Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

https://www.nature.com/articles/s41586-022-05172-4

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

So if it doesn’t travel on land, that means x = 20, so it’s just a matter of substitution. 10.44 seconds, roughly.

Second part is just year 10 calculus. Differentiate wrt x, find where it equals zero.

T’(x) = 5x / ( sqrt(36+x^2)) – 4 = 0
5x = 4 ( sqrt(36+x^2))
25 x ^2 = 16 (36 + x^2)
9 x^2 = 576
x = 8

sub back in

T(8) = 98

So time taken is 9.8 s

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

what’s the equivalent subject in Australian schools

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

x = 8
T = 98 tenths sec.

The Rev Dodgson said:

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

x = 8
T = 98 tenths sec.

I suspect dv wouldn’t approve of my method of solution.

The Rev Dodgson said:

The Rev Dodgson said:

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

x = 8
T = 98 tenths sec.

I suspect dv wouldn’t approve of my method of solution.

Mostly because of how slow it was…

dv said:

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

So if it doesn’t travel on land, that means x = 20, so it’s just a matter of substitution. 10.44 seconds, roughly.

Second part is just year 10 calculus. Differentiate wrt x, find where it equals zero.

T’(x) = 5x / ( sqrt(36+x^2)) – 4 = 0
5x = 4 ( sqrt(36+x^2))
25 x ^2 = 16 (36 + x^2)
9 x^2 = 576
x = 8

sub back in

T(8) = 98

So time taken is 9.8 s

(a)(i): it doesn’t travel on land: x=20, T(20)=10.4s;
(a)(ii): it only travels on land: x=0, T(0)=11.0s

(b): Between these two extremes is the value of x that minimises the time taken. (my emphasis)
9.8 is not between 10.4 and 11.0.

btm said:

dv said:

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

So if it doesn’t travel on land, that means x = 20, so it’s just a matter of substitution. 10.44 seconds, roughly.

Second part is just year 10 calculus. Differentiate wrt x, find where it equals zero.

T’(x) = 5x / ( sqrt(36+x^2)) – 4 = 0
5x = 4 ( sqrt(36+x^2))
25 x ^2 = 16 (36 + x^2)
9 x^2 = 576
x = 8

sub back in

T(8) = 98

So time taken is 9.8 s

(a)(i): it doesn’t travel on land: x=20, T(20)=10.4s;
(a)(ii): it only travels on land: x=0, T(0)=11.0s

(b): Between these two extremes is the value of x that minimises the time taken. (my emphasis)
9.8 is not between 10.4 and 11.0.

A close reading shows a i and a ii have the same value.

btm said:

(b): Between these two extremes is the value of x that minimises the time taken. (my emphasis)
9.8 is not between 10.4 and 11.0.

well that doesn’t seem fair, how do you get an extremum that isn’t extreme

btm said:

dv said:

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

So if it doesn’t travel on land, that means x = 20, so it’s just a matter of substitution. 10.44 seconds, roughly.

Second part is just year 10 calculus. Differentiate wrt x, find where it equals zero.

T’(x) = 5x / ( sqrt(36+x^2)) – 4 = 0
5x = 4 ( sqrt(36+x^2))
25 x ^2 = 16 (36 + x^2)
9 x^2 = 576
x = 8

sub back in

T(8) = 98

So time taken is 9.8 s

(a)(i): it doesn’t travel on land: x=20, T(20)=10.4s;
(a)(ii): it only travels on land: x=0, T(0)=11.0s

(b): Between these two extremes is the value of x that minimises the time taken. (my emphasis)
9.8 is not between 10.4 and 11.0.

8 is between 0 and 20 though.

dv said:

The Rev Dodgson said:

The Rev Dodgson said:

x = 8
T = 98 tenths sec.

I suspect dv wouldn’t approve of my method of solution.

Mostly because of how slow it was…

It was pretty quick actually.

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

It’s probably not that surprising to me as to how quickly the skills & knowledge needed to solve that fade when not used. Well, with me at least.
I can answer A1, but not the remaining two questions. Not the slightest idea. :(

Yeah I would say what they mean is “between the extremes of swimming all the way and swimming the minimum”.

