Degree/Radian question when diff/antidiff/RoC

[Greatness]

Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.
First one was in Mav 2011 exam 2 and i forgot to change from degrees to radians
And the other one was in Kilbaha 2010 exam 1. But in this question all the info they gave us was in degrees so this time i assumed we didnt have to convert to radians. But yet again im wrong
So do you have to covert to radians everytime with differentiation/rate of change/antidiff questions???

[dc302]

Do you mean when you're expressing the answer, or when you're doing the working out?

It shouldn't matter when your're doing the working out, as long as you're consistent. But if i'ts about writing the solution, then I'm not sure which format VCAA prefers (but I would guess radians).

And which question was it?

[b^3]

Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.
First one was in Mav 2011 exam 2 and i forgot to change from degrees to radians
And the other one was in Kilbaha 2010 exam 1. But in this question all the info they gave us was in degrees so this time i assumed we didnt have to convert to radians. But yet again im wrong
So do you have to covert to radians everytime with differentiation/rate of change/antidiff questions???

Everytime you diff/antidiff something you should work in radians. The definitions we have for our differentiating are in radians. e.g. d/dx(sin(x))=cos(x)
If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180

In short, convert to radians and have your calc in radian mode.

[Greatness]

Here is the question. I did everything right except for the converting to radians. The solutions says you have to convert the 1 deg per min to radians. :/ Please explain?

[dc302]

Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.
First one was in Mav 2011 exam 2 and i forgot to change from degrees to radians
And the other one was in Kilbaha 2010 exam 1. But in this question all the info they gave us was in degrees so this time i assumed we didnt have to convert to radians. But yet again im wrong
So do you have to covert to radians everytime with differentiation/rate of change/antidiff questions???

Everytime you diff/antidiff something you should work in radians. The definitions we have for our differentiating are in radians. e.g. d/dx(sin(x))=cos(x)
If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180

In short, convert to radians and have your calc in radian mode.

Don't you mean, d/dx (sin(pi*x/180)) = pi*cos(pi*x/180) /180 ?

If d/dx sinx = pi*cosx /180, then that would mean the derivative at x=0 is actually pi/180, which is clearly not the case.

[dc302]

Here is the question. I did everything right except for the converting to radians. The solutions says you have to convert the 1 deg per min to radians. :/ Please explain?

Sorry about the double post but since this is addressing another comment, I wasn't sure if I should have edited my old post or not.

Anyway, the reason why you have to convert the 1 deg/min to radians is because sin and cos take radians. If you wish to work in degrees, you'd have to use something like sin(x pi/180), which would essentially convert degrees to radians anyway.

Does this help?

[Greatness]

Here is the question. I did everything right except for the converting to radians. The solutions says you have to convert the 1 deg per min to radians. :/ Please explain?

Sorry about the double post but since this is addressing another comment, I wasn't sure if I should have edited my old post or not.

Anyway, the reason why you have to convert the 1 deg/min to radians is because sin and cos take radians. If you wish to work in degrees, you'd have to use something like sin(x pi/180), which would essentially convert degrees to radians anyway.

Does this help?

Thank you! That's what i was looking for I wasnt entirely sure about where the radians came in but i thought it had something to do with the sin/cos. lets hope there's something like this on the exam!

[b^3]

Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.
First one was in Mav 2011 exam 2 and i forgot to change from degrees to radians
And the other one was in Kilbaha 2010 exam 1. But in this question all the info they gave us was in degrees so this time i assumed we didnt have to convert to radians. But yet again im wrong
So do you have to covert to radians everytime with differentiation/rate of change/antidiff questions???

Everytime you diff/antidiff something you should work in radians. The definitions we have for our differentiating are in radians. e.g. d/dx(sin(x))=cos(x)
If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180

In short, convert to radians and have your calc in radian mode.

Don't you mean, d/dx (sin(pi*x/180)) = pi*cos(pi*x/180) /180 ?

