Understand Calculus in 35 Minutes

In this video I'm going to try to teach the fundamentals of

calculus

in a very short time, so there are three areas of

calculus

that you'll want to familiarize yourself with. The first is limits, now limits help you 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 not defined or we just can't do it right, a limit will allow you to see what happens when x approaches 2, so limits are very useful to find out. What happens to a function when we approach a certain value?

The second area of ​​calculus you want to know about 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 value of x 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 basically you are 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 are opposite of each other, for example 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 integral of f prime, so you can think of the integration as find the antiderivative.

Now let's start our discussion with limits, so let's say that f of x is x squared minus oh, that's a terrible two, let me. Do it again, so let's say it's x squared minus four over x minus two, what's f of two? Can we evaluate the function at x equals two? Well, let's figure it 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 solve this, so it is indeterminate, therefore we can't evaluate the function at x equals 2. And this is when limits become important, so if we can't find the value of function when x is 2, can we find out what the function will do when x approaches 2?

So what happens when x is, say 2.1? So if we plug 2.1 into the function, we should get 4.1. Now what happens if we plug in a number that is 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 we can use a limit expression so we could say that the limit as x gets closer to two of x squared minus four over x minus two we know it's equal to four but to show your work it What you should do is factor x squared minus four and when you factor it use the difference of the method of squares is x plus 2 minus, I mean multiplied by x minus 2.

Therefore, we could cancel this factor and we are left with the limit as x approaches 2 of x plus two, so now we can use direct substitution to be able to 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 or f value of x will get closer to a value of four, so even though f of two does not exist the limit when x approaches two of the function f of x exists and that is equal to four, so make sure you

understand

that limits help you see what happens or what happens to a function when x approaches a certain value. let's move on to derivatives number two 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 like x raised to the power n is n x raised to the power n minus 1. So how do we apply that? Well, let's say if we want to find the derivative of x squared, in this case n is 2, so it's going to be 2 times x to the power of 2 minus 1 or just 1, so we get. 2x if we want to find the derivative of say x cubed n is 3 then it will be 3 x to the power of 3 minus 1 which is 3x squared now if we want to find the derivative of x to the fourth power it will be 4 x to the power of 4 minus 1 which is 4 x cubed y that is a simple process where you can 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 It is a line that touches a curve at a point, now you need to be familiar with it. 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 have done this before in algebra using this expression is an increase over the 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 equals two, so the slope of the tangent line will 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? As you travel one unit to the right, the curve will increase by a value of 12. So each time you travel or each time the value of x increases by one, the value of y will increase by 12. So, that's what which means a slope of 12. Draw a sketch of the original function x cubed. Now the graph doesn't have to be perfect, but it does have to be good enough for instructional purposes and to keep things simple, let's say at this point x is equal to 2.

Now, granted, the graph isn't. drawn to scale, so keep that in mind now what we are going to do is 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 out a point that is to the left of where you want to evaluate the slope of the tangent line and one that is to the right, so this point is two, we must choose two points such that the midpoint of those two points is equal to two, so We can use one and three as an example, we can use a point that is 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 two points 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'll use this formula y2 minus y1 over x2 minus x1 so that y is equal to f of x by the way this is x1 and this is x2 so x2 is 3 x1 is 1. y2 will 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 insert 3 into that expression, so this will be three to the third power and f of one is one to the third power three to the third power is twenty seven a 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 are not very far apart, so I'm going to write this at the bottom. for reference, now let's choose numbers that are closer to 2, like 1.9 and 2.1, so 2 is the midpoint, so it will be f of 2.1 minus f of 1.9 over 2.1 minus 1.9, so let's plug 2.1 into that expression , so it'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 you'll notice that this is a very good approximation to the slope of the tangent line, so notice that the slope of the secant line approaches the slope of the tangent line as these values ​​get closer and closer to 2.

Now, what was it that helped us evaluate a function as x gets closer and closer? more to some value if you remember its limits? So we need to incorporate a limit expression with some kind of rate of change that looks like this, so let's use limits to evaluate the slope of the tangent line, so that x approaches 2 and it will be f of x minus f of two over x minus two now f of x we ​​know it's formula is a minus b times a squared plus a b plus b squared, so factoring 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. this is how you can factor x cubed minus 8. so now we have this limit since 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 for x, so we have the limit when x approaches two of x squared plus two x plus four and so it becomes two squared plus two times two plus four two squared is four two times two is four and if you add four times three it is the same as taking four and multiplying it by three, which gives twelve, so we could use a limit process to find the slope of the tangent line, as well as simply find the derivative and replace it so you can see, the derivative is a function that gives us the slope. of the tangent line at some value of x and there are several ways 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 the limits with derivatives in a video now the next area is integration, which is equivalent to antidifferentiation, as we said before, it is the process of finding the antiderivative.