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

Seems to be quite straightforward to me, and much easier than some questions I faced in my Higher School Certificate.

KJW said:

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

Seems to be quite straightforward to me, and much easier than some questions I faced in my Higher School Certificate.

Agree. Part a is trivial.

Part b, draw the graph of the function.

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

I have been giving this lengthy further thought, and have finally come up with a solution that requires zero knowledge of calculus, and is in retrospect bleedin obvious.

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

So, for minimum travel time sin(A) = 0.8 so A = asin(0.8)

x = 6 * tan(asin(0.8)) = 8

But if we don’t want to look up asin and tan we can note that if sin(A) = 0.8 then the sides of the triangle must be 3:4:5, so the length x is 6 *4/3 = 8.

So we don’t even need a slide rule :)

The Rev Dodgson said:

requires zero knowledge of calculus, and is in retrospect bleedin obvious.

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy

so how did we know the relation is smooth

The Rev Dodgson said:

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

I have been giving this lengthy further thought, and have finally come up with a solution that requires zero knowledge of calculus, and is in retrospect bleedin obvious.

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

So, for minimum travel time sin(A) = 0.8 so A = asin(0.8)

x = 6 * tan(asin(0.8)) = 8

But if we don’t want to look up asin and tan we can note that if sin(A) = 0.8 then the sides of the triangle must be 3:4:5, so the length x is 6 *4/3 = 8.

So we don’t even need a slide rule :)

Seems like you are inventing calculus…

I had a think about it and the answer is definitely less than six hours, as that’s how long a crocodile can hold its breath for.

dv said:

The Rev Dodgson said:

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

Seems like you are inventing calculus…

‘e doesn’t know that

SCIENCE said:

The Rev Dodgson said:

requires zero knowledge of calculus, and is in retrospect bleedin obvious.

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy

so how did we know the relation is smooth

Why wouldn’t it be?

The sine function is smooth isn’t it?

dv said:

The Rev Dodgson said:

fsm said:

Can you solve the math problem that stumped Scottish students?

A bamboozling question about a crocodile stalking its prey was one reason the pass mark for Higher maths had to be lowered, a report has found.

A question about a crocodile stalking its prey became particularly notorious – with the SQA acknowledging in its report that it had “proved to be challenging for most candidates”.

https://www.miragenews.com/can-you-solve-the-math-problem-that-stumped-scottish-students/

I have been giving this lengthy further thought, and have finally come up with a solution that requires zero knowledge of calculus, and is in retrospect bleedin obvious.

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

So, for minimum travel time sin(A) = 0.8 so A = asin(0.8)

x = 6 * tan(asin(0.8)) = 8

But if we don’t want to look up asin and tan we can note that if sin(A) = 0.8 then the sides of the triangle must be 3:4:5, so the length x is 6 *4/3 = 8.

So we don’t even need a slide rule :)

Seems like you are inventing calculus…

I don’t know about inventing it.

I suppose it uses some of the basic principles of calculus, but only stuff that would be obvious to anyone familiar with the basics of geometry and trigonometry.

the equation given in the OP relates time in tenths of a second to x, the distance from the crocodile’s location to where it exits the water (ignoring the width of the river.) Since the time is in tenths of a second, and not continuous, and calculus depends on limits of continuous functions, calculus can’t be used to solve the problem.

btm said:

the equation given in the OP relates time in tenths of a second to x, the distance from the crocodile’s location to where it exits the water (ignoring the width of the river.) Since the time is in tenths of a second, and not continuous, and calculus depends on limits of continuous functions, calculus can’t be used to solve the problem.

Where does it say the answer will be an integer number of tenths of a second?

The Rev Dodgson said:

btm said:

the equation given in the OP relates time in tenths of a second to x, the distance from the crocodile’s location to where it exits the water (ignoring the width of the river.) Since the time is in tenths of a second, and not continuous, and calculus depends on limits of continuous functions, calculus can’t be used to solve the problem.

Where does it say the answer will be an integer number of tenths of a second?

It doesn’t. Why is that relevant?

The Rev Dodgson said:

dv said:

The Rev Dodgson said:

I have been giving this lengthy further thought, and have finally come up with a solution that requires zero knowledge of calculus, and is in retrospect bleedin obvious.

