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Need help, URGENT maths question: (2 Viewers)

leehuan

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Hey leehuan. Yes, I am doing MATH1151. Do you mean the red booklet? I can't find the answers in the purple booklet anywhere....
The red one has answers before the past quiz papers to some selected problems. That was one of the few proofs it chose to give.
 

InteGrand

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Thanks. There is this question that's been bugging me for a while now.


I get part (a). The answer is . But the answer to part (b) that I get is 166.67 seconds.
For part b), we are given dV/dt in terms of h. We can also find dV/dh in terms of h using geometry. So we can use the chain rule to get dh/dt as a function of h, f(h). So dt will be dh/f(h). So you can find the required time by integrating the RHS from h = 0 to 100.
 

1008

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The red one has answers before the past quiz papers to some selected problems. That was one of the few proofs it chose to give.
Thanks, found it. I was referring to the purple booklet. But that is literally the only proof they've given in the answers.

BTW any thoughts on the cone question?
 

InteGrand

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Thanks, found it. I was referring to the purple booklet. But that is literally the only proof they've given in the answers.

BTW any thoughts on the cone question?
I put up a method above. Is that the method you were using? If not, what did you do?
 

1008

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I put up a method above. Is that the method you were using? If not, what did you do?
Sorry InteGrand, I was replying to leehuan's question. And after I posted it, I saw your reply. I was actually using a different (and incorrect method). I've put it below:



But I can see why it's incorrect now. This was one of various approaches that got me different answers.
 
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1008

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Hey everyone, I've got another question.

I get the first two parts of the question, just need help with the third part.
 
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InteGrand

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Hey everyone, I've got another question.

I get the first two parts of the question, just need help with the third part.
Note that p4'(x) = p3(x) for all x. Also, p4"(x) = p2(x) for all x. We've shown that p2(x) is a positive definite quadratic, so that means the second derivative of p4 is always positive. This means the minimum of p4 will be a global min.

Now, to find this min., set p4'(x) = p3(x) = 0. (There is exactly one solution to this, as you've shown in a previous part.)

But observe that p4(x) = p3(x) + x4/4!. So at the point where p4'(x) ≡ p3(x) = 0, say at x = a, we have p4(a) = 0 + a4/4! > 0 (strictly positive since clearly a is not zero, because p3 doesn't have its root at 0).

So the absolute minimum value of p4(x) is positive. Hence p4(x) > 0 for all real x.
 

1008

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Note that p4'(x) = p3(x) for all x. Also, p4"(x) = p2(x) for all x. We've shown that p2(x) is a positive definite quadratic, so that means the second derivative of p4 is always positive. This means the minimum of p4 will be a global min.

Now, to find this min., set p4'(x) = p3(x) = 0. (There is exactly one solution to this, as you've shown in a previous part.)

But observe that p4(x) = p3(x) + x4/4!. So at the point where p4'(x) ≡ p3(x) = 0, say at x = a, we have p4(a) = 0 + a4/4! > 0 (strictly positive since clearly a is not zero, because p3 doesn't have its root at 0).

So the absolute minimum value of p4(x) is positive. Hence p4(x) > 0 for all real x.
The way I did it was I just chose an arbitrary interval [-25,0] and just went from there. Here's my working:


So how would your explanation to part iii continue on from my working for the first two parts?
 
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leehuan

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The way I did it was I just chose an arbitrary interval [-25,0] and just went from there. Here's my working:
Thought you only needed part iii anyway?

Also, I think you ment no real zero to p2(x) or no real solution to p2(x)=0
 

1008

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Yep, so I was just wondering how would InteGrand's explanation to part iii continue on from my working for the first two parts?

EDIT: Also, @leehuan, you're right, I mean there's no real solution to P2(x) = 0
 

InteGrand

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Yep, so I was just wondering how would your explanation to part iii continue on from my working for the first two parts?
I used some stuff from your first two parts, like the fact that the second derivative is always positive.
 

1008

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Also, how would you do this one. I could only figure out one of the solutions.

