A derivative represents the instantaneous rate of change of a function at a specific point, and to interpret its meaning in context, one must include the subject (noun), the numerical value with its sign indicating increase or decrease, the appropriate units, and the specific time or input value at which the derivative was computed. For example, if w'(12) = -0.5 for a temperature function w(t) in degrees Celsius, the interpretation is: 'The temperature of the water in the pond is decreasing by 0.5 degrees Celsius per day at t = 12 days.'
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Calc Topic 4 1Added:
okay up to this point we've only dealt with uh primarily multiple choice questions but we're going to get um ready to start seeing some free response questions and also we're going to learn about the meaning behind all of the derivatives um that we have learned we've learned all the derivative rules and the basics but uh we've learned how to find derivatives but we haven't put them into context and that's what we're going to start doing so again it says in units two and three we've extensively studied the procedures for finding a derivative now in unit four we will turn our attention to what the derivative means in the context of a problem okay let's take a look at example number one this came from the 2001 calculus exam uh the temperature in degrees celsius of the water in a pond is a differentiable function w of time t the table to the right shows the water temperature as recorded every three days over a 15 day period use the data table they love tables don't they use the data table use the table to find an approximation of w prime of 12 show the computation that leads to your answer okay so w prime of 12 what is that i know it's the first derivative at t equals 12 days would that be 22 now 22 is w of 12.
that's the output but it's not the derivative it's not the instantaneous rate of change so how can we approximate the instantaneous rate of change in other words how can we approximate the slope of the tangent line well again we don't have a function here we just have a table so if you recall we had instantaneous rate of change and we had average rate of change so if we could get our average rate of change close we could approximate the instantaneous rate of change so average rate of change was the slope between two points and we were using we were calling it the slope of the secant line that's what we call it an algebra in calculate average rate of change so in your story problems average rate of change and instantaneous rate of change are the terms you should use you shouldn't use slope of the secant line slope of the tangent line you have to use calculus definitions calculus vocabulary okay so anyway getting back to the problem how can we approximate the instantaneous rate of change at 12 days well if we can get an average rate of change around that time approximate it so what we're going to do is we are going to approximate by using the values the two points on either side of 12 so that would be 24 minus 21 change in our y's over changing our x's 9 minus fifteen so not equals we can want to use the approximation sign here so i suppose we could say equals equals um that would be negative one half so that would be negative well that would be a three over negative six which is negative one half it says indicate units of measure so we have degrees celsius and days so it's degrees celsius per day okay so that would be worth two points out of nine on one of the free response questions and you can see this on the a b and the bc test so you've got two out of nine points now what if you were asked to interpret the meaning of the approximated derivative value interpreting the meaning of the value in the context of a problem is certainly going to be on the ap exam so this little acronym that's going to help you get full credit what is it that you need to have in your interpretation well the first thing you need to have is a noun or a number and or a number a noun and or number so the subject of the problem and the numerical value okay so what are we talking about up here what are we talking about degrees celsius per day we're talking about the temperature of the water so the temperature of the water in the pond is now when it's negative we're going to say we're going to describe what that means okay we're going to interpret that the temperature of the water in the pond is decreasing because remember this is a rate okay so the temperature of the water in the pond is decreasing so we have the noun okay now we need the numerical value decreasing by what decreasing by a half degree celsius now are they going to mark you down for putting degrees celsius no we're using the symbol degrees celsius no i like to be as specific as possible and as thorough as possible to write out my interpretation just to be safe okay so there's the the noun with the number okay and then do we have the units half a degree celsius well if not just celsius it's celsius per day okay so let's do this let's say this is the noun which is all of this with the number okay then we are going to say do we have the units we have the units and now we need the time so when i say time include the time at which the derivative for the change was computed what was the time so that's where we are going to say at the 12th so day color do we want to use for that oh let's say green that kind of looks like green already so why don't we just use black okay so make sure you have those three things in your interpretation okay all right so next one um is a calculator problem okay um we've done