 So applications of differential equation continued. Applications of ordinary differential equations continued. Why continued? Because we had already discussed one of the applications which was orthogonal trajectories. So we have already discussed orthogonal trajectories. So today I am going to begin with the growth decay problems. The growth decay problems. Growth and decay problems. Such kind of problems are primarily seen in chemistry also under radioactive decay. Many a times we talk about growth decay problems with respect to population, with respect to growth of money, with decay of radioactive substance. That is why the name of growth and decay has been given to it. So let's quickly start with questions based on the same because from theory point of view there is nothing very important with respect to theory because you have already done it. So let's start with the question. So let's start with this question. NPS, YPR, do you have school tomorrow? Anybody from NPS, YPR who can confirm that? Because I remember somebody told me that you have some extra classes. Okay till 12. Okay fine. Thank you. All right so here we have a question. The population of a certain country is known to increase at a rate proportional to the number of people present, presently living in the country. If after two years the population has doubled and after three years the population is 20,000, estimate the people initially living in the country. I'll give you an idea how to solve this but only after you have tried it out because I just want to give you some time to think about it, how to solve these kind of questions. So the idea is pretty simple frame a differential equation. So how is the population changing the rate at which it is changing? They have already mentioned it to you. So frame a differential equation. I think they have given certain set of boundary conditions also utilize it and then find out how many people were living at time t equal to zero in the country. Once done you can give me a response on the chat box. So let us say N is the amount of people living in a country okay number of people living at time t. Now you have to figure it out Siddhant. We have to solve the differential equation and check whether it is exponentially related. So what are the first thing that I can say? You know reading the statement the population of a certain country is known to increase at a rate proportional to the number of people living in it. So dN by dt is directly proportional to N okay and we can put a proportionality constant k okay in place of okay that proportionality constant where k is some positive number. Why positive number? Because the question has categorically mentioned that there is an increase in the population. Okay okay so I think we have already formed a differential equation. So this is a simple variable separable solve for it and see what do you obtain. So you get dN by N is equal to k dt okay integrate both the sides. So end up getting ln N is equal to k t plus a c okay. Now few things have been provided to us few things have been provided to us that after two years the population has doubled okay. So now let's do one thing first of all at t equal to zero we know that the population is N zero okay where N zero is the initial population of the country or initial number of people living in the country. So when t equal to zero N becomes N naught okay. So c becomes ln N naught correct. Now place your c as ln N naught and send it to the other side. So ln N minus ln N naught is equal to k t in short N by N naught ln of it is equal to k t okay. Now let's use the information given to us after two years the country has doubled the population of the country has doubled and after three years it is 20,000 okay great. So N is equal to 2 N naught when t is equal to 2 okay. Siddhas has got some answer okay fine. So use this fact over here and you'll end up getting ln 2 is equal to 2k that means k is equal to half ln 2 yes or no yes or no okay. So using this fact again in this expression what do we get what is k here proportionality constant because they just said that it is directly proportional to the population right. So once you put that value back here what do you end up getting you end up getting ln N by N naught is equal to half ln 2 into time yes or no. Now they also mentioned that after three years the population has become 20,000 that means now I'm using a fresh piece of information provided to me that t equal to 3 N has become 20,000 that means if I put it over here ln N by 20,000 this is half ln 2 into 3 okay. So let's try to solve this so this is 3 by 2 ln 2 and this is ln N by 20,000. So N by 20,000 is as good as 2 to the power of 3 by 2 2 to the power of 3 by 2 is 2 root 2 oh I'm so sorry other way around I'm so sorry this is going to be this is going to be 20,000 and this is going to be N naught our our agenda was to find out N naught yeah so this is going to be 20,000 and this is going to be N naught so N naught is going to be 10,000 divided by root 2 yes or no root 2 is roughly 1.4 ish so if you calculate it how much does it come out to be roughly 10,000 divided by a root 2 divided by root 2 roughly comes out to be 7 yeah 7 0 7 1 kind of a thing is this fine any questions exactly so initially somebody gives some 5,000 something no actually 7,060 okay number of people's are in decimals you have to approximate it okay it's it's a rough estimate okay so depending upon the options of course options will be significantly different okay it will not be like very close to each other but always express it to the next you can say smallest figure not the highest figure next smallest figure is it fine any questions any concerns is the approach clear for this type of question okay we'll try to take one more question this let's take another one okay I'll be I'll be giving a question from my side okay the question is five mice in a stable population of 500 are intentionally are intentionally infected with a contagious disease infected with a contagious disease okay just to test the theory of epidemic okay now the rate of change the rate of change in the infected population change in the infected population population is is twice the product of is twice the product of the number of mice the number of mice who have the disease who have the disease disease with the number that are disease free with the number data disease free okay so I hope this part of the question is clear so the change the rate of change of the infected population is the product of the number of mice who have the disease and the number of mice who are disease free okay okay now how long how long will it take will it take for 90% of the population population to contract the disease to contract the disease time is to be considered in days okay time to be taken in days is the question clear to you the question is initially they were okay uh 500 uh you know mice present in a stable population stable population means uh this was a healthy population of mice so what happened is initially five five of them were out of that five hundred five were intentionally injected with some contagious disease just to check the spread of the uh you know that epidemic so it was found at the rate at which the change of the infected mice you know uh increased was twice the product of the number of mice who had the