Maths question in exam todau

billybob89

billybob89

billybob89

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During a test with a thermocouple pyrometer the e.m.f. (E in millivolts) was
measured against the temperature at the hot junction (t in °C) and the following
results were obtained:

t (° C)
200 300 400 500 600 700 800 900 1000

E (millivolts)
6 9.1 12.0 14.8 18.2 21.0 24.1 26.8 30.2


The law connecting t and E is supposed to be E = at + b. Test if this is so and find suitable values for a and b.

Came up in maths today, was unsure how to do it any ideas as it really bugging me?
 
This surely cannot be done with 2 variables in the equation?
There is an infinite number of solutions?
(a= 0.003 b = 0 seems to be the most obvious solution)
 
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Plot a graph with E on the y axis and t on the x. Should give a straight line with gradient a. The line should intercept the y-axis at b.
 
E = at + b is just like the standard linear equation y = mx + c
(a/m is the gradient, b/c is the y intercept)

There are two ways to do this but the easiest one here is to first find the gradient using the formula:

Method 1

m(the gradient, a in this case) = y-y1/x-x1

Taking any two sets of data (using two as far apart as possible gets the most accurate result, I used the first and the last) from what you've been given you get:

(30.2 - 6)
----------- = 0.03025
(1000 - 200)

That's your gradient (or a), so, right now you have

E = 0.03025t + b

You have E and t values so substitute any one of them in to get b
E = 14.8 t = 500
b = 14.8 - (0.03025 x 500)
= -0.325

To check (use another set of data, I chose t = 900 E should = 26.9):
E = 0.03025 x 900 - 0.325
E = 26.9 (value in data is actually 26.8)

This is not as accurate as using simultaneous equations though (The other method to solve this) which yields exact answers where a = 0.03 and b = 0 shown here:

Method 2

Two unknown variables (a and b) means making two equations using the data you've been given

I used the first and the third:
(t = 200 E = 6, t = 400 E - 12)

6 = 200a + b and
12 = 400a + b

You can eliminate b by rearranging:

b = 6 - 200a and (1)
b = 12 - 400a (2)

so 6 - 200a = 12 - 400a
rearrange (take -400a to left and +6 to the right) to get:
200a = 6
a = 6/200

Substitute that back into one of the equation, say (1)
b = 6 - 200(6/200)
b = 6-6 thus b = 0

You can check this by substituting these values back into the original equation:
E = 0.03(t) + 0 (note 6/200 = 0.03)

Take any value from your data say, t = 700, you get E = 4200/200 + 0
E = 21.0 as shown in your data table.

Hope this makes sense, It's easier than it looks, I just put in all the steps so you could understand. :o
 
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It tells you that the relationship is linear so just draw a graph and work out the gradient and intercept.

sid
 
Beenom said:
E = at + b is just like the standard linear equation y = mx + c
(a/m is the gradient, b/c is the y intercept)

There are two ways to do this but the easiest one here is to first find the gradient using the formula:

Method 1

m(the gradient, a in this case) = y-y1/x-x1

Taking any two sets of data (using two as far apart as possible gets the most accurate result, I used the first and the last) from what you've been given you get:

(30.2 - 6)
----------- = 0.03025
(1000 - 200)

That's your gradient (or a), so, right now you have

E = 0.03025t + b

You have E and t values so substitute any one of them in to get b
E = 14.8 t = 500
b = 14.8 - (0.03025 x 500)
= -0.325

To check (use another set of data, I chose t = 900 E should = 26.9):
E = 0.03025 x 900 - 0.325
E = 26.9 (value in data is actually 26.8)

This is not as accurate as using simultaneous equations though (The other method to solve this) which yields exact answers where a = 0.03 and b = 0 shown here:

Method 2

Two unknown variables (a and b) means making two equations using the data you've been given

I used the first and the third:
(t = 200 E = 6, t = 400 E - 12)

6 = 200a + b and
12 = 400a + b

You can eliminate b by rearranging:

b = 6 - 200a and (1)
b = 12 - 400a (2)

so 6 - 200a = 12 - 400a
rearrange (take -400a to left and +6 to the right) to get:
200a = 6
a = 6/200

Substitute that back into one of the equation, say (1)
b = 6 - 200(6/200)
b = 6-6 thus b = 0

You can check this by substituting these values back into the original equation:
E = 0.03(t) + 0 (note 6/200 = 0.03)

Take any value from your data say, t = 700, you get E = 4200/200 + 0
E = 21.0 as shown in your data table.

Hope this makes sense, It's easier than it looks, I just put in all the steps so you could understand. :o
Click to expand...

This only takes into account the first and second points in the data set though, which makes it pretty inaccurate and unreliable.

The correct way to do it would be to use least squares regression, like talkshowhost said, but it's been ages since I've had to use any kind of regression so I can't remember how!
 
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billybob89 said:
During a test with a thermocouple pyrometer the e.m.f. (E in millivolts) was
measured against the temperature at the hot junction (t in °C) and the following
results were obtained:

t (° C)
200 300 400 500 600 700 800 900 1000

E (millivolts)
6 9.1 12.0 14.8 18.2 21.0 24.1 26.8 30.2


The law connecting t and E is supposed to be E = at + b. Test if this is so and find suitable values for a and b.

Came up in maths today, was unsure how to do it any ideas as it really bugging me?
Click to expand...

Not possible. If 200 oc 6, and 400 oc 12, then 300 cannot oc 9.1

since E = at + b is a linear equation, you simply work out (n2)E - (n1)E, and divide that by (n2)t - (n1)t to work out what a is. You then take at from E to get b.

