eh I thought 27 because the force on the right pushing it left + reaction from the force from the block on the left. I am stunned I can't answer that question.
Use the Chain Rule to find dz/dt. (answer only in terms of t.)
z = tan-1(y/x), x = et, y = 4 - e-t
I need help with this problem....I worked it out but the answer is incorrect..
tan(z) = y/x
tan(z) = (4-et)/et
tan(z) = 4e-t-e-2t
sec2(z)dz/dt = 2e-2t-4e-t
dz/dt = cos2(x)(2e-2t-4e-t)
z = tan-1(y/x), x = et, y = 4 - e-t
I need help with this problem....I worked it out but the answer is incorrect..
tan(z) = y/x
tan(z) = (4-et)/et
tan(z) = 4e-t-e-2t
sec2(z)dz/dt = 2e-2t-4e-t
dz/dt = cos2(x)(2e-2t-4e-t)
Last edited by Sheen1101 (2010-02-22 19:20:47)
You didn't actually use the chain rule...when you sub in for y and x, that means when you take the derivative there is also a dy/dt and dx/dt term respectively.
My own question: if an inductor and a capacitor are in series over a long time with a source, there is still no current right?
My own question: if an inductor and a capacitor are in series over a long time with a source, there is still no current right?
Ask FreezerFlaming_Maniac wrote:
My own question: if an inductor and a capacitor are in series over a long time with a source, there is still no current right?
I gather from a class example that I am correct, but this circuit is just wonky. I am trying to get the equation of current over time for a critically damped LRC circuit that initially and finally as a current of 0, but the energy stored in the cap still has to be accounted for...I have all the equations sitting in front of me that seem to work out to i(t) = 0, but that can't be right...
Sheen1101 wrote:
Use the Chain Rule to find dz/dt. (answer only in terms of t.)
z = tan-1(y/x), x = et, y = 4 - e-t
I need help with this problem....I worked it out but the answer is incorrect..
tan(z) = y/x
tan(z) = (4-et)/et
tan(z) = 4e-t-e-2t
sec2(z)dz/dt = 2e-2t-4e-t
dz/dt = cos2(x)(2e-2t-4e-t)
Okay so I redid the work and still got the wrong answer, what am I doing wrong?:Flaming_Maniac wrote:
You didn't actually use the chain rule...when you sub in for y and x, that means when you take the derivative there is also a dy/dt and dx/dt term respectively.
My own question: if an inductor and a capacitor are in series over a long time with a source, there is still no current right?
z = tan-1(y/x), x = et, y = 4 - e-t
dz/dt = (1/1+(y/x)2)(-y/x2)et+(1/1+(y/x)2)(1/x)e-t
dz/dt = (-y/(x2+y2))et+ (x/(x2+y2))e-t
dz/dt = (-e-t(e2ty-x))/ (x2+y2)
Surely for something that's being divided you should use the quotient rule?Sheen1101 wrote:
Sheen1101 wrote:
Use the Chain Rule to find dz/dt. (answer only in terms of t.)
z = tan-1(y/x), x = et, y = 4 - e-t
I need help with this problem....I worked it out but the answer is incorrect..
tan(z) = y/x
tan(z) = (4-et)/et
tan(z) = 4e-t-e-2t
sec2(z)dz/dt = 2e-2t-4e-t
dz/dt = cos2(x)(2e-2t-4e-t)Okay so I redid the work and still got the wrong answer, what am I doing wrong?:Flaming_Maniac wrote:
You didn't actually use the chain rule...when you sub in for y and x, that means when you take the derivative there is also a dy/dt and dx/dt term respectively.
My own question: if an inductor and a capacitor are in series over a long time with a source, there is still no current right?
z = tan-1(y/x), x = et, y = 4 - e-t
dz/dt = (1/1+(y/x)2)(-y/x2)et+(1/1+(y/x)2)(1/x)e-t
dz/dt = (-y/(x2+y2))et+ (x/(x2+y2))e-t
dz/dt = (-e-t(e2ty-x))/ (x2+y2)
I'd type my pc specs out all fancy again but teh mods would remove it. Again.
Reduce (a+b)2-2a(a+b)
The result I got was a2+b2-2aa-2ab
Looks wrong to me though (and yes, this is probably really easy to you guys, but I'm not that good at math).
I am doing something wrong, right? And if yes, what?
The result I got was a2+b2-2aa-2ab
Looks wrong to me though (and yes, this is probably really easy to you guys, but I'm not that good at math).
I am doing something wrong, right? And if yes, what?
a2 + 2ab + b2 - 2a2 -2ab
reduce
b2 - a2
reduce
b2 - a2
Last edited by Bevo (2010-02-24 08:11:37)
(a+b)2-2a(a+b)Amdi Peter wrote:
Reduce (a+b)2-2a(a+b)
The result I got was a2+b2-2aa-2ab
Looks wrong to me though (and yes, this is probably really easy to you guys, but I'm not that good at math).