disease and the number of mice who were disease free okay how long will it take how long will it take for 90% of the population to contract that disease more decisions more trading happened yesterday itself right six fifteen to seven fifteen i don't know whether it's a different calendar that you follow yes correct because once the the product will become maximum when the number of infected and non-infected are same yeah absolutely right and then we'll follow okay is it coming exactly six days because you know i'm not very sure about the exactness of the answer or you rounded it off okay many times for such questions they will give you some critical values if at all you are not using because you're not be allowed to use a calculator no so whatever numbers you will be seeing they'll give you the you know values for those numbers but as of now if you feel that you need a calculator for calculating something uh you may use it but don't worry when it comes actually in the examination hall they will give you some critical values which will be required in the problem solving example if you need some loan or something they'll give you don't worry so first frame a differential equation so i've got one answer so far we'll wait for this to respond okay should we discuss it now anybody who wants me to wait 30 seconds okay fine take one minute take one moment no worries in vectors uh have you completed a dot product cross product stp btp okay coplanarity of uh three vectors and coplanarity of four points those concerns linear dependence non-linear dependence kind of a thing right okay so here can we say that they have mentioned that the rate of change in the infected population so let n be the number of mice infected at a given time t let n be the number of infected mice at a time t okay so as per the given uh information dn by dt is twice the product of the number of mice which are infected and the mice which are not infected so out of 500 n are infected so 500 minus n is not infected so this is the situation given to you at hand in short you have been provided with something like this okay so we are separating the variables now and let's try to apply uh partial fractions here so this is going to be if i'm not mistaken this is going to be something by uh 500 n minus n plus something by n okay if i'm not mistaken i could choose both of them to be one by 500 because if i do this it conveniently gets the n cancelled from here and 500 also gets cancelled because of this one by 500 okay so using partial fractions we can basically write the term on the left hand side to be like this in short you are basically trying to let me write it like this you are basically trying to integrate this dn is equal to thousand d thousand d correct all right let's integrate both the sides so this is going to give you a negative ln uh 500 minus n plus ln n this is thousand t plus c that means ln of n upon 500 minus n thousand t plus c okay now use the fact that when t equal to zero the number of population number of mice who were infected by this infectious disease were only five okay so only five mice at t equal to zero were infected by this infectious or contagious disease so ln five by 495 which happens to be one by 99 if i'm not wrong this is equal to c correct that means c is equal to ln one by 99 okay so send the c to the other side so it's ln n by 500 minus n minus ln one by 99 is thousand t okay now this is plain and simple this is going to be 99 n by 500 minus n correct me if i'm wrong this is equal to thousand yes now what are they asking at what time almost 90 percent of the population will be infected by the disease so n should become 90 percent of 500 which is almost 450 mice are now infected so what is the time at which 450 mice are infected by the disease correct so just put here n value as 450 so it's really 99 into 450 upon 50 okay is equal to thousand t so this will go by this thing so the answer will be one by thousand of ln 495 okay so how many days it is let's check it out in time in days in days so basically you're you know writing your answer in terms of days oh i'm so sorry so sorry this is nine yeah so it'll become 981 isn't it so as per my phone's calculator sorry i'm using the phone because i've not been provided with the value of this so this will come out to be almost point ln of 691 comes out to be 679 almost 6.79 okay so it is almost like within this many days i mean you can count it in terms of seconds or hours also within within 6.79 to 10 to the power negative three days almost 90 of the population would have got infected by the disease it's a very very less time okay so this disease will grow very very fast and within this many days okay the entire population would have got 90 of the population would have acquired the disease any fine any problem everything is fine any questions any concerns i mean i have chosen it to be in days it is like up to me to choose it in seconds also because this unit i have to choose if you find it in terms of seconds so in one day there are 12 into 8600 okay into 6.79 by 1000 so almost around in 600 seconds this comes out to be somewhere around uh roughly 587 seconds okay so less than six minutes sorry less than 10 minutes the entire population will be 90 percent of the population will be infected where above manigar you are not paying attention to the question what did i write over here didn't i write in the question time is to be taken in days question center it's the prerogative of the question center to give you the time right whether he's talking about time in days or seconds or minutes okay clear above manigar yeah okay next type of question that we are going to talk about is uh newton's law of cooling now you have already done this in class 11 physics so the rate at which the temperature of a substance changes is directly proportional to the difference of the temperature with the surrounding temperature okay now rate of cooling when you talk about rate of cooling the t value is always more than the value of the surrounding right so the rate at which the temperature decreases okay will be directly proportional to the difference of the temperature of the body with the surrounding temperature okay s is the surrounding temperature now you already know this information from your newton's law of cooling correct kinshuk kinshuk is surprised what happened kinshuk anything that i've written wrong you asked something yes yes yes you can use definite integral from five to yes absolutely right that c value that you are finding out if you can do away with the c value if you put your limits of integration definitely you can do that okay now please understand here please understand here that this situation will be provided to you in your examination is in your question because they don't expect you to know this being a math student yes being a physics student you may be expected to know this beforehand so please watch out for the information given to you with respect to any cooling or heating related issues so here i'm assuming that this information would be provided to you okay in the examination you know in the question itself there will be mention this will be mentioned to you so we'll take a question based on the same okay so let's let's take a question based on the same okay now let's read this question the question