So:
n - - E - - t
n1 = 6 / 200
n2 = 9.1 / 300
n3 = 12 / 400

(n2)E (9.1) - (n1)E (6) = 3.1
(n2)t (300) - (n1)t (200) = 100

3.1 / 100 = 0.031

that means:

(n1)
200 * 0.031 = 6.2
6 - 6.2 = -0.2

So a would equal 0.031, and b would equal -0.2, right? Let's check that:

(n2)
300 * 0.031 = 9.3
9.1 - 9.3 = -0.2

Sure enough, it fits. a is still 0.031, and b is still -0.2

However, if you use n2 and n3:

(n2)E (12) - (n2)E (9.1) = 2.9
(n2)t (400) - (n1)t (300) = 100

2.9 / 100 = 0.029

that means:

(n2)
300 * 0.029 = 8.7
9.1 - 8.7 = 0.4

so a is 0.029, and b is 0.4

test that on n3 aaaand:

(n3)
400 * 0.029 = 11.6
12 - 11.6 = 0.4

So yes, that matches up.

However, 0.031 =/= 0.029, and -0.2 =/= 0.4, so the rule does not hold true. So, the question is not possible, unless you were meant to spot a pattern and identify the real values, which would be:

from:

t (° C)
200 300 400 500 600 700 800 900 1000

E (millivolts)
6 9.1 12.0 14.8 18.2 21.0 24.1 26.8 30.2

to:

t (° C)
200 300 400 500 600 700 800 900 1000

E (millivolts)
6.0 9.0 12.0 15.0 18.0 21.0 24.0 27.0 30.0

In which case, it would be:

E = at + b
6 = 0.03 * 200 + 0

Which would make sense for a maths question. Unless they wanted you to work out averages or draw a graph and get as close a result as possible to what the real equation could be.

I reckon this question should be more aimed at a science or statistics paper, showing analysis of anomalous or varying results. That is, if it is for GCSE level. I don't know if it could be for higher education, I don't know what kind of questions they have.

If it was in a science paper, they would probably ask you to round each result so the steps are equal (e.g. 6 = 200, 9 = 300, 12 = 400). If it was in a statistics paper, they'd ask you to create graphs to find the closest result, and in a GCSE maths paper, they'd probably give you "Sally thinks the equation for these values is X. Is Sally correct? Explain why."
 
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Graph is the first thing that comes to mind . . . plotting a graph will make it immediately obvious if the data fit a linear regression, and then also the graph makes finding the values of the gradient/intercept trivial as well.
 
Inquisitor said:
This only takes into account the first and second points in the data set though, which makes it pretty inaccurate and unreliable.

The correct way to do it would be to use least squares regression, like talkshowhost said, but it's been ages since I've had to use any kind of regression so I can't remember how!
Click to expand...
Given it's meant to use E = at + b, then it's safe to assume the sensor behaves linearly over the given range. I'd use the first and last points to cover the largest area when calculating the gradient, which is what he's done.
 
D D Danneh said:
Not possible. If 200 oc 6, and 400 oc 12, then 300 cannot oc 9.1

since E = at + b is a linear equation, you simply work out (n2)E - (n1)E, and divide that by (n2)t - (n1)t to work out what a is. You then take at from E to get b.

So:
n - - E - - t
n1 = 6 / 200
n2 = 9.1 / 300
n3 = 12 / 400

(n2)E (9.1) - (n1)E (6) = 3.1
(n2)t (300) - (n1)t (200) = 100

3.1 / 100 = 0.031

that means:

(n1)
200 * 0.031 = 6.2
6 - 6.2 = -0.2

So a would equal 0.031, and b would equal -0.2, right? Let's check that:

(n2)
300 * 0.031 = 9.3
9.1 - 9.3 = -0.2

Sure enough, it fits. a is still 0.031, and b is still -0.2

However, if you use n2 and n3:

(n2)E (12) - (n2)E (9.1) = 2.9
(n2)t (400) - (n1)t (300) = 100

2.9 / 100 = 0.029

that means:

(n2)
300 * 0.029 = 8.7
9.1 - 8.7 = 0.4

so a is 0.029, and b is 0.4

test that on n3 aaaand:

(n3)
400 * 0.029 = 11.6
12 - 11.6 = 0.4

So yes, that matches up.

However, 0.031 =/= 0.029, and -0.2 =/= 0.4, so the rule does not hold true. So, the question is not possible, unless you were meant to spot a pattern and identify the real values, which would be:

from:

t (° C)
200 300 400 500 600 700 800 900 1000

E (millivolts)
6 9.1 12.0 14.8 18.2 21.0 24.1 26.8 30.2

to:

t (° C)
200 300 400 500 600 700 800 900 1000

E (millivolts)
6.0 9.0 12.0 15.0 18.0 21.0 24.0 27.0 30.0

In which case, it would be:

E = at + b
6 = 0.03 * 200 + 0

Which would make sense for a maths question. Unless they wanted you to work out averages or draw a graph and get as close a result as possible to what the real equation could be.

I reckon this question should be more aimed at a science or statistics paper, showing analysis of anomalous or varying results. That is, if it is for GCSE level. I don't know if it could be for higher education, I don't know what kind of questions they have.

If it was in a science paper, they would probably ask you to round each result so the steps are equal (e.g. 6 = 200, 9 = 300, 12 = 400). If it was in a statistics paper, they'd ask you to create graphs to find the closest result, and in a GCSE maths paper, they'd probably give you "Sally thinks the equation for these values is X. Is Sally correct? Explain why."
Click to expand...

It says the results were from an experiment, so slightly incorrect results would be expected.

Good grief.
 
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