I am doing something wrong, right? And if yes, what?
for the first bit you want to use foil
so (a+b)2
becomes (a+b) (a+b)
then you follow the diagram above, the second part is self explanatory
I don't know what you did on the second line...Sheen1101 wrote:
Sheen1101 wrote:
Use the Chain Rule to find dz/dt. (answer only in terms of t.)
z = tan-1(y/x), x = et, y = 4 - e-t
I need help with this problem....I worked it out but the answer is incorrect..
tan(z) = y/x
tan(z) = (4-et)/et
tan(z) = 4e-t-e-2t
sec2(z)dz/dt = 2e-2t-4e-t
dz/dt = cos2(x)(2e-2t-4e-t)Okay so I redid the work and still got the wrong answer, what am I doing wrong?:Flaming_Maniac wrote:
You didn't actually use the chain rule...when you sub in for y and x, that means when you take the derivative there is also a dy/dt and dx/dt term respectively.
My own question: if an inductor and a capacitor are in series over a long time with a source, there is still no current right?
z = tan-1(y/x), x = et, y = 4 - e-t
dz/dt = (1/1+(y/x)2)(-y/x2)et+(1/1+(y/x)2)(1/x)e-t
dz/dt = (-y/(x2+y2))et+ (x/(x2+y2))e-t
dz/dt = (-e-t(e2ty-x))/ (x2+y2)
Write out dz/dt, dx/dt, dy/dt. Then plug in the values first to z (like you did the first time) just remember that when you take the derivative, there is a dy or a dx after those specific parts. Then plug in.
Thanks for the help, I think I got it in the end.
physics is raping me
anyone fluent?
anyone fluent?
i take a massive interest in physics but i think your level is above meBevo wrote:
physics is raping me
anyone fluent?
angular velocity and acceleration confuses the shit out of memkxiii wrote:
i take a massive interest in physics but i think your level is above meBevo wrote:
physics is raping me
anyone fluent?
statics good, dynamics bad.
I find the worst part about physics is that many physicists are poor public speakers so its difficult to get a good professor/lecturer.Bevo wrote:
physics is raping me
anyone fluent?
while my prof is a terrible speaker, i wouldnt mind if he would teach stuff related to the homework we doArc wrote:
I find the worst part about physics is that many physicists are poor public speakers so its difficult to get a good professor/lecturer.Bevo wrote:
physics is raping me
anyone fluent?
i have to look up pretty much everything in this stupid book which isnt detailed enough to account for "what ifs"
yeah, i do physics and none of the lectures help with our problems
pretty much everyone has to copy off the cleverest person they know
pretty much everyone has to copy off the cleverest person they know
Dont you have tutorials?
nopeWinston_Churchill wrote:
Dont you have tutorials?
my tutor thought we were getting a different tutor this semester so never turns up
even though everyone keeps the same one.
I have emailed him but he is away atm and will see me on the 9th march
What exactly do you need help in? I'm not an angular dynamics pro, but I did get through the kinetics course with a B, that's gotta be good for something.Bevo wrote:
angular velocity and acceleration confuses the shit out of memkxiii wrote:
i take a massive interest in physics but i think your level is above meBevo wrote:
physics is raping me
anyone fluent?
statics good, dynamics bad.
i've got a "racecar" on a raised curve problem, where I have to find the max velocity it can achieve without slipping outwards. There's friction involved too. No idea how to set it up -the book has a similar problem, but it includes no friction and therefore confuses me. I don't know at which point to include friction in the calculations, or if I should include the x component of the mg, or if that's nullified by centripedal acceleration...
also got a block at rest in the middle of a turntable, and given a friction and angular acceleration, want to know at what point it falls off the table. again no idea how to deal with angular anything because it was never taught anywhere
also got a block at rest in the middle of a turntable, and given a friction and angular acceleration, want to know at what point it falls off the table. again no idea how to deal with angular anything because it was never taught anywhere
Angular stuff is the same, all the same equations, just with radians/second instead of meters/second.Bevo wrote:
i've got a "racecar" on a raised curve problem, where I have to find the max velocity it can achieve without slipping outwards. There's friction involved too. No idea how to set it up -the book has a similar problem, but it includes no friction and therefore confuses me. I don't know at which point to include friction in the calculations, or if I should include the x component of the mg, or if that's nullified by centripedal acceleration...
also got a block at rest in the middle of a turntable, and given a friction and angular acceleration, want to know at what point it falls off the table. again no idea how to deal with angular anything because it was never taught anywhere
I misread the turntable problem, the mass is some distance from the center and we're supposed to find the time it takes for it to start movingS.Lythberg wrote:
Angular stuff is the same, all the same equations, just with radians/second instead of meters/second.Bevo wrote:
i've got a "racecar" on a raised curve problem, where I have to find the max velocity it can achieve without slipping outwards. There's friction involved too. No idea how to set it up -the book has a similar problem, but it includes no friction and therefore confuses me. I don't know at which point to include friction in the calculations, or if I should include the x component of the mg, or if that's nullified by centripedal acceleration...
also got a block at rest in the middle of a turntable, and given a friction and angular acceleration, want to know at what point it falls off the table. again no idea how to deal with angular anything because it was never taught anywhere
we don't get a mass though, which is confusing...
moment of inertia = mass