says a body at a temperature of 50 degree Fahrenheit is placed outdoors okay where the temperature is 100 degree Fahrenheit so now here the heating is happening because the surrounding temperature is more okay so if the rate of change of the temperature of the body is proportionate to the temperature difference between the body and the surrounding medium and if five minutes if after five minutes the temperature becomes 60 degree Fahrenheit then part of the question is how long will it take the body to reach the temperature of 75 degree Fahrenheit and part b how long will the sorry what will the temperature of the body after 20 minutes there's no this sorry sit down i missed your question yes as i already told you in the examination room and the exam in the exam they will be providing you with the required log values of those miscellaneous you know a lot expression that we're coming up okay they don't expect you to remember log and go don't worry okay so as per the information given to us there has to be an increase in the temperature of the body because it is kept in a it is kept in a medium which is at a higher temperature right so let's say t is the temperature of the body at a given time t temperature at a time t so because it is kept in a medium which is at a higher degree temperature the there will be an increase in the temperature of the body with respect to time so k being the proportionality constant here k has to be positive okay or if you want to write t minus 100 like the way I've written over here then you have to keep your k as a negative because overall there's an increase in the temperature so once you have obtained this it's a variable separable so you can write it like this integrate both sides so this is going to give you negative ln 100 minus t is equal to kt plus c now as per the as per the question at t equal to 0 the temperature of the body was 50 degree Fahrenheit that means negative ln 50 is equal to c put it over here back okay so negative ln 100 minus t is equal to k yeah is equal to kt minus ln 50 okay so that's going to give you ln of 50 divided by 100 minus t is equal to kt now I need one more piece of information to get my k value and I think they have already provided us that after five minutes okay the temperature of the body is 60 degree Fahrenheit so I'm not taking my unit to be in minutes so when t is equal to 5 uh capital t that is the temperature of the body becomes 60 okay so ln of 50 by 40 is equal to 5k so k is one fifth of ln 5 by 4 put it back over here and I think that will give you a final relationship between the temperature and the time so instantaneous temperature and time would be related to each other by distillation in fact t by 5 ln 5 by 4 now what is the thing that they require let's try to figure it out they're asking us how long will the body take to reach a temperature of 75 okay so what is t when capital t is 75 so when you put it over here you get ln 2 if I'm not mistaken because 50 divided by 25 is this okay and t by 5 ln of 5 by 4 okay so let's see what we can do with this figure so as per our expression it should be 5 ln 2 upon ln of 5 by 4 again they have not mentioned as the values without that it'll be difficult for us to figure it out ideally they should have given us the values so again I'll use my uh mobile device to find it out so 5 ln 2 okay whole divided by ln of 1.25 it comes out to be roughly 15.53 aditya your answer is coming very very different shittish and mine is matching you got the same figures this figure only came out okay if this figure has come out then it's fine no need to worry much okay so this comes out to be roughly 15.53 minutes 15 and a half minutes you can say okay all right so we'll move on to the next part of the question next part of the question says the temperature of the body after 20 minutes so now this time small t is given to us that is 20 and capital t is unknown for us so let's put that these values here so ln 50 by 100 minus t is equal to 20 by 5 ln of 5 by 4 now this is actually simpler because this is 4 times ln 5 by 4 is equal to ln of 50 by 100 minus t so ln of 50 by 100 minus t is equal to ln of 5 by 4 to the power of 4 so 50 by 100 minus t is roughly 1.25 to the power of 4 so t will be 100 minus 50 divided by 1.25 to the power of 4 whatever it comes out to be roughly how much is it let's check it so 50 divided by 50 divided by 1.25 to the power of 4 that comes out to be roughly 20 and minus 100 will give you 79.52 okay roughly 79.52 degree Fahrenheit is it fine now don't worry about these calculation part for that all the required information would be provided to you here I don't know I was surprised to see that information not being given but normally they give that information is it fine any questions any concerns with respect to this so it's all about you know framing the differential equation and solving it anywhere any place you feel left out disconnected please highlight what was the second part answer again yes and second part answer was this much 79.52 degree Fahrenheit okay all right so with this we now move on to the geometry based problems okay 2d and 3d geometry based problems geometry based problems as you can see we are not taking many problems of a type because I think if you're aware of the process more or less you have to follow the same thing for almost all type of questions so one one example is good enough okay let's take this question find the equation of the curve which is such that the area of the rectangle constructed the area of the rectangle constructed on the abscissa of any point and the intercept of the tangent at this point on the y-axis is equal to 4 okay read the question once again if you're not very sure about what this question is saying please read it couple of times the equation of the curve which is such that the area of the rectangle constructed on the abscissa of any point and the intercept of the tangent at this point on the y-axis is equal to 4 please send it to me I'll clarify them so that's you can you can message to me what are the questions where you want clarification either I will send you a voice message or I will solve it and send it to you okay so that did you see the solution I mean first all the DPPs we provide with the solution so did you see the solution first okay still you didn't get it then you can send it to me aditya has a question what does the constructed on the abscissa mean okay now see see let's say there's a curve I'll just draw a rough figure over here let's say this is here this and you're drawing a tangent at this point and this point is cutting the y-axis at some point so let me draw a tangent let's take the point here instead of there okay let's say the point is here now basically what they're saying is that if you make a trying rectangle like this this rectangle this rectangle has area of four units is it fine now from here on you have to proceed further I think I got an answer from shittage okay shittage now one of the major concerns that people show initially while solving the problem is that sir we normally write our equation in terms of small x and small y right that's the norm that we follow but here the point is also taken as a small x and small y so it looks very you know odd and incorrect also to write something like this y minus y is equal to dy by dx x minus x