derivatives on our calculator and we've also talked about the trig functions on our calculator so you should know how to take a derivative on your calculator and you should also know that in calculus you need your calculator in radian mode when you're doing trig derivatives okay for those of you in physics those are the people who get it mixed up so always make sure on a calculus test your calculator is in radian mode okay so it says here's g of t we're talking about the rate at which unprocessed gravel arrives and where tea is measured in hours um the plant has 500 tons of unprocessed gravel during the hours from zero to eight the plant processed gravel at a constant rate of 100 tons per hour find g prime of 5 well here is g and let's go ahead and find the derivative of g at 5 on your calculators so pause the video and find your calculators and give it a try in your calculator all right so when you're using a calculator the rule for your answer has to be three or more decimal places three or more three or more not less if you do two you're going to miss that point okay all right so g prime 5 and again it's approximately equal to negative 24.5875 okay now i like to put four just to be safe um you could put three you can either round or you can truncate so if you rounded your answer it could be 0.588 if you truncated your answer which means just drop everything after the third one it could be point five eight seven i'm just going to keep it as 0.5875 just to be safe i'm going to go four decimal places you do not get marked down for doing more okay all right so it says use the correct units and interpret your answer so what is the unit for this well g is a rate in tons per hour g is a rate in tons per hour so if you take its derivative up here w was in degree celsius and its derivative was degrees celsius per day here our g is tons per hour so when you take its derivative and t is measured in hours it's going to be tons per hour that's the rate and when you take its derivative it's going to be tons per hour per hour okay so for those of you in physics we can actually call that acceleration or in this case velocity the rate this is the derivative of a rate which is except we'll be talking about soon okay so now we have to write up our little description again i would pause the video and practice writing your description so what are we talking about we're talking about the rate at which the gravel is arriving at the plant not just the rate but be specific the rate at which the gravel is arriving okay at the plant now if you just said the gravel's arriving i'm sure you're fine but again this is for college credit so the more thorough you can be the better better your score might be okay the rate at which the gravel is arriving at the plant okay well what does this tell us that's the noun what about the number the noun and the number so this negative tells us that the rate is decreasing by 24.5875 and tons per hour per hour you can write tons per hour squared a lot of the scoring guides it will save tons per hour per hour so either way and now our time at use the word at time t equals five five what five hours don't you say five be specific five hours so up here up here you could have said at time t equals 12 days at the 12th days is good as well okay all right last one why don't you go ahead and give it a try yourself pause the video and come on back okay w prime of 20 is approximately equal to negative 0.2855 and yes i always put an approximation when you're using a table and when you're using your calculator your calculator is approximating the derivative we'll talk about how and how you get approximations of derivatives and anti-derivatives um but it is an approximation okay and again i went to four decimal places and we are talking about what find w prime find the windchill the windchill is given by w of v and what would be the unit for this so we have degrees fahrenheit per what when you take a derivative it's always going to be what you started with here your domain your range in a sense here so up here it was degrees celsius per days here it was tons per hour per hour and this one is going to be degrees fahrenheit per our input what is our input velocity so it's going to be per miles per hour okay so let's go ahead and explain the meaning explain the meaning in terms of the windchill okay so the wind chill is what we're talking about the windchill decreasing it was positive you would say increasing is decreasing by now this is miles per hour so this is what it's decreasing by miles per hour so that is a rate it is decreasing by a rate and it's a derivative so all derivatives are rates even the second derivative is a rate so decreasing um by a rate of 0.2855 um let me squeeze that in here degrees fahrenheit per miles per hour so there is our noun there is our number there is the units now at at what at the velocity again that's our input velocity equals 20 miles per hour so that one was a little bit different with the units see if you can go through and figure out the units and how they are computed again um one more time we started with the derivative of w is degrees per days okay and this one g of t g of t is measured in tons per hour so tons per hour per what your input is which is time per hour and then this one right here our w okay what was our w our w was wind chill so degrees fahrenheit degrees fahrenheit per what's the unit of your input the unit of your input is miles per hour so it's degrees fahrenheit per miles per hour and that's the start of the interpretation of a derivative
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