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

So, for minimum travel time sin(A) = 0.8 so A = asin(0.8)

x = 6 * tan(asin(0.8)) = 8

But if we don’t want to look up asin and tan we can note that if sin(A) = 0.8 then the sides of the triangle must be 3:4:5, so the length x is 6 *4/3 = 8.

So we don’t even need a slide rule :)

Seems like you are inventing calculus…

I don’t know about inventing it.

I suppose it uses some of the basic principles of calculus, but only stuff that would be obvious to anyone familiar with the basics of geometry and trigonometry.

Well okay but honestly the easiest way would be to use calculus, basically 1 line of working, and the students taking this class would have been well past calculus.

The Rev Dodgson said:

SCIENCE said:

The Rev Dodgson said:

requires zero knowledge of calculus, and is in retrospect bleedin obvious.

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy

so how did we know the relation is smooth

Why wouldn’t it be?

The sine function is smooth isn’t it?

we mean this bit

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

The Rev Dodgson said:

SCIENCE said:

dv said:

The Rev Dodgson said:

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

Seems like you are inventing calculus…

‘e doesn’t know that

I don’t know about inventing it.

see

btm said:

The Rev Dodgson said:

btm said:

the equation given in the OP relates time in tenths of a second to x, the distance from the crocodile’s location to where it exits the water (ignoring the width of the river.) Since the time is in tenths of a second, and not continuous, and calculus depends on limits of continuous functions, calculus can’t be used to solve the problem.

Where does it say the answer will be an integer number of tenths of a second?

It doesn’t. Why is that relevant?

You said “the time is in tenths of a second, and not continuous” so I assumed you thought the time had to be an integer number of seconds.

If the time can be any value, why isn’t it continuous?

dv said:

The Rev Dodgson said:

dv said:

Seems like you are inventing calculus…

I don’t know about inventing it.

I suppose it uses some of the basic principles of calculus, but only stuff that would be obvious to anyone familiar with the basics of geometry and trigonometry.

Well okay but honestly the easiest way would be to use calculus, basically 1 line of working, and the students taking this class would have been well past calculus.

But you need to know how to find the derivative of whichever function you choose, which my solution doesn’t.

And if the students taking the class were well past calculus, how come the question caused such a problem?

SCIENCE said:

The Rev Dodgson said:

SCIENCE said:

so how did we know the relation is smooth

Why wouldn’t it be?

The sine function is smooth isn’t it?

we mean this bit

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

The time difference is shown to be proportional to the sin of the angle, so it must be continuous.

The Rev Dodgson said:

dv said:

The Rev Dodgson said:

I don’t know about inventing it.

I suppose it uses some of the basic principles of calculus, but only stuff that would be obvious to anyone familiar with the basics of geometry and trigonometry.

Well okay but honestly the easiest way would be to use calculus, basically 1 line of working, and the students taking this class would have been well past calculus.

But you need to know how to find the derivative of whichever function you choose, which my solution doesn’t.

And if the students taking the class were well past calculus, how come the question caused such a problem?

maybe they learnt to fly before they learnt to crawlw

SCIENCE said:

The Rev Dodgson said:

dv said:

Well okay but honestly the easiest way would be to use calculus, basically 1 line of working, and the students taking this class would have been well past calculus.

But you need to know how to find the derivative of whichever function you choose, which my solution doesn’t.

And if the students taking the class were well past calculus, how come the question caused such a problem?

maybe they learnt to fly before they learnt to crawlw

Could well be.

I mean from my own experience in a far-off land about 55 years ago, the way maths was taught was not a good way for many of the students to understand the basics, and quite possibly not much has changed.

The Rev Dodgson said:

btm said:

The Rev Dodgson said:

Where does it say the answer will be an integer number of tenths of a second?

It doesn’t. Why is that relevant?

You said “the time is in tenths of a second, and not continuous” so I assumed you thought the time had to be an integer number of seconds.

If the time can be any value, why isn’t it continuous?

Because the original question, as quoted in the OP, in introducing the equation, says,
“The time taken, T, measured in tenths of a second is given by”

The Rev Dodgson said:

SCIENCE said:

The Rev Dodgson said:

But you need to know how to find the derivative of whichever function you choose, which my solution doesn’t.

