Understand Calculus in 35 Minutes

in this video i'm going to attempt to teach the fundamentals of calculus in a very short time so there's three areas of calculus that you want to be familiar with the first one is limits now limits help you to evaluate a function let's say if we want to evaluate f when x is equal to 2 but for some reason a function is undefined or we just can't do it well a limit will allow you to see what happens as x approaches 2. and so limits are very useful to find out what happens to a function as we approach a certain value now the second area of calculus that you want to know is derivatives so what are derivatives well basically derivatives are functions that give you the slope of an original function at some value so let's say if i have some function f of x the derivative of f of x is known as f prime of x now this function here will tell me the slope of this function at some x value it will give me the slope of the tangent line so derivatives are very useful for calculating rates of change now the third area is integration integration is basically the opposite of derivatives with integration you're basically finding the area under the curve integration is very useful for calculating how much something accumulates over time and what you need to know is that derivatives and integration they're opposites of each other for instance we said that the derivative of the function f of x is equal to f prime of x so this is the derivative of the original function f integral of f prime of x is f of x so this is the antiderivative or the integral of f prime so you can think of integration as finding the anti-derivative now let's begin our discussion with limits so let's say that f of x is x squared minus oh that is a terrible looking two let me do that again so let's say it's x squared minus four over x minus two what is f of two can we evaluate the function at x equals two well let's find out so if we plug in two two squared is four four minus four is zero and 0 over 0 what is that well we can't figure this thing out so that's indeterminate therefore we can't evaluate the function at x equals 2. and so this is when limits become important so if we can't find out the value of the function when x is 2 can we find out what the function will do as x approaches 2.

so what happens when x is let's say 2. 1 so if we were to plug in 2. 1 into the function you should get 4.

1 now what if we plug in a number that's even closer to 2 let's say 2. 01 well if we do that this will give us 4. 01 so notice that as x gets closer and closer to 2 the value of the function gets closer and closer to 4.

so thus we can use a limit expression so we could say that the limit as x approaches two of x squared minus four over x minus two we know it's equal to four but to show your work what you need to do is factor x squared minus four and when you factor it using the difference of squares method it's x plus 2 minus i mean times x minus 2. and so we could cancel this factor and so we're left with the limit as x approaches 2 of x plus two so now we can use direct substitution so we can replace x with two so this becomes two plus two and this gives us the limit which is four so this tells us that as x gets closer and closer to two the y value or f of x will approach a value of four so even though f of two does not exist the limit as x approaches two of the function f of x that exists and that's equal to four and so make sure you understand that limits help you to see what happens or what happens to a function as x approaches a certain value now let's move on to number two derivatives so how can we find the derivative of a function the most basic rule we need to know is the power rule the derivative of a variable raised to a constant such as x raised to the n power is n x raised to the n minus 1. so how do we apply that well let's say if we want to find the derivative of x squared so in this case n is 2 so it's going to be 2 times x raised to the 2 minus 1 or just 1 so we get 2x if we want to find the derivative of let's say x cubed n is 3 so it's going to be 3 x to the 3 minus 1 which is 3x squared now if we want to find the derivative of x to the fourth power it's going to be 4 x raised to the 4 minus 1 which is 4 x cubed and so that's a simple process in which you could find the derivative of a function but now what does it mean so we said that the derivative is a function that gives us the slope of the tangent line of some original function so you might be wondering what is a tangent line so let's talk about that let's say if we have some curve and a tangent line is a line that touches a curve at one point now you need to be familiar with a secant line a secant line is a line that touches a curve at two points so the slope of a tangent line is equal to the derivative of a function at some value to calculate the slope of a secant line you've done this before in algebra using this expression it's rise over run y2 minus y1 over x2 minus x1 now let's say that f of x is x cubed using the power rule the first derivative is 3x squared so let's say if we wanted to calculate the slope of the tangent line when x is equal to two so the slope of the tangent line is going to be f prime of two which is 3 times 2 squared so it's 12.

so this means that at x equals 2 the slope is 12. so what does that mean that means that as you travel one unit to the right the curve will increase by a value of 12. so every time you travel or every time the x value increases by one the y value will increase by 12.

