Math Is Fun Forum
You are not logged in.
- Topics: Active | Unanswered
#51 Help Me ! » Taylor series » 2014-05-27 18:52:51
- gourish
- Replies: 25
hi i am interested in learning Taylor series but i still haven't come across integration to a good depth can i still learn it and can anybody help me learn it because i didn't get any good sites to learn it from....
P.S. i don't even know when is the Taylor series used... i am just curious to learn about it
#52 Re: Science HQ » volume content » 2014-05-27 18:46:34
you guys still in school?
i thought you are in college @agnishom and niharika
#53 Re: Science HQ » work and energy problem » 2014-05-27 18:43:18
thanks again shivam your explanation is much better then that available online.... 
P.S. if i want to find the the velocity of the particle from the frame of earth then? and can you tell me what is your last diagram for? it looks like a parabolic motion of some body i don't understand it because the particle is supposed to move "on" the surface of sphere
#54 Re: Science HQ » work and energy problem » 2014-05-24 03:18:18
i found the solutions online but they are hardly clear it' all getting straight to answer i need to know how to get to the solution... people on this forum usually explain why a certain step is taken to find the answer so i wanted an answer on the forum but i can wait for the answer... thanks Shivam...
#55 Science HQ » work and energy problem » 2014-05-24 02:44:10
- gourish
- Replies: 5
a smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero velocity with respect to the sphere. find the speed of the particle with respect to the sphere as a function of the angle θ it slides..
i hope i will get a quick reply i am breaking my head over this problem....
#56 Re: Help Me ! » (Less than) 21 Trig Questions from Amrita » 2014-05-17 04:56:49
the equality sin(pie sinθ ) = cos(pie cosθ )
on applying arc sine on both sides becomes pie sinθ = pie/2 -(pie cosθ ) now shift the terms on the same side and divide throught by pie
you get sinθ + cosθ =1/2 and then square both sides to get 1+2sinθ cosθ =1/4 and then solving 2sinθ cosθ =-3/4
and we already know that 2sinθ cosθ = sin2θ
#57 Re: Help Me ! » (Less than) 21 Trig Questions from Amrita » 2014-05-17 04:52:11
Agnishom,
try using the inverse trignometric functions on the equality maybe it will help....
#58 Re: Help Me ! » (Less than) 21 Trig Questions from Amrita » 2014-05-17 04:08:57
shouldn't sin2θ= -3/4 cause that's quite easy to solve by using inverse trignometric functions (i assumed the angles to be in the principal branches) then what's the problem in finding it?
#59 Re: Help Me ! » a good book can solve my problems » 2014-05-10 18:32:51
well i used the mac graw hill's book but it's just having bunch of equations and no proof's... plus it's not helping me visualize a geometrical interpretation of problems on complex numbers.... any suggestions on how to deal with this?
#60 Help Me ! » a good book can solve my problems » 2014-05-10 05:13:48
- gourish
- Replies: 2
hi,
i find it hard to deal with complex numbers maybe because they are just "complex" i do not want to attempt any great depth into the topic but i need to know the complex numbers right from their basics to using them in argand plane and other geometries like conic sections can you guys suggest me the right book to read?
P.S. i remember things only when i derive the result on my own or read the complete explained derivation so i would like "that" kind of book..
#61 Science HQ » electrostatics... » 2014-05-10 05:09:53
- gourish
- Replies: 9
potential inside a hollow conductor is the same at all points and is equal to that at its surface. This is true for all hollow conductors whether spherical, cylindrical or of any arbitrary shape. can anybody give me the proof of this statement?
#62 Re: Science HQ » vector » 2014-05-04 03:05:20
hey which chapter is it form... i.e. from HC VERMA?
#63 Re: Science HQ » mirrors... » 2014-04-22 02:21:52
i didn't understand your question.... the "What transformation takes the original shape onto the final image?" part is bit confusing what do i need to find here...?
#64 Re: Science HQ » mirrors #2 » 2014-04-22 02:20:24
but it can take a lot of time to find the exact number of images being formed isn't there a better way then this that i can use? other then the formula that i mentioned because it doesn't work when the mirrors are the 3 sides of an equilateral triangle....
#65 Re: Science HQ » root mean square value » 2014-04-21 06:43:42
well i=3 sin Δ t + 4 cos Δ t differentiate this with respect to time (after all it's a wave which is variable in space with time) doing this i get diff(i) w.r.t.time = 3cosΔt-4sinΔt
for maxima diff(i) w.r.t.time=0 so 3cosΔt-4sinΔt=0
which gives tanΔt=3/4 now we can put it back into the original equation to find the answer to my question which is what i did
#66 Re: Science HQ » mirrors #2 » 2014-04-21 05:44:58
hi bob,
i still don't get it... since the image can be placed anywhere between them i am unable to make any relation with the number of images formed can you please explain me how to analyse the diagram that you have posted i am still having trouble dealing with the resulting proof
#67 Re: Science HQ » mirrors... » 2014-04-21 05:41:52
thanks that was enough to find the answer... any ideas where i can practice more of these kind of optics related problems?
#68 Re: Help Me ! » functions related to geometry » 2014-04-21 05:39:33
it was my mistake the actual question was about a cone which i posted already and found the answer
#69 Re: Science HQ » root mean square value » 2014-04-21 05:38:30
bob bundy the same thing is achieved even by calculus and bobbym i differentiated the current with respect to time and set the resulting equation to zero and found tan theta = 3/4
#70 Re: Science HQ » root mean square value » 2014-04-20 04:08:14
well maxima was at tan θ =3/4 but the link that you gave had the required formula to find the actual rms of the equation so it's really necessary
#71 Re: Science HQ » mirrors... » 2014-04-20 04:06:45
well 1=3=4 but what about 2 and 5?
#72 Re: Science HQ » root mean square value » 2014-04-20 04:02:41
thanks bobbym... your link helped i used the rms total=sqroot {rms1^2+rms2^2} where rms1 is 3/root(2) and rms2 is 4/root(2) so i ended up with the same answer 5/root(2)...
#73 Re: Science HQ » root mean square value » 2014-04-20 03:58:56
well that's the problem i don't know it's a combination of both sine wave and cosine wave... if we could maybe make them turn into either a sine wave or cosine wave we will find the rms value... i differentiated the equation and found that the current is maximum at tan θ =3/4 and plugged in the values into the original equation i am guessing the rms value is 5/root(2) but i am not sure....
#74 Re: Science HQ » root mean square value » 2014-04-20 03:46:10
well it depends on the time "t" so it's not a constant value the current is a function of time... generally if current i was given by
i= I cos (omega)t i know that I is the maximum value of current and find the RMS as I/root(2) which is not the case for this function.... i seem bit silly but i feel differentiating might help but i am not sure about it...
#75 Re: Help Me ! » Lagrange's interpolation polynomial » 2014-04-20 03:41:38
the formula needs a bit unnecessary thing if we can just the use cubic to plug in the given data and find the answer don't you think?