Y12 U3 Maths Assignment help

[jasmine383]

Hi, last year in grade 11, I basically failed methods PSMT, I didn't know what to do and gave up and the teachers could only give minimal assistance but I really had no clue what to do. This year, the teachers keep asking us to refer back to our last year's assignment but the thing is, its a complete disaster . So therefore, I would really appreciate any assistance with my task this year 

Basically, we were given data looking at temperature per min of the black/milk before/milk after tea and were asked to investigate the claim that the tea will not cool as quickly as if you add the milk after adding the hot water.
The topic is basically about further calculus and differentiation but my teacher has told us that we need to do linearisation which comes in two forms: semi-log and log-log (if I'm correct).

e.g. for semi log, I used the exponential equation y=Ae^kx and made it to a semi-log equation ln⁡(y)=kx+ln(⁡A)
If you'll have a look at the attatchment I make a graph based on the time and the log of temperature and then added the linear regression trendline which gave me a nice line of y=-0.0115x+4.2456
I then used the linear stat regression to find a k (m) and A (b) value and get the equation: y=4.2456×e^(-0.0115x) which can also be graphed against the data.

How would I know which graph to use/what determines a good graph?
Additionally, when I made a residual plot from the data, I got something like... (open attachment 2) but when I searched it up, a u-shape correlation means that a linear model is NOT fit for the graph so I'm just really confused in all 

And also where does the differentiation takes place? I assume that's after I generate a good model and then differentiate it to determine the maximum?   

Hope you understand what I mean and thank you so much for reading, I really appreciate it
~Jas383

[Bri MT]

Hey,

Ok, so if you're investigating a claim about cooling quickly, what does this mean in terms of things you can measure/graph? What does differentiation tell you about?

The point of transforming your x and y axis was to try and make a linear line a good fit for your graph. The reason you are seeing the u-shape residuals is because after transformation your curve is closer to linear but it still isn't linear.

The best graph is one that is informative. I.e. it represents your data well and you can use it to learn something

if you make a linear model then the maximum will just be at t=0 (assuming negative slope, which is the case)


Hope this helps! Please feel free to ask follow up questions or clarification, I know this can be a bit confusing

[jasmine383]

Ohhh wow I get it...its newton's law of cooling!

So does this mean I still need to linearise since we get the equation T(t) = C*e^-kt + Ta
where T = temp, t = time, Ta = ambient temp and C = constant

So I derive the equation I get to obtain the max with temp=0, is that what you mean?

Thanks 

[Bri MT]

Ohhh wow I get it...its newton's law of cooling!

So does this mean I still need to linearise since we get the equation T(t) = C*e^-kt + Ta
where T = temp, t = time, Ta = ambient temp and C = constant

So I derive the equation I get to obtain the max with temp=0, is that what you mean?

Thanks 

No worries!

You should still linearise as per your teacher's instructions.

What I'm saying is that there isn't much point in differentiating the linearisation (let's say this is l(x)= mx + c) since you'll just get l'(x) = m right? It's not very informative.  The maximum of a linear function is always just at one end of its domain; solving for the maximum of it doesn't require differentiation [remember that the differentiation step when solving for the max of a function tells you the x value of interest]. 

I might be misinterpreting you here and you might be thinking of using differentiation to find the max rate of change rather than the max temperature. This would tell you information you can't get from the graph

[jasmine383]

Ok, so I don't use Newton's hahah.... So when I'm trying to find the rate of change with a set of data, do I use the first and last set?

Ohh I get what you're saying, but then I don't understand why my teacher would still want us to linearise the data if it doesn't really show anything  because I asked my teacher if 'we use rate of change to analyse the data (derivatives) instead of linearizing it, or do we have to linearise it?' and she said that you need a model to perform a derivative so does what does that mean?

Sorry if I'm sounding demanding! I just have my draft due on Tuesday and I'm not quite sure whaat I'm doing
Thanks 

[Bri MT]

Ok, so I don't use Newton's hahah.... So when I'm trying to find the rate of change with a set of data, do I use the first and last set?

Ohh I get what you're saying, but then I don't understand why my teacher would still want us to linearise the data if it doesn't really show anything  because I asked my teacher if 'we use rate of change to analyse the data (derivatives) instead of linearizing it, or do we have to linearise it?' and she said that you need a model to perform a derivative so does what does that mean?

Sorry if I'm sounding demanding! I just have my draft due on Tuesday and I'm not quite sure whaat I'm doing
Thanks 

Nothing to apologise for!

It's not a case of either linearising or using Newton's law of cooling. The point of then fitting a line of best fit is that this tells you something about the data you actually had whereas the model tells you about the data you should theoretically have. Generally, you want to relate the data you had actually with the model's predictions and analyse it in that way. You took the log of temperature for your graph, try converting Newton's law of cooling into a log form (hint: taking the log of T-Ta might be better than using T, in which case also try graphing your results using T - Ta). This should allow you to find what the value of k and C is for each of your conditions (black, milk before & milk after) - keep in mind that you already know what Ta is. Once you have those values, you can use them for your model of each tea.

Finding the line of best fit isn't pointless (it lets you get those values for your model), it's differentiating the line of best fit which is pointless

What your teacher is saying is that differentiating is something you do to a function, right? So then to differentiate, you need a function which models the relationship you're interested in. You have discovered the generic version of the model you want to use, and you can use the line of best fit to get numbers that make it specific to your situation.


I hope this has cleared things up a bit, and again, nothing to apologise for

[jasmine383]

Thank you so much Bri!  I totally get what you're saying and thank you so much for being so understanding but I've just asked my teacher if we were allowed to do newton's and she said No. That really sucks cos I started doing all my models so I was wondering, is there a way to calculate a model with rate for a set of data without simultaneous equations?

Kind regards,
jasmine

[Bri MT]

Thank you so much Bri!  I totally get what you're saying and thank you so much for being so understanding but I've just asked my teacher if we were allowed to do newton's and she said No. That really sucks cos I started doing all my models so I was wondering, is there a way to calculate a model with rate for a set of data without simultaneous equations?

Kind regards,
jasmine

Yeah you can definitely construct your own model! Consider a few different families of equations (e.g. quadratic, exponential etc.) and see what fits your data best. You can use the same kind of process as you would've for Newton's law of cooling but you make the generic equation yourself.

[jasmine383]

Yeah you can definitely construct your own model! Consider a few different families of equations (e.g. quadratic, exponential etc.) and see what fits your data best. You can use the same kind of process as you would've for Newton's law of cooling but you make the generic equation yourself.

Hi Bri, thanks for replying 

I think I get it!

Kind regards,
Jasmine