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Non-uniform acceleration

malli07

An objects displacement in metres at t seconds is given by the function: s= sin(2t)-t^3+4t^2+2
The input of the wine function is in degrees
A) Find the objects velocity at time t=0
B) Fund the objects acceleration at t=1

Reply 1

mathstutor24

Can you post your working so far and what you need help with please? :smile:

Reply 2

malli07

Original post by mathstutor24

Can you post your working so far and what you need help with please? :smile:

I don’t know how to go about it... the main struggle is to differentiate sin(2t)

Reply 3

mathstutor24

When differentiating

sin(2t)

, you need differentiate it as if it was

sin(u)

(so this would be

cos(u)

) and then multiply it by the differentiation of "u" i.e. what is in the brackets.

So, if

y=sin(2t)

then,

\frac{dy}{dt}=2cos(2t)

Reply 4

malli07

Original post by mathstutor24

I've put together a graphic for you, as it's difficult to type everything here in code. Hope this helps:

Ok thank you 😃

Reply 5

davros

Original post by mathstutor24

When differentiating

sin(2t)

, you need differentiate it as if it was

sin(u)

(so this would be

cos(u)

) and then multiply it by the differentiation of "u" i.e. what is in the brackets.

So, if

y=sin(2t)

then,

\frac{dy}{dt}=2cos(2t)

Note that according to the OP, the question states that: "The input of the wine function is in degrees" (sic)

This makes the differentiation "interesting" :smile:

Reply 6

mathstutor24

Original post by davros

Note that according to the OP, the question states that: "The input of the wine function is in degrees" (sic)

This makes the differentiation "interesting" :smile:

Must confess - did't read that bit :smile:

Just helping with basic differentiation of trig

Reply 7

davros

Original post by mathstutor24

Must confess - did't read that bit :smile:

Just helping with basic differentiation of trig

My point being: the derivative of sin x is only cos x provided that x is measured in radians...

Reply 8

mathstutor24

Original post by davros

My point being: the derivative of sin x is only cos x provided that x is measured in radians...

Thanks for pointing it out :smile:

As I said, I missed that part of the question.

OP - do you need help differentiating in degrees, or are you okay with that bit?

Reply 9

davros

Original post by mathstutor24

Thanks for pointing it out :smile:

As I said, I missed that part of the question.

Sorry - I wasn't trying to be 'sharp' by repeating myself. I re-read ,my original post and realised that you might have thought I was just drawing attention to the silly typo, so I thought I'd better add clarification :smile:

I wonder if this sort of question is going to arise more frequently - I know it's always been stated at A level that trig differentiation relies on radian measure, but I'm not sure if students are prepared to handle the adjustment needed if an independent variable is specified in degrees!

Reply 10

mathstutor24

Original post by davros

No, no! I didn't think you were being sharp at all :smile:

I think differentiation of trig using degree measure is a good thing to be teaching. It's something that I can certainly see Edexcel sneakily throwing in, and as proved in this thread, it's really easy to overlook it and go ahead with "usual" differentiation (i.e. using radian measure).