Post by: Greatness on November 07, 2011, 01:51:31 pm
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???
Post by: dc302 on November 07, 2011, 01:58:29 pm
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?
Post by: b^3 on November 07, 2011, 02:02:08 pm
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???
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.
Post by: Greatness on November 07, 2011, 02:09:29 pm
Post by: dc302 on November 07, 2011, 03:26:42 pm
Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.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)
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???
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.
Post by: dc302 on November 07, 2011, 03:34:59 pm
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?
Post by: Greatness on November 07, 2011, 03:40:45 pm
Thank you! That's what i was looking for :D 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! :PHere 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?
Post by: b^3 on November 07, 2011, 04:03:55 pm
Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.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)
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???
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.
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,
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
Post by: dc302 on November 07, 2011, 04:23:27 pm
"If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180" - this is still right.Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.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)
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???
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.
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
Quote
Don't you mean, d/dx (sin(pi*x/180)) = pi*cos(pi*x/180) /180 ?
Post by: b^3 on November 07, 2011, 04:27:28 pm
"If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180" - this is still right.Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.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)
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???
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.
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,
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
While what you said now is correct, this is not what you said earlier.
Earlier, you did d/dx sinx, not d/dy sinx.
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.
Post by: dc302 on November 07, 2011, 04:29:07 pm
Yes but earlier I was saying in the second line that x is in degrees, while on the first line x was in radians."If you diff it in degrees you have to put d/dx(sin(x))=pi*cos(x)/180" - this is still right.Ok so i've come across these questions twice now. Theyre a rate of change question with something moving and the angle changing.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)
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???
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.
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,
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
While what you said now is correct, this is not what you said earlier.
Earlier, you did d/dx sinx, not d/dy sinx.
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.
Post by: dc302 on November 07, 2011, 04:30:45 pm
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.
Post by: dc302 on November 07, 2011, 04:36:24 pm
Quote
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.
Post by: b^3 on November 07, 2011, 04:37:57 pm
Let me just check though.
Post by: b^3 on November 07, 2011, 04:46:33 pm
QuoteAnd 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.
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.
Post by: haddzy on November 07, 2011, 04:48:42 pm
Post by: dc302 on November 07, 2011, 04:51:26 pm
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.QuoteAnd 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.
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.
Yeah, the calculator does a lot of things 'behind the scenes' haha. Of course, every problem is solved as long as you use radians, as you said. :)
Post by: Greatness on November 07, 2011, 05:03:41 pm
Looks like radians is the way to go! thanks again guys :D
Post by: dc302 on November 07, 2011, 05:07:40 pm
intense maths-debate :P
Looks like radians is the way to go! thanks again guys :D
I see what you did there.