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secondary 4 | A Maths
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glittr
Glittr

secondary 4 chevron_right A Maths chevron_right Singapore

must i assume that abc are positive ? what if they happen to be negative ? will i get 2 sets of answers ?

Date Posted: 5 years ago
Views: 314
J
J
5 years ago
What is t? The number of hours after 0300?
glittr
Glittr
5 years ago
i got a= -4 vs a=4. Then sketch . Then reject positive 4 becos it is supposed to be lower temp at the start .
glittr
Glittr
5 years ago
am i right to say that it is best to sketch in step 1 ?
J
J
5 years ago
What's the value of c you obtained ? 30? If yes, then the temperature is 26 for both t = 300 and t = 1500, when a = -4
glittr
Glittr
5 years ago
yes c=30 for both a= posi or neg 4. thanks !!
J
J
5 years ago
Then there's something odd about this question. When a = 4, both temperatures are 34°C. When a = -4, both temperatures are 26°C.
glittr
Glittr
5 years ago
i just posted a pic . i used elimination method to eliminate positive 4
J
J
5 years ago
Yup I got your point. What I'm saying is b can't be π/12


why a has to be -4 :

if the graph starts from 0300, then it should be increasing (upward sloping) from 0300.

I.e the coefficient of cos(bt) should be positive. So -a has to be positive, which means a has to be negative
glittr
Glittr
5 years ago
i just posted pic about b
J
J
5 years ago
If you actually plotted the graph of T(t) = 4 cos(π/12 t) + 30 using a software (eg. Desmos), you will get T = 26 for both t = 300 and t = 1500. So this question is flawed
glittr
Glittr
5 years ago
i posted pic of desmos . it looks like vertical value = 26 when horizontal = 0 and 24 hr .
J
J
5 years ago
Yup. So the graph can't just have b = π/12 without any translation.


Try plotting T(t) = -4(π/1200 t - π/4) + 30

Using b = π/12 implies that the temperature fluctuates too rapidly throughout the day
glittr
Glittr
5 years ago
i’m confused . what is wrong with y= 4cos (pi/12 x ) + 30 as per your previous previous comment ? it seems to fit the storyline of temp =26 , then 34 in middle , then back to 26 .
J
J
5 years ago
The problem is, your t starts from 0. It should start from 3. But the minimum temperature isnt at 12 midnight. It is at 3am.

The question says t is the time in hours. So at 3am, it would be 0300 hours, i.e t = 300. At 3pm, it would be 1500 hours, i.e t = 1500.
glittr
Glittr
5 years ago
is it the 3am refers to t=0 ? like the equation starts counting from 3am , means t=0 is the same thing as 3am ?
J
J
5 years ago
That's what the question was implying. But the definition of t is poorly worded. It should have been phrased as something along the lines of

't is the number of hours elapsed from 0300 hours of the present day'
glittr
Glittr
5 years ago
i would interpret 3am as clock reading , not hours . Hours is the time lapsed from the clock
glittr
Glittr
5 years ago
i agree the english is lacking . haha . will update this space when the student gets back to me with her sch explanation
glittr
Glittr
5 years ago
thanks for helping although i’m not a bona fide student
J
J
5 years ago
No problem.

Anyway, the way t is equated in the parts of the question also don't make sense.

We never see things like 't = 0300' in Cambridge papers. t is variable of a value ,not a clock reading , similar to your mentioning of interpreting 3am as a reading and not hours.

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1st
this is actual qn
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answer for i
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to find b value
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when t=0 and 24 , vertical value = 26
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sch answer part 1
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sch ans part 2