correct because the point is also taken to be small x and small y so please don't do this while you're writing it else you will get yourself badly confused okay so the better way here is instead of using see you have already designated your unknown point to be small x small y so don't use the same notation while you're writing the equation of a tangent so use capitals okay that is what what I suggest so use capital x and capital y to represent your generic x comma y or you can say your equations x comma y okay now you are basically you know finding what you are finding mod of right the y intercept whatever the y intercept of this with the x coordinate over here so this length is x coordinate isn't it so x into whatever is the y intercept coming out so what is the y intercept coming from from this you can obtain the y intercept by putting your capital x as zero okay so when you put your capital x as zero you'll end up getting capital y minus small y as dy by dx into negative x in short your y intercept becomes y minus x dy by dx so as per the given problem x times y minus x dy by dx is equal to four units is equal to four units is this fine any questions any concerns all right so from here I would end up actually getting x times y minus x dy by dx as plus minus four okay let's account for both of them together by using plus minus four wherever we can so this gives us x y minus x square dy by dx is equal to plus minus four okay so far so good no issues all right so can I divide with a negative x square throughout so when I divide with negative x square throughout I'll get minus y by x and this will become you know you can keep writing plus minus also doesn't make a difference because both of them you know are taken care of some people do it separately also some people can some people you know do not remove that plus minus thing so that they don't do it twice the same thing okay now what kind of a differential equation is this if I put a plus sign here and I put a negative one by x doesn't it look like dy by dx is equal to py equal to q where p and q both are functions of x or constants correct so this is clearly an LD okay so what is the if if it will be e to the power integral of minus one by x dx which is e to the power minus ln x which is actually one upon x so this is as good as y into one by x is integral of plus minus four by x cube is it fine okay so this will give you plus minus if I'm not mistaken two by x square plus c so y by x is equal to plus minus two by x square plus c any questions any questions any concerns to highlight aditya how did you get that hyperbola rectangular hyperbola and I think very similar shittish also got some kind of a weird expressions any mistake that you did while you were forming the differential equation oh okay okay okay oh it's a differential equation itself was incorrect okay anyways is this fine any questions okay we'll take one more but before that if you have any questions if you want to copy down anything please do so yes shittish tell me any questions you have rectangle with area two send me the screenshot okay send me the screenshot we'll check on my whatsapp yeah okay so we'll take a another question let's take this almost you know our physics question that you probably would have done in your fluid dynamics chapter find the time required for a cylindrical tank of radius small r and height capital H to empty through a round hole of area a at the bottom the flow through the hole is according to the law the velocity is k times under root 2 g ht ht is a function height function with respect to time I mean it's a height at a given time t okay vt and ht are respectively the velocity of flow through the hole and height above the hole at a time t g is the acceleration due to gravity okay so this shittish has sent me a photograph one second yeah so this that intercept the tangent at a point is meeting at three right okay and the width of the rectangle is almost 1.3 ish I mean if I just look at my naked eyes so doesn't it become a four how are you saying it to no no you you didn't get the question properly see you you made a tangent here right it met here right okay so basically you have to find out this area that is four no as per the diagram also you were looking at some different area I think I think you're looking at abcd s s correct sir can you explain the last question last question means this question the one which was before that this one the square one square one no it's a rectangle one by the way yeah see what it says at a generic point x comma y if I draw a tangent it meets it meets the y axis at some position okay from there if you make a triangle like this which is basically having its you can say width to be x and height of the triangle or the you can say this is the width and this is the length so length of the triangle is the x coordinate of that given point p where you are sketching the tangent and width is where this tangent is meeting the y axis so what I did I first assume the equation of the tangent to be this so where does it meet the y axis for that capital x is made zero which happened here so I got y value as y minus small x dy by dx so area of a triangle is length into width which is x into this and this is equal to four of course I've taken a mod because I don't want a negative answer to come out so from there I got this and that led to a linear differential equation which I solved by using my idea of linear differential equation where is the doubt in this case which part you did not understand the part where you took mod this part length into breadth is equal to area of the triangle what is the doubt in this length is x width is y minus x dy by dx I did not take mod so if you do not take mod that doesn't mean your my method is wrong and your method is right you have to take mod that's the learning why you didn't take mod what is the product was negative so where is the part you didn't understand don't say I didn't do it the same way you said you did not understand where is the part you didn't understand that you did not take mod that is your mistake but where you did not understand that is something else coming back to this question no you said no you want me to explain some parts so explain means you want to know where is where is the lacking in your understanding right okay wanted to see the solution okay okay so I hope everybody would have experienced such kind of a question in your physics you know fluid dynamics topic now many of us know that many are solving such kind of a question you can use your equation of continuity isn't it everybody knows equation of continuity a1v1 is equal to a2v2 right that comes from the fact that the mass of liquid at any junction is conserved right mass is neither created nor destroyed so let's use whatever concept we know to you know crack this problem okay so aditya has given some response okay anybody else okay okay pradyan only till six yes today only classes till six once you're done with the application we are done with the class also because you already had a class three hours in the morning so I didn't want you to sit for another three three and a half hours okay so uh think that okay no need to write type out the answer some of you are you know kind enough to type it out it's okay don't take that don't bother yourself typing so much so