so that's what a slope of 12 means now let's draw a rough sketch of the original function x cubed now the graph doesn't have to be perfect but just good enough for instructional purposes and just to keep things simple let's say at this point x is equal to 2. now granted the graph is not drawn to scale so keep that in mind now what we're going to do is we're going to estimate the slope of the tangent line using the slope of the secant line now to calculate the slope of the secant line you need to choose two points a point that's to the left of where you want to evaluate the slope of the tangent line and one that's to the right so this point is two we need to choose two points such that the midpoint of those two points is equal to two so we can use one and three for an example we can use one point not in 2. 1 or we could use 1.

99 and 2. 01 in each of those cases 2 is the midpoint but we're going to see what happens to the slope of the secant line as the two points as they get closer and closer to the slope of the tangent line so first let's calculate the slope on the interval from one to three so we're going to use this formula y2 minus y1 over x2 minus x1 so y is the same as f of x by the way this is x1 and this is x2 so x2 is 3 x1 is 1. y2 is going to be equal to the function when x is 3 y 1 is equal to the function when x is 1.

so basically to evaluate f of 3 we need to plug in 3 into that expression so this is going to be three to the third power and f of one is one to the third power three to the third is twenty seven one cube is one three minus one is two so we get twenty six over two which is thirteen so this is not a bad approximation for the slope of the tangent line as you can see 12 and 13 they're not too far apart so i'm just going to write this down on the bottom for reference so now let's choose numbers that are closer to 2 like 1. 9 and 2. 1 so 2 is the midpoint so it's going to be f of 2.

1 minus f of 1. 9 over 2. 1 minus 1.

9 so let's plug in 2. 1 into that expression so that's going to be 2. 1 to the third power minus 1.

9 to the third power 2. 1 minus 1. 9 that's 0.

2 so if you plug this in you should get 12. 01 and notice that this is a very good approximation to the slope of the tangent line so notice that the slope of the secant line approximates the slope of the tangent line as these values get closer and closer to 2. now what was that something that helped us to evaluate a function as x gets closer and closer to some value if you remember its limits so we need to incorporate a limit expression with some sort of rate of change that looks like this so let's use limits to evaluate the slope of the tangent line so x is going to approach 2 and it's going to be f of x minus f of two over x minus two now f of x we know it's x cubed f of two if you plug in two into that expression 2 to the third is 8 and you get this now we need to factor x cubed minus 8.

so to factor differences of perfect cubes you could use this formula it's a minus b times a squared plus a b plus b squared so to factor x cubed minus eight x is a b is the cube root of eight which is two a squared is x squared a b is 2x b squared is 4. so this is how you can factor x cubed minus 8. so now we have this the limit as x goes to two of x minus two times x squared plus two x plus four over x minus two so now these two factors will cancel and now we can substitute 2 with x so we have the limit as x approaches two of x squared plus two x plus four and so becomes two squared plus two times two plus four two squared is four two times two is four and if you add four three times it's the same as taking four and multiplying it by three which gives you twelve and so we could use a limit process to find the slope of the tangent line as well as just finding the derivative and plugging this in so as you can see the derivative is a function that gives us the slope of the tangent line at some x value and there's multiple ways in which you can calculate the slope of the tangent line you can estimate it using the slope of the secant line or you can calculate it using a limit process so i wanted to connect limits with derivatives in one video now the next area is integration which is equivalent to anti-differentiation as we said before it's the process of finding the anti-derivative now if you recall we said that the derivative of x to the fourth is 4x cubed using the power rule well the integral of 4 x cubed should give us x to the fourth and here's the formula for finding the anti-derivative or the integral so you're going to add 1 to n and then divide by that result and you're going to add some constant of integration the derivative of any constant is 0 so when you integrate you always have to add a constant to your expression so this is going to be 4 x to the 3 plus 1 divided by 3 plus 1.