If d/dx sinx = pi*cosx /180, then that would mean the derivative at x=0 is actually pi/180, which is clearly not the case.

"If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180" - this is still right.
Lets say that x is in radians, and y is in degrees. Then .
So lets diff it in degrees. So we want
Using chain rule,


(even check it on the calculator)

It works, just using the other method, you have converted the x.
What I was saying in the above (previous post) was that if x was in degrees then you can do d/dx(sin(x))=pi*cos(x)/180.

And with x=0, using the x-axis with degrees, (i.e. to 360 e.t.c) the gradient at x=0, is , even check using the calculator. If you were to convert that back into radians then it would be 1, as expected.

[dc302]

Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.
First one was in Mav 2011 exam 2 and i forgot to change from degrees to radians
And the other one was in Kilbaha 2010 exam 1. But in this question all the info they gave us was in degrees so this time i assumed we didnt have to convert to radians. But yet again im wrong
So do you have to covert to radians everytime with differentiation/rate of change/antidiff questions???

Everytime you diff/antidiff something you should work in radians. The definitions we have for our differentiating are in radians. e.g. d/dx(sin(x))=cos(x)
If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180

In short, convert to radians and have your calc in radian mode.

Don't you mean, d/dx (sin(pi*x/180)) = pi*cos(pi*x/180) /180 ?

If d/dx sinx = pi*cosx /180, then that would mean the derivative at x=0 is actually pi/180, which is clearly not the case.

"If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180" - this is still right.
Lets say that x is in radians, and y is in degrees. Then .
So lets diff it in degrees. So we want
Using chain rule,


(even check it on the calculator)

It works, just using the other method, you have converted the x.
What I was saying in the above (previous post) was that if x was in degrees then you can do d/dx(sin(x))=pi*cos(x)/180.

And with x=0, using the x-axis with degrees, (i.e. to 360 e.t.c) the gradient at x=0, is , even check using the calculator. If you were to convert that back into radians then it would be 1, as expected.

While what you said now is correct, this is not what you said earlier.

Earlier, you did d/dx sinx, not d/dy sinx.

edit: and notice how what you have just asserted is exactly what I wrote, if you sub in x = pi y /180

Don't you mean, d/dx (sin(pi*x/180)) = pi*cos(pi*x/180) /180 ?

[b^3]

Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.
First one was in Mav 2011 exam 2 and i forgot to change from degrees to radians
And the other one was in Kilbaha 2010 exam 1. But in this question all the info they gave us was in degrees so this time i assumed we didnt have to convert to radians. But yet again im wrong
So do you have to covert to radians everytime with differentiation/rate of change/antidiff questions???

Everytime you diff/antidiff something you should work in radians. The definitions we have for our differentiating are in radians. e.g. d/dx(sin(x))=cos(x)
If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180

In short, convert to radians and have your calc in radian mode.

Don't you mean, d/dx (sin(pi*x/180)) = pi*cos(pi*x/180) /180 ?

If d/dx sinx = pi*cosx /180, then that would mean the derivative at x=0 is actually pi/180, which is clearly not the case.

"If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180" - this is still right.
Lets say that x is in radians, and y is in degrees. Then .
So lets diff it in degrees. So we want
Using chain rule,


(even check it on the calculator)

It works, just using the other method, you have converted the x.
What I was saying in the above (previous post) was that if x was in degrees then you can do d/dx(sin(x))=pi*cos(x)/180.

And with x=0, using the x-axis with degrees, (i.e. to 360 e.t.c) the gradient at x=0, is , even check using the calculator. If you were to convert that back into radians then it would be 1, as expected.

While what you said now is correct, this is not what you said earlier.

Earlier, you did d/dx sinx, not d/dy sinx.

Yes but earlier I was saying in the second line that x is in degrees, while on the first line x was in radians.

e.g. d/dx(sin(x))=cos(x) - x in radians
If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180 - x in degrees, thats why it says if you diff it in degrees.

