# Differential equation questionWatch

#1
Hi, for this question which is from the 2015 june C4 paper I don't get why the answer can't be in terms of degrees in the last part of the question to find t. Surely it makes more sense that t=27 years rather than 0.473( in radians). The mark scheme says specifically that only 0.473 is accepted but the question does not state this.

http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf Question 7c
Thanks
0
2 years ago
#2
(Original post by coconut64)
Hi, for this question which is from the 2015 june C4 paper I don't get why the answer can't be in terms of degrees in the last part of the question to find t. Surely it makes more sense that t=27 years rather than 0.473( in radians). The mark scheme says specifically that only 0.473 is accepted but the question does not state this.

http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf Question 7c
Thanks
Surely you can see how unrealistic 27 years is?

Anywho, the reason you have to use radians is that all your derivatives and integrals depended on the fact that t was is radians. d/dt(sin t) = cos t only if t is in radians, if t is in degrees then that's not true anymore. So you solved your differential equation using things like integral of cos t = sin t, etc... which all made use of the fact that t was in radians.

So now you can't suddenly go back and use t as degrees, since the solution you have is one that depends on t in radians.

A lesson to learn here is that radians are infinitely better in most respects to degrees in calculus. So in an A-Level paper if you come across trigonometry in a calculus setting, it is 100% going to be in radians.
0
2 years ago
#3
(Original post by coconut64)
Hi, for this question which is from the 2015 june C4 paper I don't get why the answer can't be in terms of degrees in the last part of the question to find t. Surely it makes more sense that t=27 years rather than 0.473( in radians). The mark scheme says specifically that only 0.473 is accepted but the question does not state this.

http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf Question 7c
Thanks
In any calculus of trig, (differentiation, integration, differential equations, etc.) the argument for the trig function is always in radians.
d(cos(x))/dx = - sin(x), etc. is only true when x is in radrians.
1
2 years ago
#4
(Original post by coconut64)
Hi, for this question which is from the 2015 june C4 paper I don't get why the answer can't be in terms of degrees in the last part of the question to find t. Surely it makes more sense that t=27 years rather than 0.473( in radians). The mark scheme says specifically that only 0.473 is accepted but the question does not state this.

http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf Question 7c
Thanks
The integration identities that you have assumedly used are dependent on t being measured in radians.
0
#5
That makes sense, thanks for all the help
0
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