so that's 4 x to the fourth over four and let's not forget to add plus c and so as you can see we get x to the fourth plus some constant so integration is the opposite process of differentiation so you can call it anti-differentiation now i want to take a minute to compare derivatives and anti-derivatives or integration side by side just so you could see a summary of the key differences between the two so when dealing with derivatives you're looking for a function that will tell you the slope of the tangent line which touches the curve at one point at some x value so derivatives are useful for calculating the instantaneous rate of change now when dealing with integration this will help you to determine how much something accumulates over a period of time whereas derivatives will tell you how fast something changes per unit of time so integration is very useful for calculating the area under the curve so when dealing with derivatives it will give you the slope of the tangent line which you can calculate it by dividing y by x to calculate the area you need to multiply y by x so notice the difference so in its simplest form when you're differentiating you're dividing the y values by the x values whereas when you're integrating you're multiplying the y values by the x values and multiplication and division they're opposite processes and that's essentially what you're doing when you're differentiating and integrating now of course is a little bit more complicated than that but that is the gist of it let's consider this problem so we have a function a of t which is point zero one t squared plus point five t plus a hundred and this function represents the amount of water in gallons that is inside a tank at any time t and t is in minutes so what we need to do is calculate the amount of water at the following times so when t is 0 when it's 9 10 11 and 20. so go ahead and do that when you get a chance so feel free to pause the video and so here is our original function it's 0. 01 t squared plus 0.

5 t plus 100 so let's go ahead and plug in zero if we plug in zero it's going to give us this value which is a hundred now if we plug in 9 it's going to give us 105. 31 now let's plug in 10. so it's going to be 106 and if we plug in 11 we're going to get 106.

71 now let's plug in 20. so you should get 114. now keep in mind t is in minutes and a of t is in gallons so let's focus on part b how fast is the amount of water changing in the tank when t is equal to 10.

so do we need to find the derivative or do we need to integrate how fast is the amount of water changing or we're dealing with rates of change or accumulation so because we're dealing with rates of change we're dealing with derivatives so let's begin by finding the first derivative a prime of t so what is the derivative of t squared so using the power rule it's going to be 2 t to the 2 minus 1 which is 2t to the first power or simply 2t now what is the derivative of t or t to the first power it's going to be 1 t raised to the 1 minus 1 which is 1 t to the 0 anything raised to the zero power is one so this is going to be point five times one now what about the derivative of a constant like a hundred the derivative of any constant is zero if you were to use the power rule this would be a hundred and then times zero because you need to bring this to the front and then it's going to be t to the zero minus one a hundred times zero the whole thing is zero so therefore the derivative function is going to be a prime of t which is point zero one times two that's point zero two t plus point five so this is the derivative it's going to tell us how fast the amount of water is changing in the tank at any time t so we want to find out how fast is changing when t is 10. so that's 0. 02 times 10 plus 0.

5 0. 02 times 10 that's 0. 2 and 0.

2 plus 0. 5 it's 0. 7 so this tells us that the amount of water is changing by 0.

7 gallons every minute when t is 10. now if we were to graph the original function this value will represent the slope of the tangent line but how do we know if our answer is correct well one way to determine that is to calculate the slope of the tangent line by approximating it using the slope of the secant line now it's important to understand that the slope of the tangent line represents the instantaneous rate of change it's the rate of change exactly when t is 10 that's at one point the slope of the secant line represents the average rate of change which you can calculate it using two points so to calculate the slope of the secant line or the average rate of change our two points will be 9 and 11 because 10 is the midpoint of 9 and 11. so remember to calculate the slope which is associated with derivatives you need to divide the y values by the x values so t would be along the x axis a of t would be along the y axis so a 11 is a y value and the same is true for a of 9.

11 and 9 those are x values or t values so at 11 it's 106. 71 and at 9 it's 105. 31 11 minus 9 is 2.

so if we were to plug this in to the calculator this will give us 0. 7 and so in this case the slope of the secant line is the same as the slope of the tangent line now another way in which you could see the answer is by looking at the table so remember this is the amount of water that changes every minute so 0. 7 gallons per minute tells us that every minute the amount of water in the tank is increasing by 0.

7 gallons and so looking at the table going from 9 minutes to 10 minutes in one minute the increase in the amount of water was 0. 69 gallons and going from 10 to 11 or the next minute the increase in the amount of water is 0. 71 gallons so if we were to average the changes from 9 to 10 and 10 to 11 it will give us 0.