And if the students taking the class were well past calculus, how come the question caused such a problem?

maybe they learnt to fly before they learnt to crawlw

Could well be.

I mean from my own experience in a far-off land about 55 years ago, the way maths was taught was not a good way for many of the students to understand the basics, and quite possibly not much has changed.

The regular beatings have ceased.

btm said:

The Rev Dodgson said:

btm said:

It doesn’t. Why is that relevant?

You said “the time is in tenths of a second, and not continuous” so I assumed you thought the time had to be an integer number of seconds.

If the time can be any value, why isn’t it continuous?

Because the original question, as quoted in the OP, in introducing the equation, says,
“The time taken, T, measured in tenths of a second is given by”

Why does that imply T isn’t continuous?

Witty Rejoinder said:

The Rev Dodgson said:

SCIENCE said:

maybe they learnt to fly before they learnt to crawlw

Could well be.

I mean from my own experience in a far-off land about 55 years ago, the way maths was taught was not a good way for many of the students to understand the basics, and quite possibly not much has changed.

The regular beatings have ceased.

:)

Actually the maths teacher had the strange ability to maintain an attentive well-behaved class without even raising his voice, let alone regular beatings.

The Rev Dodgson said:

btm said:

The Rev Dodgson said:

You said “the time is in tenths of a second, and not continuous” so I assumed you thought the time had to be an integer number of seconds.

If the time can be any value, why isn’t it continuous?

Because the original question, as quoted in the OP, in introducing the equation, says,
“The time taken, T, measured in tenths of a second is given by”

Why does that imply T isn’t continuous?

By what possible stretch of the imagination does it imply that it is continuous? T, by the definition in the OP, is explicitly stated to be in tenths of a second, and so must be granular. It’s therefore not continuous.

btm said:

The Rev Dodgson said:

btm said:

Because the original question, as quoted in the OP, in introducing the equation, says,
“The time taken, T, measured in tenths of a second is given by”

Why does that imply T isn’t continuous?

By what possible stretch of the imagination does it imply that it is continuous? T, by the definition in the OP, is explicitly stated to be in tenths of a second, and so must be granular. It’s therefore not continuous.

Would it be granular if it said T was measured in seconds?

The Rev Dodgson said:

btm said:

The Rev Dodgson said:

Why does that imply T isn’t continuous?

By what possible stretch of the imagination does it imply that it is continuous? T, by the definition in the OP, is explicitly stated to be in tenths of a second, and so must be granular. It’s therefore not continuous.

Would it be granular if it said T was measured in seconds?

Yeah I’m not sure whether you’re having a laugh, btm. The formula is scaled in deciseconds, that’s all. Nothing is implied about granularity.

The Rev Dodgson said:

Witty Rejoinder said:

The Rev Dodgson said:

Could well be.

I mean from my own experience in a far-off land about 55 years ago, the way maths was taught was not a good way for many of the students to understand the basics, and quite possibly not much has changed.

The regular beatings have ceased.

:)

Actually the maths teacher had the strange ability to maintain an attentive well-behaved class without even raising his voice, let alone regular beatings.

so as long as there’s beatings at a rate of 0.01 Bq instead of 0.01 Hz it’s all right then

The Rev Dodgson said:

Witty Rejoinder said:

The Rev Dodgson said:

Could well be.

I mean from my own experience in a far-off land about 55 years ago, the way maths was taught was not a good way for many of the students to understand the basics, and quite possibly not much has changed.

The regular beatings have ceased.

:)

Actually the maths teacher had the strange ability to maintain an attentive well-behaved class without even raising his voice, let alone regular beatings.

One of our maths teachers worked out early in the year who was good at maths and who wasn’t and paired us up. Those who were good at maths got better (yes, I was one of them, back then) because we had to think about how to explain it. Those who weren’t so good got better because they had help.

Horses for courses. I think it would take an especially bright or insightful child to come up with Rev’s method de novo.

The article says that the problem was with the overall test, not just this particular problem. Perhaps they overstuffed the test with somewhat tricky problems and students just ran out of time.