much see first of all can I say the rate at which the volume is decreasing here is the rate at which okay so the the decrement of the volume from here plus the volume that comes out from here that should be equal to zero isn't it okay or in short can I say pi r square dh by dt is under root of k under root of 2 gh I'm not writing t in the bracket and all because it just confuses the viewers because many times we think it's a product of h and t okay into the area of the hole but remember this is there will be a negative sign over here okay why because this is the amount of water volume of water flowing out per second from the hole right so here if you compare it directly with pi r square dh by dt and dh by dt will be a negative quantity because height is reducing or decreasing with time so you cannot say the volume of water gushing out of the hole where everything is positive over here okay is equal to a negative quantity that would be you know as an equation it would be an incorrect way to represent so you have to put a negative sign over here now this is a clear cut variable separable once again I can say dh by root h is is k into under root 2 g into a by pi r square you can put a negative sign and dt of course now here you can do a simple activity remember the height was initially capital h and you want it to become zero and here let's say t equal to zero and at time capital t it basically mt is completely that means the the cylindrical tang is totally empty at capital t time okay so you can directly use your definite integrals concept over here right so this is you know simple let's do this so this is h to the power minus half plus one by minus 0.5 plus one is 0.5 so sorry about that so this becomes if I'm not mistaken two under root of h okay and you can observe the negative sign here and put it from 0 to capital h this will be capital t minus 0 okay so to under root h you can you know write it like k under root 2 g into area by pi r square t so it becomes a super easy question it becomes 2 under root h into pi r square by by k under root 2 g is it fine now the very same question let us say somebody gives you in the form of not as a cylinder but gives you a form of a semi spherical tank or hemispherical tank okay and he gives that the height of the tank initially height of water in the tank initially is equal to the radius and again there is a similar type of a hole there from where the water is basically you know leaking okay so the question can be framed in any structure all you need to do is carefully figure out this differential equation that you have formed initially here it was quite simple but if I make the structure of a cylindrical tank to be a hemispherical tank then what all changes will happen please try this out as a question so think as if this hole is of area a okay again the water is coming out at a rate of let's say you know k under root of 2 g h h being the height instantaneous height at time t okay so let's say this is your small h fine not now as a homework okay and then figure out what is the time required for tank to empty for tank to empty so please try this out for a homework question as a homework uh siddhant when I am writing something you should focus on copying that part down rather than worrying about something which has already been written right so I'm giving you a question to work out for your homework isn't it any less importance is this any less important than what you have done in fact you will be surprised to know this question came I think in j 2001 this question which I'm giving you okay yeah please please when you when I'm saying something when I'm explaining something don't worry about copying something which has been already written so I'm giving you a question that you have to solve for a homework so just note that down first then we'll come back to whatever you want to copy no 2001 there was no mains yeah you can say the mains that time was j advanced and near about 2006 there used to be uh separate brilliance round and the advanced round so advanced round used to be called as j main exam yes so it has gone a lot of I don't know name changes it j exam then they made it the j advanced exam so currently the name that we have we got it post I think 2008 or 2009 I believe before that they used to say j at j it j exam and even before that they used to call it as you know uh j main exam and there used to be a prelims exam so what what main exam you have written used to be the prelims and j j main exam used to be the advance of now okay so this came in j advanced equivalent to now yes they they have made the name slightly sounds scary okay fine copied everybody done any questions so now we'll move on now we'll move on in fact I think we can take one more question one more question on this because geometry based questions are very important type let's take this one I hope you can read this written in small font but I think you can get it given the curves y equal to f of x passing through 0 comma 1 and y equal to minus infinity to x f of t passing through 0 comma half the tangent drawn to both the curves at the point with equal abscissa intersect on the x axis then you have to find the f of x curve do you want me to run the pole okay let me draw launch the pole at least so that that I keep the idea of the time also gone the tangent drawn to both the curves at the point with equal abscissa intersect on the x axis okay excellent I've got one response in near about one minute yeah yeah if you want to change your answer you can change it by typing it out on the chat box which answer you think is correct if at all by mistake you have pressed something which you feel is incorrect and you figure it out later you can message me on the chat box that I feel this is the right answer because anyway I don't get to see who has voted for which option okay so far nine people have answered it's almost four minutes I can give you one more minute as of now and then we'll discuss it out because we have few more concepts left to be covered I don't want to run myself out of time giving too much time to this question okay five four two one so most of you have said option number c okay most of you have said option number c let's check it out see let's talk about the first curve y is equal to f of x curve okay and let's say I'm drawing the tangent at a point x comma y so what's the equation of the tangent equation of tangent at x comma y so let's write down the equation of the tangent which is going to be capital y minus small y is equal to f dash x capital x minus small x okay now for the other curve for the other curve I'm drawing the tangent at such a point which has got equal abscissa okay so let's say the curve that has been given to me I'm calling it by the name of okay let's keep it this only so let's say this is my curve sorry to see that they have missed out a dt over there okay so now you are basically drawing a tangent at a point x comma let's say y one okay so equation of tangent drawn at x comma y one okay is what capital y minus y one okay equal to derivative of this function what is dy by dx at x comma y one can I say it'll be f of x only correct by Leibniz rule so if you differentiate this with respect to x right we have to put x in place of t into derivative of x then minus minus infinity in place of t or at least t