hey friends my name is and you're watching you miss easy welcome to a new video for igcse at massive today we have rules and examples for rates of change and we'll start off with some basics but before you get into it don't forget to leave a like subscribe and you win nothing about selling them in any future videos so now we have some basics so we have an equation for the rate of change which is always d over d t like always d something over dt that for example dv over dt like v stands for volume to say it verbally it will be the rate of change of volume with respect to time so whatever d is something over dt is the rate of change of that thing with respect to time so one thing to note is that if the rate of change is positive then the item is increasing like the by that thing but the rate of change of that thing is increasing but if the rate of change is negative it is decreasing so this really one nice thing to know is that if dt over dx equals u you can actually reciprocal both sides to become dx over dt equals one over u then we have some basis for the chain rule so we have the chain rule for the rate of change so let's say if we have d of v or d y over d is basically equal to dy over dx times by dx over dt because we can transfer these two out to remain with dy over dt so here we have an extended rate of change which is basically with three different variables like three different things so that d y over d t could be d y over d s times d x over d u times d u over d t that will cancel out can sound like a telescope in series so here we have small changes so we have this symbol right here which sentence for delta it's a degree letter delta so here's a smaller case and here's a big bit like bit letter like um capital letter but we use a smaller smaller capital so delta right here delta is equal to the small change and so delta x right here will be the small change in x so here's how you would like formulate your like um your response so you will first write that delta the change in y over the change in x is roughly equal to d y over dx then you can go on to say delta y over delta x equals to y over g x and just solve for whatever that you want so once you have in this form right here you have two choices either d uh delta y equals d y over dx times by delta x by just times everything or you could reciprocal everything reciprocal to get delta x equals delta y times by dx over d y so notice the difference right here d y and dx like so here we have an example so the radius r in cm of a circle changes with time t in seconds and they are related by the equation of r equals t square plus two find the rate of change of r during the interval t equals two and t equals uh 2.1 so the rate of change of the radius the uh delta r over delta t the change in radius of the change in like the r will be we have this equation so when t equals 2.1 minus when t equals two so it would be two point one squared plus two minus two squared plus two all over the change in time two point one minus two which 0.41 over 0.1 they're both in the the r is in cm per second like this and therefore you divide both uh you divide the expression by 0.1 which gives us 4.1 cm per second right so so that's an easy example and now we're looking to percentage change so we have these two nodes right here if x changes from x to x plus delta x then the percentage the percentage change in x is equal to delta x over x times 100 or 100 therefore the same thing happens to y the percentage change in y will be equal to delta y over y times a hundred percent and it will be a percent so basically this right here is basically from if let's say this is x it will be changed to delta s so this would be delta x because we know that if this is the endpoint minus the starting point this will be delta x and now we'll look at some examples so the question just now the reduced rcm of a cycle changes with time blah blah blah when they're related by the equation r equals t squared plus two and from this time we saw that it's 4.1 cm per second so for question a it'll be 4.1 cm per second number two find the rate of change of r during the interval from t to t times t plus delta t so we know that from the eq from the question delta r will be equal to the second half of the interval t plus delta t squared plus two minus t squared plus two and if we expand on the bracket out and like times t times a t plus delta t like squared plus two minus t squared minus two you will get two t delta t minus sorry plus delta t squared and therefore the rate of change for uh for r will be equal to the delta r will be right here this right here so two t delta t plus delta t oops t plus delta t squared over the time interval so the time interval basically dealt with that basically delta t because we're changing from t to t plus delta t and t plus delta t minus t will be equal to delta t so we can basically just simplify this from this equation right here and i'll just write it on top so and if we were to simplify it you can um like divide both sides top and bottom by delta t so therefore delta r over delta t will be equal to t two t plus delta t and it's in change in radius of the change the rate of change of r which is cm should be 2t plus delta t cm per second like so and number c see hence find the rate of change of r at the answer where t equals two so we have r equals t squared plus two and therefore d r over d t we differentiate r with respect to t will be equal to two t and when t equals 2 it will be equal to 4 that means that the rate of change of r with respect to with respect to t at the instant where t equals 2 will be 4 cm per second so 4 cm per second so the point of this right here is to show that in the rate of change question you don't necessarily have to use delta r in every single part because sometimes it has a rate of change of something which is basically d something over d t so it doesn't involve any delta then we have question two the radius of the circle increases at a rate of 3 cm per second find the rate of change of from the rate of change of the area when the radius is 5 cm so we can deduce some information from the question first so the radius of a circle is increasing at a rate of 3 cm per second so the rate of the rate of change of the radius dr over dt will be equal to 3 and we're aiming to find the rate of change of the area so we have to find dna over dt so how to find the a over dt when you're given dr over dt you will use the chain rule so it'll be da over dr times by dr over dt we and we only have this so we have to find this right here so this basically means uh like differentiating a with respect to r so we have to find an equation that links area and radius and we have the area of a circle so basically it's a equals pi r squared differentiating it will get d a over d r equals two pi r and when you find the increasing of area when the width is 5 cm so where this is 5 cm will be equal to 10 pi cm squared per second like so so therefore we have some information d a over dt equals the a times by dr times by dr over dt so this d8 over dr so so we know that the a over dr from here is 10 pi so 10 pi times by dr over dt is given in the question 3 like so so it will be 10 pi times 3 which is just 30 pi and the unit is a change in the area so cm c m squared per second like so and here we have a last question so given that y equals three x squared minus two x minus three and that when s equals two there's a small increase in x of p percent use the use calculus to determine the approximate percentage percentage change in y so it looks like a complicated question so i'll just break it down so let's just find the corresponding value for x for y first so when x equals two y will be basically three times two squared minus two times two minus three like so right here and that will get you y equals five so let's just differentiate one with with respect to x first so d y over d x will be equal to six x minus two and when x is two and so it will be 6 times 2 12 12 minus 2 will be 10 so it will be 10 first like so so let's just uh put it aside first so let's just look at the change in x so delta x right so we know that there's a small increase in x of p percent so the change in x will be p percent times by x right here which is just p over 100 because p percent is physically p over 100 times by x and we know that x can be written as x equals 2 because at that instant that x is equal to 2 so when x equals 2 the delta x will be 2p of 100 and i will simplify first so just over here 2p over 100 and we know that delta delta y is over delta x is rather equal to d y over dx and to find delta y you just times both set by delta x so delta y equals delta x times u y over dx like so and delta x is just from just now 2 p over 100 times by d y over d x where we found just now from here so times by 10. so this let's just simplify this so be this and times by this so it'll be 2p over 10 or 0.2 p like so so we have delta y equals 2p so we're not done yet but we're almost done to find the percentage change in um the position the percentage in y there's a formula from just now remember this right here the percentage in the percentage change in y will be equal to delta y over y times 100 so delta y over y times 100 delta y is from here 0.2 p and y from the third from the question is five five times a hundred this would be equal to 20p over five which is a percent already and 20 divided by five would be four so it would be four p percent and that's the final answer and that's it for this short video for rules and examples for rates of change and i hope you'll find it useful and helpful and if it please leave a like and subscribe and you win the notification about selling numbers on any future videos and if you have any comments or contracting feedbacks about my channel or my youtube or my website you can drop them off in the comment section and reply to them or you can email me when my 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