The article also says that most students struggled with this one, implying some didn’t. This was a Higher Maths class so they’ve at least done calculus to the level that is learned by grade 10 kids in Australia.

dv said:

Horses for courses. I think it would take an especially bright or insightful child to come up with Rev’s method de novo.

The article says that the problem was with the overall test, not just this particular problem. Perhaps they overstuffed the test with somewhat tricky problems and students just ran out of time.

The article also says that most students struggled with this one, implying some didn’t. This was a Higher Maths class so they’ve at least done calculus to the level that is learned by grade 10 kids in Australia.

so uh basically none

SCIENCE said:

dv said:

Horses for courses. I think it would take an especially bright or insightful child to come up with Rev’s method de novo.

The article says that the problem was with the overall test, not just this particular problem. Perhaps they overstuffed the test with somewhat tricky problems and students just ran out of time.

The article also says that most students struggled with this one, implying some didn’t. This was a Higher Maths class so they’ve at least done calculus to the level that is learned by grade 10 kids in Australia.

so uh basically none

I don’t think I’d done any calculus in grade 10. Did what was called pure and applied maths back in the day and I really didn’t think we got to calculus until form 5.

Wtf is form 5?

dv said:

Wtf is form 5?

Grade 11.

dv said:

Wtf is form 5?

I think it was the old term for the fifth year of secondary. back in the time terrordactyls went to school.

Cheers

dv said:

Wtf is form 5?

Form 5 comes before form 6 which is the last year of secondary school.

fsm said:

dv said:

Wtf is form 5?

Form 5 comes before form 6 which is the last year of secondary school.

Except in England, where it is the penultimate year.

It is followed by form 6-II.

sibeen said:

SCIENCE said:

dv said:

Horses for courses. I think it would take an especially bright or insightful child to come up with Rev’s method de novo.

The article says that the problem was with the overall test, not just this particular problem. Perhaps they overstuffed the test with somewhat tricky problems and students just ran out of time.

The article also says that most students struggled with this one, implying some didn’t. This was a Higher Maths class so they’ve at least done calculus to the level that is learned by grade 10 kids in Australia.

so uh basically none

I don’t think I’d done any calculus in grade 10. Did what was called pure and applied maths back in the day and I really didn’t think we got to calculus until form 5.

That sounds about right to me too. High school in Melbourne in the 1970s.

buffy said:

sibeen said:

SCIENCE said:

so uh basically none

I don’t think I’d done any calculus in grade 10. Did what was called pure and applied maths back in the day and I really didn’t think we got to calculus until form 5.

That sounds about right to me too. High school in Melbourne in the 1970s.

Did absolutely none of it where I went to school.

Spiny Norman said:

buffy said:

sibeen said:

I don’t think I’d done any calculus in grade 10. Did what was called pure and applied maths back in the day and I really didn’t think we got to calculus until form 5.

That sounds about right to me too. High school in Melbourne in the 1970s.

Did absolutely none of it where I went to school.

I think we did Maths I and Maths II in form 5 (now known as year 11) and Pure Maths and Applied Maths in form 6 (now known as year 12). I’m not old enough to have done Matriculation (although Mr buffy is), I did HSC (Higher School Certificate) – everything hung on that 3 hour exam at the end of form 6 for each subject. I think it’s now called VCE.

buffy said:

Spiny Norman said:

buffy said:

That sounds about right to me too. High school in Melbourne in the 1970s.

Did absolutely none of it where I went to school.

I think we did Maths I and Maths II in form 5 (now known as year 11) and Pure Maths and Applied Maths in form 6 (now known as year 12). I’m not old enough to have done Matriculation (although Mr buffy is), I did HSC (Higher School Certificate) – everything hung on that 3 hour exam at the end of form 6 for each subject. I think it’s now called VCE.

Sounds different to what I did here in Queensland. I did 12th grade in 1982.

Spiny Norman said:

buffy said:

Spiny Norman said:

Did absolutely none of it where I went to school.

I think we did Maths I and Maths II in form 5 (now known as year 11) and Pure Maths and Applied Maths in form 6 (now known as year 12). I’m not old enough to have done Matriculation (although Mr buffy is), I did HSC (Higher School Certificate) – everything hung on that 3 hour exam at the end of form 6 for each subject. I think it’s now called VCE.