tending to minus infinity but derivative of minus infinity being a constant will be zero correct so in place of sorry in place of t I'll be putting an x and that gives me the expression for the dy by dx at x comma y one isn't it any questions here okay now the question says the tangent drawn they both meet each other they both meet each other on the x axis okay great so let us first put capital y equal to zero over here and see what is my capital x coming out so when I put capital y is equal to zero my capital x will come out to be this divided by this and x will go there okay yeah so it'll become x minus y upon f dash x right similarly if I put capital y equal to zero here let's say what what does capital x come out to be everything will come out to be the same is just that you have a y one in place of y and you will have f of x in place of f dash okay now since they're meeting at the same point these two values are equal right so they should be same because they're meeting at the same point on the x axis so the x coordinate of the point of intersection would be the same in short what are you saying you're saying small x minus y by f dash x is small x minus y one okay by f of x in short minus minus also goes in short you'll end up something like this f dash x by f of x is equal to y by y one am I right yes or no okay now y by y one what do you understand by y by y one let's say I call this by some name instead of writing a y I call it as gx I call this as gx so can I say y by y one y by y one is as good as saying g dash x by gx because this this y is also satisfying this condition see this y one is what we this y one is satisfying this right no I think once again this will become once again once again this small correction here see this y one is coming from g of x and this y is coming from f of x isn't it correct so it is y is f of x by y one g of x sorry about that yeah correct yes or no see this y that you had taken this y is basically satisfying this property right at that x y is f of x isn't it so this y that I had over here this y that I had over here after I remove everything let's say I remove these yeah this y I change it to f of x okay and y one is this curve right y one is a point which is satisfying this curve right so y one is actually your g of x so this y one is your g of x okay so far so good any questions so far okay and yes you were right f of x is g dash x now it has come like this okay initially I jumped onto this but now it is clear okay so once you have got it you can now apply integral on both the sides with respect to x okay so you can integrate both sides with respect to x and this gives you ln of f of x is equal to ln of g of x okay any any confusion or any question so far okay and we need to add a constant of integration correct okay so after this what should we do we already know we also know the fact that g dash x is your f of x right g dash x the derivative of this is your f of x can we use that can we use that can we use this information somewhere fine so okay this this step is obvious then what else what else I like so from here I can say g dash x by g of x is again a constant correct which means if I integrate it once again can I not say ln of g of x becomes c of x plus a k in short I can say I can say g of x becomes sorry for that crooked handwriting once again yeah can I say g of x will become k e to the past c x okay now please note many many of us we fail to see that this is actually this that is e to the power k into e to the past c x where e to the power a can be written as a k it does anyways a constant of integration yes or no so if this is g of x what is f of x f of x is the derivative of this term which is k c e to the power k x now let us use some information provided to us with respect to the point we already know that the function f of x passes through 0 comma 1 right which means this is 1 when x is 0 that means k c is equal to 1 let's call it as the one set of information and we also know that g passes through 0 comma half 0 comma half so here in this if I put half and this is 0 k becomes half isn't it so if k becomes half c becomes 2 as per this condition okay so when c becomes 2 and k becomes half your f of x that is what they're asking us your f of x will actually become 1 e to the power 2 sorry k is yeah k is half oh sorry this is this is c not k this is c so it becomes 1 e to the power c x c x is 2 x which means your function f of x let me write it as a side your function f of x becomes e to the power of 2 x which is clearly option number c option number c is correct okay so I'll repeat once again what all I did yeah yeah you could have scammed it but my purpose is not to tell you a shortcut my purpose is to rigorously solve it see once we got this stage okay what we did we basically integrated both the sides and we ended up getting this again let me remind you instead of c I could have written l and c also doesn't make a difference fine and now I use the fact that f of x is g dash x so here I replace g dash once again brought this g of x down integrated both sides with respect to x once again which led to this expression okay many a times we can write l and k also here so from here I ended up g of x as k times e to the power c x so f of x being derivative of g f of x will come out to be this of course I've used differentiate differentiation of an exponential function now once we have got this we got our k and c from the boundary conditions provided to you in the question that means your g was passing through 0 comma half so when you put x as 0 g of 0 becomes half so half is equal to k and from here I ended up getting again 1 is equal to kc so c became a 2 so k is half c is 2 so kc is 1 e to the power 2 x that finally leads your function to be e to the power 2 x good enough any questions here all right so now we take up a dilution question dilution you can say application so another application of differential equation in dilution based problem dilution now let me first give you a scenario okay I'll give you a scenario and we will try to frame a differential equation on that scenario we'll not solve it we'll solve it when there is an actual figure provided to us okay so let us say there is a tank okay this tank is filled up with some solution let's say it is filled up with brine okay we all know brine is a concentrated solution of sodium chloride right is my chemistry good enough okay so this is a this is a tank which is full of you know it is filled it has brine in it and let's say there are v0 liters of okay so this volume of this tank is v0 liters of brine in it okay and you can say the amount of salt which has been put into this is let's say AKG's okay so putting AKG of salt in v0 liters of water you have created a brine solution like this okay so AKG of salt is being put fine now this tank there is a hole at the bottom of this tank and through this the brine solution leaves at the rate of you can say f liters per minute okay so through this hole in the tank the brine solution which was already in the tank leaves at the rate of f liters per minute okay now at the same time you are pouring in fresh you know brine solution which has been you know which has got the concentration as let's say b kg's per liter okay and you are pouring it at a rate of let's say e liters per minute okay so there