REVISION 2

Added: 2026-05-15

[00:00:21]Good evening ladies and gentlemen.

[00:00:26]We are going to start right away.

[00:00:31]Please, if you can if you can hear me, you can hear me say yes in the comment box.

[00:01:00]If you can hear me say yes in the comment box so that we can uh continue.

[00:01:14]Can you hear me? Just say yes if you can.

[00:01:18]Okay. Yes.

[00:01:21]Okay. I'm going to share a document that um I believe you can u see the document I've shared.

[00:02:15]So, we're going into the business.

[00:02:35]Okay.

[00:02:38]Um I have about 15 about.

[00:02:45]So and um one may ask you define vector or unit vector.

[00:03:00]We know that a unit vector is a vector that has a magnitude equal to one and is used to specify direction.

[00:03:12]If capital letter A is a vector, the unit vector will be that equal to A with the arrow divide by the the magnitude of that vector.

[00:03:29]And we can also one can also ask you or you may be asked to define a dot product.

[00:03:37]And we know that dot product is also called a scalar product.

[00:03:45]And the dot product of two vectors is defined as a with the arrow dot capital letter B of A modus of B cos theta and it produces a scalar quantity.

[00:04:06]So quantity and um you may also be asked to cross products.

[00:04:27]Another name for cross productduct is what we call the vector product.

[00:04:35]The cross product is a cross bal of a modus of b sin theta multiply by the unit vector called n.

[00:04:52]So question two, if vector A is given at Axi + A Y + A Z K and vector B is given as B X I + B Y + B Z K.

[00:05:14]We can now define what is the dot product.

[00:05:19]So using the distributive property dot product a dot axi + a y + a z k in bracket dot bx i + b y + bz k.

[00:05:40]Remember if you have something like this if you want to multiply taking the first term to multiply by all the terms in the bracket is really a because know the you are familiar with the some identities like I do. J gives dots. We know that. So we simply multiply the I B X I J B Y J A Y S Z K B Z K and that is what you have in the next line. By the time you do this, we know that I do I is one, J. J is one K is one and this will amount A do B that is a dot product is equal to A X Y that is a dot product then the crossroduct will also be the product of the cart form using the matrix approach I JK and by the time you expand it you have this expression So looking at if this vector question three given that vector A is equal to 3 I + 2 J minus K and vector B is 2 I - 4 J + 5 K and vector C is given as - I + 6 J + 2 K.

[00:07:20]We can find the resultant of the three vectors.

[00:07:27]we can find the resultant of the three vectors.

[00:07:34]I repeat, vector A is given as 3 I + 2 J - K. Vector B is given as 2 I - 4 J + 5 K and vector C is given as minus I + 6 J + 2 K.

[00:07:55]Find the resultant of the three vectors and what is resultant? The total effective was simply to add the three of them. So the resultant R is equal to vector A + B + C. So by the time you do that you have rx is equal to 3 + 2 - 1.

[00:08:25]Remember the i's are the x which you have for vector a we have three for vector b we have two for vector c you have minus one. So for RX that is the resultant of the X that is the I's we have 3 + 2 - 1 which will give us 4 and then for the R Y are the J's you have 2 J that is 2 + - 4 + 6 you have 2 - 4 + 6 and that will give us four and then for Z that We're talking about the case - 1 + 5 + 2 that will give us 6. So the total resultant becomes 4 I + 4 J + 6 K. And what is the magnitude? If you asked to find the magnitudes that is the modulus of that square root of 4 + 4 that is the 4 I + 4 J + 6. If you take the square root of this it will give you roo<unk> 68.

[00:09:47]And if you find the the square root of 68 that will give you approximately 28.25.

[00:09:58]So if you are asked to differentiate assuming you are given vector a = t² i + 3t g.

[00:10:12]Find differentiate and that will be the ad.

[00:10:17]We know how to differentiate 2t i + [snorts] 3.

[00:10:24]We are differentiating with respect to t.

[00:10:28]So this will be 2 t + 3 g.

[00:10:37]So question four, what is the position vector?

[00:10:54]Vector that specifies the location of a particle relative to an origin. That is what we call the position vector.

[00:11:02]Position vector is a vector that specifies the location of a particle relative to an origin.

[00:11:12]And what is a velocity vector? The rate of change of position with time or change of displacement with time.

[00:11:23]So if you have question five, if you have the position vector r of t = to xti + yt + z tk and you are asked to find the the velocity is to differentiate that with respect to t which is equal to the x dt t plus d y d thet t plus the z dt and what will be the acceleration?

[00:12:01]The rate of change of velocity.

[00:12:04]Acceleration becomes the v dt which is equal to second derivative of the position vector the t².

[00:12:16]Question six.