 

InteGrand

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Also, how would you do this one. I could only figure out one of the solutions.

We can parameterise a point in the ellipse like x = t, y = sqrt(1 - (t^2/4)), for -2 <= t <= 2. (This only gets the top half of the ellipse but by symmetry, the distance will be same as considering the whole ellipse). We can then consider the squared distance of the point (a, 0) to a point on the ellipse (t, sqrt(1 - (t^2/4))), -2 <= t <= 2. We can then use calculus to minimise and maximise this. When we do the minimum, for some values of a, the answer is of a different shape than for other values of a. This is because for after some value of a, the actual point in the ellipse corresponding to the minimal distance with the other values of a would become a point with t having to be greater than 2, which means the point isn't on the ellipse (and the y-value would be imaginary). So a slightly different form of answer will be needed. The whole Q. is a bit tedious to write out now, so I've just given a method for now.
 
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1008

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We can parameterise a point in the ellipse like x = t, y = sqrt(1 - (t^2/4)), for -2 <= t <= 2. (This only gets the top half of the ellipse but by symmetry, the distance will be same as considering the whole ellipse). We can then consider the squared distance of the point (a, 0) to a point on the ellipse (t, sqrt(1 - (t^2/4))), -2 <= t <= 2. We can then use calculus to minimise and maximise this. When we do the minimum, for some values of a, the answer is of a different shape than for other values of a. This is because for after some value of a, the actual point in the ellipse corresponding to the minimal distance with the other values of a would become a point with t having to be greater than 2, which means the point isn't on the ellipse (and the y-value would be imaginary). So a slightly different form of answer will be needed. The whole Q. is a bit tedious to write out now, so I've just given a method for now.
Thanks! Considering you gave a response at 3 in the morning!

But yeah, there are like 3 answers to the question. Here they are:
.

I could only get (1-a^2/3) by considering
 
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1008

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Just another question:



Is this right, or am I missing something
 

1008

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Also, how would you do this one. I could only figure out one of the solutions.

Thanks! Considering you gave a response at 3 in the morning!

But yeah, there are like 3 answers to the question. Here they are:
.

I could only get (1-a^2/3) by considering
Thanks! If you don't mind, could you also give a proof for this one InteGrand?
 

InteGrand

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Thanks! If you don't mind, could you also give a proof for this one InteGrand?
So the ellipse condition is that the point P on the (upper-half of the) ellipse has .

The squared distance as a function of x is then .

This can be differentiated to help find the minimum:



.

So the minimal squared distance is

From simplifying this, you get the top part of the answer you posted.

To see the reason for the split in answers for different values in a, note that (since it's under a square root). Substituting yields in order for the top half of the answer to be feasible. In other words, we need -2 ≤ x ≤ 2 for the point to be on the ellipse and make sense, but if a > 3/2, this no longer happens, as x becomes bigger than 2.

(For , the point on the ellipse we worked out as being closest, namely , is no longer a real point, which is why we need the condition on a.)

To see why using when is optimal, note that the squared distance function represents a concave up parabola. Therefore, it is minimised at its vertex, and as we saw before, this globally minimising x-value is as . So the parabola is minimised at a point xvert. where xvert. > 2, so the closest we can get to this minimal value by taking a point on the ellipse (where x needs to be less than 2) is to make x closest to xvert.. To make x closest to xvert. and have x be on the ellipse, we need to take x = 2, as xvert. > 2. This point (the point (2,0)) on the ellipse has distance to the point (a, 0) of as required.

(So basically I'm saying that if a quadratic is minimised at an x-value of xvert., if we are restricted to x-values less than some maximal allowable value X, then the minimal attainable value for the quadratic is its value at this X (because the quadratic will be monotone decreasing on (-∞, X], as X < xvert.). In our case, the maximal allowable value of X is X = 2 (this constraint coming from the ellipse). So taking x = 2 gives us the place of the min.)

Edit: Realised I didn't prove the maximal distance part. But this part is easier than the minimal part, so I'll leave it as an exercise for the reader. :)
 
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