Sounds different to what I did here in Queensland. I did 12th grade in 1982.

When I was in year 10, I had a motorbike prang and missed the last half of the year, then did two years at Agricultural College, so I missed out on all of those Acronyms. In my early twenties I did an adult entry exam to get into Uni where I did advanced maths, but I haven’t used it for so long that I forget most of it.

Spiny Norman said:

buffy said:

Spiny Norman said:

Did absolutely none of it where I went to school.

I think we did Maths I and Maths II in form 5 (now known as year 11) and Pure Maths and Applied Maths in form 6 (now known as year 12). I’m not old enough to have done Matriculation (although Mr buffy is), I did HSC (Higher School Certificate) – everything hung on that 3 hour exam at the end of form 6 for each subject. I think it’s now called VCE.

Sounds different to what I did here in Queensland. I did 12th grade in 1982.

Primary school was Prep, then Grades 1-6. High school was Forms 1-6. Total of 13 years. I did HSC in 1977.

I was a croc at math.

I did Years 11 & 12 in NSW, 1977-78. I did 4 Unit Maths which was the 3 Unit Maths course plus an extra 4th Unit that was a separate subject to the 3 Unit course. In my HSC year, the 3 Unit exam was notoriously difficult, making it to the news, and causing fellow students I knew to freak out. Fortunately, the 4th Unit exam the next day was much easier (relatively speaking).

As for calculus, I started doing that in Year 11. I do remember a table of integrals at the back of the trigonometry tables book that we used before Year 11. I was always intrigued by them until I eventually found out what they were about.

So it is only deevs who did calc in form 4 (year 10)?

sibeen said:

So it is only deevs who did calc in form 4 (year 10)?

I think so.

I didn’t do any of it till mid-way through Year 11. (in 1988).

party_pants said:

sibeen said:

So it is only deevs who did calc in form 4 (year 10)?

I think so.

I didn’t do any of it till mid-way through Year 11. (in 1988).

well we suppose we can’t all be baby geniuses

but then again

https://www.theage.com.au/national/victoria/more-schools-fast-tracking-students-despite-push-to-seal-them-off-20220913-p5bhmk.html

Just reading about “coffin problems”, which were problems set in oral exams which had obvious but difficult to do in your head analytic solutions, but also non-obvious but simple solutions. Apparently they were used in the USSR to keep Jews (and other “undesirables”) out of university.

So does this count as a coffin problem?

Perhaps more of a 5 day self-isolation problem.

The Rev Dodgson said:

Just reading about “coffin problems”, which were problems set in oral exams which had obvious but difficult to do in your head analytic solutions, but also non-obvious but simple solutions. Apparently they were used in the USSR to keep Jews (and other “undesirables”) out of university.

So does this count as a coffin problem?

Perhaps more of a 5 day self-isolation problem.

More about coffin problems

The Rev Dodgson said:

Just reading about “coffin problems”, which were problems set in oral exams which had obvious but difficult to do in your head analytic solutions, but also non-obvious but simple solutions. Apparently they were used in the USSR to keep Jews (and other “undesirables”) out of university.

So does this count as a coffin problem?

Perhaps more of a 5 day self-isolation problem.

Oh, what is the non-obvious but simple solution in this case?

The Rev Dodgson said:

The Rev Dodgson said:

Just reading about “coffin problems”, which were problems set in oral exams which had obvious but difficult to do in your head analytic solutions, but also non-obvious but simple solutions. Apparently they were used in the USSR to keep Jews (and other “undesirables”) out of university.

So does this count as a coffin problem?

Perhaps more of a 5 day self-isolation problem.

More about coffin problems

I found 3.1 a bit odd. All points on a quadrilateral lie on a plane. Done.

dv said:

The Rev Dodgson said:

Just reading about “coffin problems”, which were problems set in oral exams which had obvious but difficult to do in your head analytic solutions, but also non-obvious but simple solutions. Apparently they were used in the USSR to keep Jews (and other “undesirables”) out of university.

So does this count as a coffin problem?

Perhaps more of a 5 day self-isolation problem.

Oh, what is the non-obvious but simple solution in this case?