[00:12:19]Given that the position vector is this, we can if you want to t the position vector r = t ra^ 3 i + 2t² g. If you are asked to find the velocity the differentiate with respect to the rt and differentiating 3 t^ 3 will be 3 t² and um 2 t² will be 4 t. We know how to differentiate.

[00:12:59]And at t = 2, we now substitute the value of t.

[00:13:08]If t is 2, 2² because we have 3 t². So substitute the value of uh t² which is 4 * 3 that is 12. Well, that is what we have here. and t for 2 that is 2 * 4 that is 8. We now have 12 i + 8 g. Then for acceleration we are talking about the v dt that's the rate of change of velocity with respect to time. So if you now differentiate 3 t² that will now give us 60 I. If you differentiate 4 t it to give us four. So at t = to 2 we also substitute that is uh 2 * 6 that will give us 12 i and uh 2 * uh at this point there's nothing to substitute. Why? Because we have 4 j. So this will be 12 I + 4 J.

[00:14:21]I hope we are getting along.

[00:14:26]So question seven, the first law, Newton's law of motion.

[00:14:40]Second law, Newton's second law of motion and third law. The first law state that a body remain at rest or uniform motion unless acted upon by an external force. We all know that Newton's first law of motion a body remains at rest or in uniform motion unless acted upon by external um forces.

[00:15:13]Then the Newton's second law force equal to the rate of change of momentum and we know that F is equal to ma then the third law Newton's third law of motion for every action there's an equal and opposite reaction.

[00:15:36]So question eight, if you are asked to find the magnetic force, magnetic force F is equal to Q VB. But if angle is involved, F now will be equal to QVB sin theta.

[00:16:01]We we know that. But if angle is not involved F will be equal to QVB and then the centrial force is equal to MV² / R.

[00:16:16]So if you now equate the centrial force and the magnetic force we can now solve for R. that is saying that QVB is equal to MV² / R and R now will be equal to MV / QB.

[00:16:40]So given the values we can substitute the mass of the electron and um the charge of the electron.

[00:16:54]So this we can determine the value of R.

[00:17:02]Then the momentum conservation in an isolated system.

[00:17:14]Momentum conservation in an isolated system.

[00:17:19]We know that action is equal to reaction and that's we know then for work done work done we have w = to the integral from 0 to x kx dx and w is equal to integral from 0 to x kx dx and w =/ k by integration we have/ kx².

[00:18:00]So if the values are given we can substitute and you calculate the work done.

[00:18:16]We can also define a simple harmonic motion of a restoring force.

[00:18:24]We can also define the damping.

[00:18:32]This have been explained reduction in amplitude over time.

[00:18:37]Simple harmonic motion. motion where restoring force is directly proportional to displacement and resonance the maximum amplitude at natural frequency.

[00:18:52]So this can be defined and then the energy conservation is equal to/ MV² = to G capital M small M all over R. That is the case of escape velocity. And you can make V the subject to be V = to square root of 2 GM all over R where all parameters have their usual meaning.

[00:19:24]Then if you substitute we can get the value of escape velocity because all these parameters are constants.

[00:19:38]They are all constants and they will always be provided.

[00:19:44]So these are the things you should get familiar with them and um you be good [snorts] to go. So if there's any question you can ask ask your questions because it is getting late and um I wouldn't want to keep you here for a long time.

[00:20:37]So if if you don't have any question I will call it today for the recap.

[00:21:03]Unit vector dot products cross productducts.

[00:21:12]We should be able to calculate the dot products cross products given these values.

[00:21:21]Calculate the resultance of two or more vectors.

[00:21:28]What is another name for resultant? The total effective You should be able to calculate the magnitude.

[00:21:36]You should be able to differentiate vectors.

[00:21:42]The position vector, velocity vector, you are familiar with the differentiation given the values at t = to so of values.

[00:21:56]You'll be able to that first law, Newton's first law. Second law, third law, magnetic force, momentum conservation, work done, simple harmonic motion damping, resonance, escape velocity and so on.

[00:22:23]So I wish you guys [snorts] good luck.

[00:22:33]If there's any question, I'm still waiting. But if there's none, ask your questions.

[00:23:12]Okay, [snorts] somebody's asking how can I calculate Energy conservation.

[00:23:31]How can you calculate energy conservation?

[00:23:42]We know that the kinetic energy is equal to that is half mv² is equal to capital g m small m all over r² that is for escape velocity you can always solve for v.

[00:24:08]You can always solve for V. Have MV²= to G M s capital G M small M / R².

[00:24:16]So you can make V the subject [snorts] that is saying that V is equal to square root of 2G M all over R.

[00:24:26]and all this uh uh somebody's asked a question here.

[00:24:42]Um I am confused about Q E I A. I don't understand that.

[00:24:54]Q E I A. You can put that question across again.

[00:25:03]Okay.

[00:25:10]If there's no any other question, we'll end the stream.

[00:25:18]Thank you.