is a fresh brine another brine solution another brine solution and this brine solution basically has the concentration of brine as b kg's b kg's per liter and you are pouring in this is the rate at which you are pouring in rate of pouring which is e liters per minute okay write a differential equation write a differential equation what will be the ordinary differential equation for q q being the amount of salt in the tank at a given time t amount amount of salt at time t so write an od for a rate of let me write it like this for a rate of change of q which is the amount of salt q is the amount of salt okay so what is the how would you write an expression for dq by dt yes only for the amount of salt so you have to write the differential equation for the rate of change of the amount of salt okay in terms of the given parameters given to you in terms of whatever v naught a f b e whatever you want to use from there so remember two things are happening there was already a solution of brine okay v naught volume containing a kg of salt and f you can say f liters per minute brine solution is lost from the tank but at the same time you are pouring in e liters per minute of another brine solution whose concentration is b kg's per liter let's write down okay we'll check out kinshuk whether your response is correct or not see let's talk about let let q be the amount of salt present in the solution amount of salt at time d what will be the volume what will be the volume at time d let's say v is the volume at time d volume of brine in the tank at time d so what will be the volume so you can see that v naught was the initial volume you are losing water at the rate of you're losing brine at the rate of f liters per minute so f t will reduce but you are still pouring in at the rate of e t so this will be the volume at a given time t so can i say can i say the amount of salt that you are adding is be per unit time but you are losing at a rate of the amount of salt per unit volume into f do you all agree with me on this or not okay so your final expression would look like this d q by d t is b e minus q f by v zero plus e minus f times t so this is the differential equation that will get formed now of course i will provide you with a question where you would be given b e you know all those f v naught etc then you can solve this equation by whatever technique you feel like so this is how see what is my purpose here my purpose here is to just you know explain you how to frame the differential equation for this dilution process which is happening dilution or concentration depends upon which is more b is more or e is more so we will talk about we'll talk about you know this in terms of numbers okay any question with respect to framing it y plus y minus everything is clear so this is this is helping to increase the concentration this is helping us to decrease the concentration because they are losing this amount of salt you're gaining this amount of salt see when you are when you are you know pouring in the solution this b term is clear right kinshuk okay now let's say q is the amount of salt so if you divide salt by volume so this will be the kgs per liter salt and you're losing it at a rate of f liters per minute so this amount of salt will leave the tank so this amount of salt enters the tank enters the tank per unit time okay and this amount of salt whole thing will leave the tank the salt leaves the tank so you have to subtract okay we'll take one small question based on this very simple one q by that is the density yes you can say the concentration and f is the volume that is that you're losing out so overall it gives you a you know this this becomes amount of salt per unit time let's take a question i hope you can read this question maybe it's written in a small font okay i'll put it here see this question says a 50 gallon tank okay they have used units of gallon no issues gallon and okay my my volume was in liter this is in gallon doesn't make any difference so a 50 gallon tank initially contains 10 gallon of fresh water at t equal to zero a brine solution containing one pound of salt per gallon is poured into the tank at the rate of four gallons per minute while the well stirred mixture leaves the tank at the rate of two gallons per minute uh they have asked two parts in this question one is the amount of time required for overflow to occur overflow to occur and the amount of salt in the tank at the moment of overflow now many of the student asked me said in this differential equation there is no a being used you also talked about the a being present here how do you plan to use this a or is this a immaterial to us can you tell me why where that a would be you know taken into consideration right so basically that a is actually the amount of concentration or q value at t equal to zero so please you know use that to get your c value exactly as it yeah absolutely right so okay so without much waste of time let's figure it the data just plug in the data here what is b e can i say as per this data b is one this is your b value so let me put those expressions here b is one e is four because you're pouring the uh brine solution the freshly prepared brine solution at a rate of four gallons per minute what is f what is f f is two because two gallons per minute is leaving the solution okay and v naught initially there was 10 gallons of water so v naught is your 10 only okay so as per the expression will be v minus q f f is two q by 10 plus e minus f e minus f is two t e minus f t okay in short you ended up getting a solution you ended up getting a differential equation not solution differential equation which is actually a what kind of a differential equation is this can you let me write it like this uh one by five plus t q equal to four what kind of a differential equation is this okay good enough all right so if this is an le uh let's find out our if integrating factor is e to the power e to the power integral of one by a five plus t dt that's going to be e to the power ln of five plus t which is five plus t okay so q into q into five plus t is equal to integral of four five plus t dt so this gives us q into five plus t is equal to 20 t plus 2 t square am i right put a constant of integration now uh put the value of q as 0 because a is 0 as a is 0 okay at t equal to 0 so this gives us this gives us c value as 0 itself if i'm not mistaken that means q expression is 20 t plus 2 t square upon five plus t now they're asking us uh two things here what is the time that it'll take for the tank to overflow okay that part is very simple you have 50 gallons of 50 gallon tank which 10 10 gallon of fresh water right and you can see the net influx the net influx is two right so already there is 10 gallon and there is a net influx of two so when will this touch 50 so answer is 20 correct so in 20 minutes your tank will start overflowing so first part of the answer is done is it fine because you need to have 50 gallons altogether initially there was 10 gallons and the water was uh you know coming the in in rush of water was at a rate of two gallons per minute because four was the one with four gallons with four gallons you're putting in two gallons is leaving out so there's a surplus of two gallons per minute okay so now once t is 20 minutes they're asking you what is the volume of the