If x/distance in water = inverse speed on land/inverse speed in water then that gives the minimum travel time, so x = 8.

dv said:

The Rev Dodgson said:

The Rev Dodgson said:

Just reading about “coffin problems”, which were problems set in oral exams which had obvious but difficult to do in your head analytic solutions, but also non-obvious but simple solutions. Apparently they were used in the USSR to keep Jews (and other “undesirables”) out of university.

So does this count as a coffin problem?

Perhaps more of a 5 day self-isolation problem.

More about coffin problems

I found 3.1 a bit odd. All points on a quadrilateral lie on a plane. Done.

But they don’t have to.

The Rev Dodgson said:

dv said:

The Rev Dodgson said:

Just reading about “coffin problems”, which were problems set in oral exams which had obvious but difficult to do in your head analytic solutions, but also non-obvious but simple solutions. Apparently they were used in the USSR to keep Jews (and other “undesirables”) out of university.

So does this count as a coffin problem?

Perhaps more of a 5 day self-isolation problem.

Oh, what is the non-obvious but simple solution in this case?

If x/distance in water = inverse speed on land/inverse speed in water then that gives the minimum travel time, so x = 8.

Can you prove that succinctly?

The Rev Dodgson said:

dv said:

The Rev Dodgson said:

More about coffin problems

I found 3.1 a bit odd. All points on a quadrilateral lie on a plane. Done.

But they don’t have to.

Yeah they do.

From WP:

In geometry a quadrilateral is a four-sided polygon, having four edges (sides) and four corners (vertices). The word is derived from the Latin words quadri, a variant of four, and latus, meaning “side”.

In geometry, a polygon (/ˈpɒlɪɡɒn/) is a plane figure that is described by a finite number of straight line segments connected to form a closed polygonal chain (or polygonal circuit). The bounded plane region, the bounding circuit, or the two together, may be called a polygon

dv said:

The Rev Dodgson said:

dv said:

Oh, what is the non-obvious but simple solution in this case?

If x/distance in water = inverse speed on land/inverse speed in water then that gives the minimum travel time, so x = 8.

Can you prove that succinctly?

Sure:

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

So, for minimum travel time sin(A) = 0.8 so A = asin(0.8)

x = 6 * tan(asin(0.8)) = 8

But if we don’t want to look up asin and tan we can note that if sin(A) = 0.8 then the sides of the triangle must be 3:4:5, so the length x is 6 *4/3 = 8.

So we don’t even need a slide rule :)

dv said:

The Rev Dodgson said:

dv said:

I found 3.1 a bit odd. All points on a quadrilateral lie on a plane. Done.

But they don’t have to.

Yeah they do.

From WP:

In geometry a quadrilateral is a four-sided polygon, having four edges (sides) and four corners (vertices). The word is derived from the Latin words quadri, a variant of four, and latus, meaning “side”.

In geometry, a polygon (/ˈpɒlɪɡɒn/) is a plane figure that is described by a finite number of straight line segments connected to form a closed polygonal chain (or polygonal circuit). The bounded plane region, the bounding circuit, or the two together, may be called a polygon

Oh well, I suppose they are redefining a quadrilateral to be any four lines connected to form a single closed area then.

The Rev Dodgson said:

dv said:

The Rev Dodgson said:

If x/distance in water = inverse speed on land/inverse speed in water then that gives the minimum travel time, so x = 8.

Can you prove that succinctly?

Sure:

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

So, for minimum travel time sin(A) = 0.8 so A = asin(0.8)

x = 6 * tan(asin(0.8)) = 8

But if we don’t want to look up asin and tan we can note that if sin(A) = 0.8 then the sides of the triangle must be 3:4:5, so the length x is 6 *4/3 = 8.

So we don’t even need a slide rule :)

Your quick and cheeky shortcut seems about 5 times longer than the normal way.

The Rev Dodgson said:

dv said:

The Rev Dodgson said:

But they don’t have to.

Yeah they do.

From WP:

In geometry a quadrilateral is a four-sided polygon, having four edges (sides) and four corners (vertices). The word is derived from the Latin words quadri, a variant of four, and latus, meaning “side”.

