the outcome in the final answer. Starting to think of it, I reckon it was a question written by a teacher? And they may have accidentally took mismatched the minutes 37 and 38.

 

[leehuan]

Re: HSC 2016 3U Marathon

I have a question.
A body of 200 degrees is put inside a room with temperature 5 degrees. 25 minutes later, the body is 50 degrees. The cooling stops at 37 minutes. If the cooling follows Newton's Law of Cooling, what is the temperature at 38 minutes

We know that T = 5 + Ae^(-kt)

The body is initially at 200deg C
200 = 5 + Ae^0
195 = A

T = 5 + 195e^(-kt)

When t=25, T=50
50 = 5 + 195e^(-25k)
3/13 = e^(-25k)
k = -1/25 ln(13/3)

Note: 3/13 = 45/195

T = 5 + 195e^(-t/25 ln(13/3))

When t=37

T = 5 + 195e^(-37/25 ln(13/3))
T = 27.26...
______________

If the cooling stops at t=37, then that deadlocks the temperature from then on. Because we are not told if the substance gets moved into a new environment (which is ridiculously stupid). we assume that the surrounding temperature must be equal to that of the body after 37mins.

If otherwise, give us the answer.

 

[KINGOM 885]

Re: HSC 2016 3U Marathon

We know that T = 5 + Ae^(-kt)

The body is initially at 200deg C
200 = 5 + Ae^0
195 = A

T = 5 + 195e^(-kt)

When t=25, T=50
50 = 5 + 195e^(-25k)
3/13 = e^(-25k)
k = -1/25 ln(13/3)

Note: 3/13 = 45/195

T = 5 + 195e^(-t/25 ln(13/3))

When t=37

T = 5 + 195e^(-37/25 ln(13/3))
T = 27.26...
______________

If the cooling stops at t=37, then that deadlocks the temperature from then on. Because we are not told if the substance gets moved into a new environment (which is ridiculously stupid). we assume that the surrounding temperature must be equal to that of the body after 37mins.

If otherwise, give us the answer.

I don't have an answer, but you should be correct. Logically, the temperature will increase after t=37, but as the question doesn't give any further information, i think we have to assume that the temperature is deadlocked.
Also, is Wilhelmy's law still tested in the HSC. It was in one of my old school trials, but was never tested after.
Thanks
PS: Integrand rules

 

[Paradoxica]

Re: HSC 2016 3U Marathon

 

[wu345]

Re: HSC 2016 3U Marathon

is it meant to be 1+2sinxcosx in the denominator?