ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: horizon on September 20, 2011, 01:08:31 pm
-
(http://img155.imageshack.us/img155/4386/kil2007exam2q4mainq.png)
Q) The total number of hours of daylight during the month of January can be
expressed as a definite integral. Find the total number of hours of daylight
during the entire month of January, giving your answer in hours and minutes
correct to nearest minute.
The answer says that the total number of hours of daylight required can be expressed with this integral...
(http://img192.imageshack.us/img192/618/kil2007exam2q4ans.png)
I don't get why this integral will give the total number of hours?
Someone care to explain?
Thanks heaps in advance.
-
Also this question too...
(http://img685.imageshack.us/img685/8459/kil2007exam2q17.png)
(http://img64.imageshack.us/img64/5001/kil2007exam2q17ans.png)
Don't get how they went from that in the highlighted part!
-
It's difficult to explain in text (and on my phone), but its a little bit like taking te average value f that function from t=1 to t=31, and the number that comes out is the averag number of hours of daylight per day for January. So if u multiply that by 31, it will give you the total amount of daylightfor January. Do some simplifying and it's just equal to that integral
-
Q.17 - They've just taken the four from the front of 4e^(-4x) and the four from d(4x) out in front of the integral sign, so you get 4*4 out the front of the integral, ie. 16.
-
Ok let me attempt to explain question 1.
you have h(t), and h is equal to the number of daylight hours.
so when you substitute in:
day 1, you have 14.339
day 2, you have 14.354 etc etc etc.
Now, it is going to be pain staking to doing day 1, plus day 2, plus day 3 etc.
So this is where the integral comes in.
(ok this doesn't actually seem right because the way the question is worded, it should be a sigma I reckon, NOT an integral, but that is another argument).
If you have a small amount of time, and you multiply it by what h is, you will get a very small slice of the graph. In the situation before, we were using a time of 1, and if you multiply it by h, it would be the same.
What the definite integral will do is add up all those itsy bitsy tiny little areas together, and give you a nice number.
So without actually rewriting the whole thing.....
int(h(t).dt) between the days we want. (basically h(t) multiplied by all the little times we want
int(h(t).dt,1,31) [because the question is asking from 1st January to 31st January]
int[1/2*(24+5*cos(pi*(t-22)/183)),t,1,31]
move the 1/2 inside the brackets
so you end up with int[12+5/2*cos(pi*(t-22)/183),t,1,31]
Sorry about the syntax, I used a TI-89T when I did vce =D
And I hope that I didnt confuse you more.
-
Thanks guys, I get the first question now!
But second question...err.. why can we move the 4 from the d(4x) outside? Is this some theorem or formula?
-
The 4 is a constant.
d(4x) is the same as 4d(x)