Maths question

anksta

anksta

anksta

Soldato
Joined
19 Jan 2005
Posts
2,722
Can someone please do a maths question for me? If you've read my other thread from yesterday you'll see that I'm suffering ridiculously from anxiety at the minute and this maths is the main catalyst right now.

It's not cheating I'm not doing an exam.

A person's body temperature y(degrees C) has a maximum calue of 37.3 at 6.00pm and a minimum value of 36.7 at 6.00am.

It can be modelled by an equation of the form y = A cos (kt) + C where t is the number of hours after 6.00pm and A, k and c are all constants

a) find the values of the constants A k and c
b) Find the temperature predicted by the model at 11pm
c) find the values of t at which the temperature would be 36.8 degrees C
d)write down an equation that could be used to model the body temperature if t is the number of hours after midnight instead of 6pm.

I've been doing this for ages now and want to kill myself, if anyone can just give me the answers, I don't need the working or anything.
 
Well maximum value is going to be achieved when cos(kt)=1
And at minimum cos(kt)=-1

37.3=A+C
36.7=-A+C
Add them together
So you get
74=2C
C=37
So 37.3=A+C
37.3=A+37
A=0.3

cos(kt)=-1
cos(12k)=-1
Since cos(12k) cross -1 at pi radians
12k=Pi
k=pi/12

Part B)
y=Acos(kt)+C
So at t=5
y=0.3cos(5*pi/12)+37
So temperature = 0.3cos(5*pi/12)+37

Part C)
36.8=Acos(kt)+C
36.8=0.3cos((pi/12)*t)+37
-0.2=0.3cos((pi/12)*t)
-2/3=cos(pi/12*t)
I don't have a calculator, you should know what to do now.

Part D)

Think what you need to change the T term by to get it the same as the previous model..
Y=Acos(k(t+6)) +C
 
Last edited:
i dont want to work through it myself, but i can tell you that there are two ways of doing this. the 'proper' way would be to the whole differentiate it twice to get the minimum and maximum value equations and work from there. however, you can do things much more simply.

For part a:

first you need to work out what k is. you should know that the cos function is a wave that repeats itself every 360 degrees (or 2*pi radians). you also know that t cannot get bigger than 24. if you are working with degrees you need to make it so that kt goes between 0 and 360. ie. 24k = 360.

the cos wave very conveniently peaks when t = 0 and is at its lowest when t = 180 degrees or 2*pi radians, so no addition or subtraction to the angle is needed

the next piece of info you need to use is the range of the cos wave you are finding. normally the range of the cos wave is 2 (goes from 1 to -1). your wave has a range of 0.6 (37.3 - 36.7) so you need to find a constant A that will give you a cos wave that has a range of 0.6. so A = 0.6/2 i think

the last piece of information is the middle of the wave. for a normal cos wave this is 0, but in yours it is the middle of the highest point and the lowest point. so your constant C will be (37.3 + 36.7)/ 2.


this will get you a lovely equation where Y and t are the only unknowns, perfect for plugging in the numbers of the next two questions


Part b:

remember that t is the number of hours after 11pm. plug the value of t into your newly found formula and you're off

Part c:

you have the value of Y but need to find the value of C. this will involve rearranging the equation to be t = ????. without doing it myself i can tell you it will involve an inverse cos and a division by k
 
oops, i totally forgot about part d but Roman covered it :)

you could also be a clever git and change the original equation from a cos to a sin and that should work. dont quote me on that though
 
Last edited:
You must log in or register to reply here.
Back
Top Bottom