Thread: Maths.....

Is any one out there a maths teacher? or some one who is studying maths,

I have a terrible maths teacher who really dosent have the time of day for her students, if you don`t get it first time round you don`t get it.
I`ve lernt more from youtube and google than i have from her.

but i have a couple of questions i`m struggleing with, and was wondering if any one would be willing to help me with them?

first question is....the temperature in a room is kept constant at a maximum value (Tmax)whe the heating system is switched off for maintenance the temperature falls untill the work is completed and the system is switched on again.
The temperature then increases untill it once again reaches Tmax. the temperature at ant time (t hours) during the maintenance period is given by:

Tmax = t²-9t+61/4

a) what is the value of t max.

b Use the method of completing the square to determine:

i)the minimum temperature reached during the maintenance period.

ii)the amount of time the system was switched off for maintenance.

Sorry, can't help you. I'm an English teacher not a maths teacher. I passed what we called veggie maths.

Part A is easy, t=0, so:

Tmax = 0^2 - 9*0 + 61/4
Tmax = 0 - 0 + 15.25
Tmax = 15.25

I have no idea what completing the square means, but if I sub in a random time for t, let's say 1000 hours, I get:

Tmax = 1000^2 - 9*1000 + 61/4
Tmax = 1000000 - 9000 + 61/4
Tmax = 991015.25

Which doesn't make sense in the given scenario, the temperature should keep going down. It bottoms out at 4.5 hours at -5 degrees and then starts rising again, what happened, the sun come out?

Are we sure it ain't t squared, minus 9 times t, plus 61, all over 4? Oh, and delta t, is that graphed as a straight line or curve?

If THAT is supposed to be a temperature graph, your teacher should be sacked for failing to come up with decent examples that reflect reality. There is no way that temperature becomes a 2nd order function if the heating is turned off. It will be a function following the natural logarithm, dropping 'fast' and then petering out.

Making the temperature follow a squared function would be quite an interesting challenge and require active heating. The same goes for the heating cycle. And at the temp when the heater is turned back on, the temperature also follows a natural log function and be asymmetrical, meaning the temperature would start increasing rapidly, rather than slowly and then gaining momentum.

I understand that she used the heating question as a coathanger to shove the square function into for the purpose of learning to use square functions, but it really is horribly wrong at so many levels. Questions like this should be physically accurate to at least the very basic degree. Otherwise you'll confuse many people about the entire thing.

In this example, the shape of the curve is not only wrong because the curvature would be different, but it would also be asymmetrical. She could have taken another example that would have been accurately described with a square function, like a ball being kicked upwards or a cannon firing a ball.

Air friction would still make the function non-squared, but at least the 'ideal' situation (ignoring air friction) would result in a perfect squared function.

Completing the square - Wikipedia, the free encyclopedia

t^2 - 9t +61/4 -> (t-(9/2))^2 + (-5) completing the square. Check the formula how these were calcumulated.

"ƒ(x − h) + k = (x − h)² + k is a parabola shifted to the right by h and upward by k whose vertex is at (h, k), as shown in the bottom figure."

so, it reaches it's lowest at t=4.5, which is -5, like hydraral said. However, without the theory behind it, math is just guessing.

I could be wrong though.

Originally Posted by ursus

Completing the square - Wikipedia, the free encyclopedia

t^2 - 9t +61/4 -> (t-(9/2))^2 + (-5) completing the square. Check the formula how these were calcumulated.

"ƒ(x − h) + k = (x − h)² + k is a parabola shifted to the right by h and upward by k whose vertex is at (h, k), as shown in the bottom figure."

so, it reaches it's lowest at t=4.5, which is -5, like hydraral said. However, without the theory behind it, math is just guessing.

I could be wrong though.

I agree with Ursus. I'm a math teacher and I came up with the same answer.

Originally Posted by JRMx3

I agree with Ursus. I'm a math teacher and I came up with the same answer.

I came up with that only after making the assumption that the cooling period ends at the lowest point of the cooling curve. Unfortunately, the problem does not actually state when "the work is completed"

I misinterpreted the question a little, the equation does not determine the temperature at time t, it descibes the temperature variation during the outage.

For those of us who didn't know (me included) it looks like the "completing the square" method was instructed because it will actually pop out the highest or lowest point while you are doing the calculations (see the example near the bottom):

Completing the Square

It's late, so I'm cheating and using an online calculator:

Algebra.Help -- Completing the Square Calculator

And solving using the quadratic equation will give you the two values for t when the temperature is 15.25, ie the start and end of the outage.

Algebra.Help -- Completing the Square Calculator

Originally Posted by Hydaral

the two values for t when the temperature is 15.25, ie the start and end of the outage.

not quite... the temperature at the end of the outage will be cooler than the start