salt or sorry what is the quantity of the salt so you can put your in the value of t as a 20 so two into 20 square upon 25 how much does it give you I think we can cancel off a lot of terms here without evaluating it so this is 400 by 25 400 by 25 is going to be 16 and this is 2 into 400 by 25 16 again 32 48 48 pounds of salt will be there at the time of overflow there will be 48 pounds of salt in the tank is it fine any questions any concerns okay so I would like to do one question on electrical circuits with you uh that you are only aware of I don't have to tell you the Kirchhoff's voltage law etc for any kind of a circuit related problem if this is fine I can move on to the next so let's take this question there is an RC circuit an RC circuit has an emf given in volts as 400 cos 2t a resistance of 100 ohm and a capacitance of 10 to the power minus 2 farad okay so r and c values are given and it has got an emf also connected in the circuit initially there is no charge on the capacitor find the current in the circuit at any time t find the current in the circuit at any time t so the situation is basically like this you have a resistance you have a capacitance and you have a time dependent voltage source which is 400 cos 2t r is 100 ohm and this is 10 to the power minus 2 farad which is actually very large value farads normally capacitance are in micro farads anyhow so we have to find out what is the current in the circuit at any time t okay i being the current ir okay is equal to q by c is equal to the emf correct can i write it like this okay r itself is 100 and q itself sorry i itself is dq by dt so basically we have this so this is going to give us dq by dt plus q by 100 c 100 c is equal to 4 cos 2t this will also get cancelled off because they are same values and this gives you a plain and simple i mean linear differential equation now why did i write it in terms of q because i know the initial condition on charge there was no charge on the capacitor that means at t equal to 0 q was 0 okay so instead of writing an integral sign over here i wrote i in terms of dq by dt right so instead of integral i dt i wrote q by c now this is simple to solve if over here will be what e to the power integral of 1 dt which is e to the power t okay so q into e to the power t is integral of 4 e to the power t cos 2t is it fine any questions any concerns so far now this is of the type e to the power x cos bx so what is the solution here e to the power t by 5 okay correct me if i'm wrong it's 2 sin 2t plus cos 2t is it fine any questions of course put a c at the end now use the fact that q is 0 q is 0 when t is 0 sorry q is 0 when t is 0 right so you have put t is 0 and q is 0 both so that gives us c value as negative 4 by 5 so put it back over here so i end up getting q as by the way i'll just write it in one shot i divide by e to the power t so if i divide by e to the power t i'll get something like this and c was minus 4 by 5 so e to the power minus 2 minus t will be okay correct me if i'm wrong correct now they're asking us for the current current will be nothing but derivative of this charge so that's going to be 4 by 5 this is going to be a 4 cos 2t minus 2 sin 2t and this will be plus 4 by 5 e to the power minus t is it fine any questions any concerns so this will be your answer for the charge at any time t in the circuit sorry a current at any time t in the circuit any questions any concerns with this normal method means what normal method which normal method you're talking about see have you used have have we ever solved question where there is a forced emf okay in that you don't get a exponential function yeah of course phasers and all will come into picture exactly yeah so exponential function will come anyhow so before i leave i would like you to take a question as a homework because time is almost over so one question i will give you for a homework maybe okay you can try this out taken that question into account let's let me just check on the system okay i'll just write down the question here itself simple question i mean you would have done many of these questions as your physics question so there is a mass 2 kg mass which is dropped okay with no initial velocity no initial velocity so it's a free fall okay this mass experiences an air resistance this mass experiences an air resistance which is directly proportional to the square of the velocity so v is the velocity at a time t okay so square of the instantaneous velocity directly proportion to that or let me let me just remove the this thing it experiences air resistance which is equal to v square okay let me make the life simple find the velocity of the body at time t at time t find the velocity of the body at time t or you would like to do it now or would you like to do at a homework you can tell me now okay i didn't give you that proportionality constant and just i removed it so that your answer doesn't come out in terms of the proportionality constant that's it okay if you want to do it now you can do it so everybody first frame a differential equation and let me know once you're done with the framing of the differential equation quickly take g as 10 if you want okay take acceleration due to gravity exactly as 10 meter per second square so what is the differential equation that you will be forming m dv by dt is equal to mg minus v square m is already given to you as a 2 so dv by dt is equal to g g is 10 minus v square by 2 am i right now this is this is not even and i you can you can do this as a variable separable that's you know so easy so this is going to be if i'm not mistaken 2 dv by 20 minus v square is equal to dt hope i'm not missing out anything okay so this is 2 dv root of 20 square minus v square is equal to dt so integrate both sides so this is of the nature a square minus x square so a square minus x square is 1 by 2 a ln a plus x a minus x equal to t plus c i use the fact that t equal to 0 velocity was also 0 okay so t equal to 0 velocity also 0 will make it ln 1 so everything will become a 0 so c will also become a 0 so 1 by 2 root 5 ln 2 root 5 minus v by 2 root 5 sorry 2 root 5 plus v by 2 root 5 minus v is equal to t okay now the task is on you to now convert this right v in terms of t that's it because they have asked i think what is the find velocity at time t so you have to write v as a function of time okay so with this we happily conclude the chapter and before you disperse i have a very important announcement to make last week again we will have the last session on our regular class so thursday next thursday and friday okay that is the only two days when we will have a regular session i think since you will have school and school exams will be going on we cannot have any extra session one of the days maybe i think friday or thursday i will let you know so this is a combined batch everybody will be there so even hsr people who used to come on just on friday for maths you will be coming on thursday and rajah ji nagar people who used to come on thursday only you will come for friday as well so both the days will be the combined classes okay and in these two days we will try as much as possible to finish off auc area under curves and vectors 3d by the way this can be done easily because it requires only one class this requires more classes as told by kiran sir to me okay