I probably should have said that the top line, x was in radians.

[dc302]

Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.
First one was in Mav 2011 exam 2 and i forgot to change from degrees to radians
And the other one was in Kilbaha 2010 exam 1. But in this question all the info they gave us was in degrees so this time i assumed we didnt have to convert to radians. But yet again im wrong
So do you have to covert to radians everytime with differentiation/rate of change/antidiff questions???

Everytime you diff/antidiff something you should work in radians. The definitions we have for our differentiating are in radians. e.g. d/dx(sin(x))=cos(x)
If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180

In short, convert to radians and have your calc in radian mode.

Don't you mean, d/dx (sin(pi*x/180)) = pi*cos(pi*x/180) /180 ?

If d/dx sinx = pi*cosx /180, then that would mean the derivative at x=0 is actually pi/180, which is clearly not the case.

"If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180" - this is still right.
Lets say that x is in radians, and y is in degrees. Then .
So lets diff it in degrees. So we want
Using chain rule,


(even check it on the calculator)

It works, just using the other method, you have converted the x.
What I was saying in the above (previous post) was that if x was in degrees then you can do d/dx(sin(x))=pi*cos(x)/180.

And with x=0, using the x-axis with degrees, (i.e. to 360 e.t.c) the gradient at x=0, is , even check using the calculator. If you were to convert that back into radians then it would be 1, as expected.

While what you said now is correct, this is not what you said earlier.

Earlier, you did d/dx sinx, not d/dy sinx.

Yes but earlier I was saying in the second line that x is in degrees, while on the first line x was in radians.

e.g. d/dx(sin(x))=cos(x) - x in radians
If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180 - x in degrees, thats why it says if you diff it in degrees.

I probably should have said that the top line, x was in radians.

I'm sorry but you are still wrong; I don't know how to explain it any better: and also read my edit in the last post.

[dc302]

Ok let me try again:

The problem you have is that you're trying to differentiate sinx with respect to x, while saying that the x inside the sin is in degrees, but the x outside is in radians. You can't do that: ie you have to use different variables, which you did correctly the second time.

[dc302]

And with x=0, using the x-axis with degrees, (i.e. to 360 e.t.c) the gradient at x=0, is , even check using the calculator. If you were to convert that back into radians then it would be 1, as expected.

Again, you are mistaken. There is no such thing as gradient in radians or degrees -- gradient has nothing to do with angles!

If you had a right angled triangle, with lengths 1, 1 and sqrt(2).

The angle of the slope is, you could say, 45 degrees. What is the gradient? The gradient is rise/run = 1. It is not 180/pi or whatever you claim it to be.

[b^3]

Wait I did make a mistake, some of those x's should have been y's.
Let me just check though.

[b^3]

And with x=0, using the x-axis with degrees, (i.e. to 360 e.t.c) the gradient at x=0, is , even check using the calculator. If you were to convert that back into radians then it would be 1, as expected.

Again, you are mistaken. There is no such thing as gradient in radians or degrees -- gradient has nothing to do with angles!

If you had a right angled triangle, with lengths 1, 1 and sqrt(2).

The angle of the slope is, you could say, 45 degrees. What is the gradient? The gradient is rise/run = 1. It is not 180/pi or whatever you claim it to be.

That is true, I think I know what is going wrong here. When you change the mode on the calculator, it changes the scaling of the graphs and the way the derivatives come out. Thats why the it says the gradient is pi/180=0.02 instead of 1. Effectively what the calculator is doing is evaluating the derivative and converting it back. The gradient thing is wrong. But what I'm trying to get at is that if you put the calculator in degrees mode, then it will give you a derivative with the extra junk in it.

So in short, put it in radian mode and convert everything to radians. It won't fail you and it won't confuse the hell out of you.

Sorry about all this mess above.