In geometry, a polygon (/ˈpɒlɪɡɒn/) is a plane figure that is described by a finite number of straight line segments connected to form a closed polygonal chain (or polygonal circuit). The bounded plane region, the bounding circuit, or the two together, may be called a polygon

Oh well, I suppose they are redefining a quadrilateral to be any four lines connected to form a single closed area then.

Sucks to be them because I already answered.

dv said:

The Rev Dodgson said:

dv said:

Can you prove that succinctly?

Sure:

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

So, for minimum travel time sin(A) = 0.8 so A = asin(0.8)

x = 6 * tan(asin(0.8)) = 8

But if we don’t want to look up asin and tan we can note that if sin(A) = 0.8 then the sides of the triangle must be 3:4:5, so the length x is 6 *4/3 = 8.

So we don’t even need a slide rule :)

Your quick and cheeky shortcut seems about 5 times longer than the normal way.

Shrug

The point is, you don’t need to know what the differential of (a^2+x^2)^0.5 is, or the differential of 1/sin or cot.

You just need to know that if x increases by 1 very small unit, then the hypotenuse increases by sin(theta) very small units, and that a 3:4:5 triangle is a right angle, both of which are commonly known with a much lower level of maths education.

The Rev Dodgson said:

dv said:

The Rev Dodgson said:

Sure:

Suppose x is as shown in the diagram and y is the length of the path through water.

As x increases from 0 the travel time starts to reduce but at some stage it will start to increase again, so we need to find the point where the rate of change in travel time is zero.

If for a small increase in the angle to the vertical of the path through water the increase in x is dx and the increase in y is dy then:
The increase in travel time in water = 5dy
The reduction in travel time on land = 4dx

So for zero change in travel time: 4dx = 5dy dy/dx = 0.8

If the angle of the path through water to vertical is A, then for a very small change of angle: dy/dx = sin(A)

So, for minimum travel time sin(A) = 0.8 so A = asin(0.8)

x = 6 * tan(asin(0.8)) = 8

But if we don’t want to look up asin and tan we can note that if sin(A) = 0.8 then the sides of the triangle must be 3:4:5, so the length x is 6 *4/3 = 8.

So we don’t even need a slide rule :)

Your quick and cheeky shortcut seems about 5 times longer than the normal way.

Shrug

The point is, you don’t need to know what the differential of (a^2+x^2)^0.5 is, or the differential of 1/sin or cot.

You just need to know that if x increases by 1 very small unit, then the hypotenuse increases by sin(theta) very small units, and that a 3:4:5 triangle is a right angle, both of which are commonly known with a much lower level of maths education.

(Shrugs) I guess it really depends on what the assumed knowledge is. This is an easy problem that can be done quickly using normal methods using the chain rule.

dv said:

The Rev Dodgson said:

dv said:

Your quick and cheeky shortcut seems about 5 times longer than the normal way.

Shrug

The point is, you don’t need to know what the differential of (a^2+x^2)^0.5 is, or the differential of 1/sin or cot.

You just need to know that if x increases by 1 very small unit, then the hypotenuse increases by sin(theta) very small units, and that a 3:4:5 triangle is a right angle, both of which are commonly known with a much lower level of maths education.

(Shrugs) I guess it really depends on what the assumed knowledge is. This is an easy problem that can be done quickly using normal methods using the chain rule.

That’s why it is a 5 day self-isolation problem, rather than a coffin problem.

But the fact remains, it can be solved even more quickly without even any knowledge of what calculus is, let alone how to apply calculus in this case.

The Rev Dodgson said:

dv said:

The Rev Dodgson said:

Shrug

The point is, you don’t need to know what the differential of (a^2+x^2)^0.5 is, or the differential of 1/sin or cot.

You just need to know that if x increases by 1 very small unit, then the hypotenuse increases by sin(theta) very small units, and that a 3:4:5 triangle is a right angle, both of which are commonly known with a much lower level of maths education.

(Shrugs) I guess it really depends on what the assumed knowledge is. This is an easy problem that can be done quickly using normal methods using the chain rule.

That’s why it is a 5 day self-isolation problem, rather than a coffin problem.

But the fact remains, it can be solved even more quickly without even any knowledge of what calculus is, let alone how to apply calculus in this case.

I disagree.

dv said:

I’ll have to think about that one.