7 gallons per minute and so hopefully this example illustrates how derivatives can be used to calculate an instantaneous rate of change it can tell you how fast something is changing per unit of time now let's work on this problem the rate of water flowing in a tank can be represented by the function r of t which is equal to 0. 5 t plus 20 where r of t represents the number of gallons of water flowing per minute and t is the time in minutes so here's the question how much water will accumulate in the tank from t equals 20 to t equals 100 minutes so should we use integration or differentiation remember derivatives will help you to find the rate at which something is changing but integration is the process by which you can determine how much something accumulates over time and so this is an integration problem now we can calculate the net change in the volume of water using a definite integral from a to b of the function r of t dt the difference between a definite integral and an indefinite integral is that a definite integral have the lower limit and the upper limit and it gives you a number whereas the indefinite integral which is on the bottom this gives you a function it doesn't give you a number so make sure you understand that key difference between definite integrals and indefinite integrals so going back to this problem to calculate how much water will accumulate in the tank in those 80 minutes we can see that a is 20 that's the lower limit b is a hundred r of t is point five t plus twenty so let's go ahead and find the anti-derivative of point five t and twenty so the anti-derivative of t to the first power is going to be t to the second power divided by two just add one to the exponent and then divide by that result now for 20 you can imagine 20 as being 20 t to zero so if you add one to zero you'll get one and then divide by that number and then we're going to evaluate the result from 20 to 100 so the first thing we're going to do is plug in a hundred so it's 0. 5 times 100 squared divided by 2 plus 20 t or 20 times 100 and then we're going to plug in 20 into this expression so it's minus 0.

5 times 20 squared divided by two plus twenty times twenty so now let's work out the math a hundred squared is ten thousand and ten thousand divided by two is five thousand half of five thousand is twenty five hundred and then twenty times a hundred is two thousand here we have twenty squared so 20 times 20 is 400 divided by 2 that's 200 and then half of 200 that's a hundred and here we have 20 times 20 which is 400 so now we have forty five hundred minus five hundred so the net change should be four thousand so from t equals twenty to t equals 80 the tank will gain 4 000 gallons of water earlier in this video we said that integration is associated with finding the area under the curve so what we're going to do is we're going to graph this function so the y-axis corresponds to r of t the x-axis is t now when t is 0 the y value will be 20. this is the y intercept now the points of interest are 20 and 100 so when t is 20 if you plug it into this expression 0. 5 times 20 is 10 plus 20 that's 30.

so we have this point so r of 20 is 30 and r of 0 let me use a different color is 20. now what about r of 100 so 0. 5 times 100 is 50 plus 20 that's going to give us 70.

so r of 100 is 70. and now we can draw a straight line so our goal is to calculate the area under the curve from 20 to 100 this area let me shade it represents the increase in the volume of water that's already in the tank it doesn't represent the total amount of water that's in a tank at a hundred it simply represents the change in the volume of water in those 80 minutes so let's calculate the area under the curve now using geometry it's helpful to break this region into a rectangle and a triangle and so to find the area we need to multiply the x values by the y values the area of a rectangle is left times width so the length is the difference between 120 which is 80. the unit for this is 80 minutes and the height of the rectangle is 30 and the unit for that is gallons per minute so if you multiply minutes by gallons per minute the unit minutes will cancel giving you the unit gallons so 80 times 30 8 times 3 is 24 and then if we add the two zeros that will give us twenty four hundred now let's focus on the triangle the area of a triangle is one half base times height in this case we could say the base is associated with the x-axis the height is associated with the y-axis so whenever you're dealing with integration you're going to be multiplying x by y when you're dealing with differentiation you're dividing y by x now the base of the triangle is still 80 is still the difference between 120 but the height of the triangle is the difference between 70 and 30 so it's 40.

now 80 times 40 8 times 4 is 32 and if you add the two zeros that's 3200 but we need to take half of it because we no longer have a rectangle we have a triangle and a triangle is half of a rectangle so that's why it's a half base times height so half of 3200 is 1600 thus if we add 1600 and 2400 that's going to give us our answer of 4 000.