Now, if you remember, we said that the derivative of x to the fourth power is 4x cubed using the power. rule well, the integral of 4 x cubed should give us derivative of any constant is 0, so when you integrate you should always add a constant to your expression, so it will be 4 x to the power of 3 plus 1 divided by 3 plus 1. So that's 4 x to the fourth over four and let's not forget add more c and as you can see we get x to the fourth power 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 antiderivatives or integration side by side so you can see a summary of the key differences between the two, so when it comes to derivatives you look for a function that tells you the slope of the tangent line that touches the curve at a point at some value of x, so derivatives are useful for calculating the instantaneous rate of change. Now when it comes to integration, this will help you determine how much something accumulates over a period of time, while 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 it comes to derivatives it will give you the slope of the tangent line, which you can calculate by dividing y by x to calculate the area you need to multiply by x so notice the difference, so in its simplest form Simple, when you are differentiating, you are dividing the y values ​​by the x values, while when you are integrating, you are multiplying the y values ​​by the x values ​​and multiplication and division are opposite processes and that is essentially what you are doing. when you're differentiating and integrating now of course it's a little more complicated than that, but that's the gist of the matter, let's consider this problem so that we have a function a of t which is point zero one t squared plus point five t plus one 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 must do is calculate the amount of water in the following times, so when t is 0 when it is 9 10 11 and 20. so go ahead and do it when you get a chance so feel free to pause the video and here is our original function: it's 0.01 t squared plus 0.5 t plus 100 so go ahead and plug in zero if we plug in zero it will give us this value which is hundred now if we plug in 9 it will give us 105.31 now let's plug in 10. then it will be 106 and if we plug in 11 we will get 106.71 now let's plug in 20 so we should get 114.

Now note that t is in

minutes

and a of t is in gallons, so let's focus on part b, how fast does the amount of water change? 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 the amount of water changes or are we dealing with rates of change or accumulation, because we are dealing with rates of change we are dealing with derivatives, so let's start by finding the first derivative, a prime of t, so what is the derivative of t squared? So using the power rule, it will be 2t to the power of 2 minus 1, which is 2t to the first power.power or just 2t now what is the derivative of t or t to the first power is going to be 1 t to the power of 1 minus 1 which is 1 t to the power of 0 anything raised to the zero power is one so this is going to be point five for one now, what about the derivative of a constant like hundred?

The derivative of any constant is zero. If you were to use the power rule, this would be one hundred and then multiplied by zero because you need to bring this to the front and then it's going to be t to the power of zero minus one hundred times zero, everything is zero, so the derivative function is going to be prime of t, which is the zero point one times two, that is, the zero point two t plus the point five, so this is the derivative that will tell us how quickly the amount of water in the tank changes at any time t , so we want to know how fast it changes when you 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 is 0.7 so this tells us that the amount of water changes 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? One way to determine this is to calculate the slope of the tangent line by approximating it using the slope of the secant line. Now it is important to

understand

that the slope of the tangent line represents the instantaneous rate of change is the rate of change exactly when t is 10 that is at a point the slope of the secant line represents the average rate of change that you can calculate using two points 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 that to calculate the slope associated with the derivatives you must divide the values ​​of and for the values ​​of x, so t would be along the x-axis, a of t would be along the y-axis, so a of 11 is a value of y and the same is true for a of 9. 11 and 9 those are x values ​​or t values, so in 11 is 106.71 and in 9 is 105.31 11 minus 9 is 2. So if we plug this into the calculator, this will give us 0.7 and, In this case, the slope of the secant line is the same as the slope of the tangent line now.

Another way you can see the answer is by looking at the table, so remember that this is the amount of water that changes each minute, so 0.7 gallons per minute tells us that each minute the amount of water in the tank increases 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 would give us 0.7 gallons per minute and hopefully this example illustrates how we can use the derivatives to calculate an instantaneous rate of change and can tell 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 is 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 that derivatives will help you find the rate at which something is changing, but integration is the process by which you can determine how much something accumulates over time, 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 has the lower limit and the upper limit and gives you a number, while the indefinite integral at the bottom 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 is the lower limit b is one hundred r of t is the point five t plus twenty, so let's go ahead and find the antiderivative of point five t and twenty, so the antiderivative of t to the first power will 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 20 t to zero, so if you add one to zero you will get one and then you will divide it by that number and then we will evaluate the result from 20 to 100, so the first thing we are going to do is replace one hundred so that let it be 0.5 times 100 squared divided by 2 plus 20 t or 20 times 100 and then we are going to replace 20 in this expression so that it is minus 0.5 times 20 squared divided by two plus twenty times twenty so now let's do the calculations hundred squared is ten thousand and ten thousand divided by two is five thousand half of five thousand is twenty-five and then twenty times one 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 one 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 of t is equal to twenty and t is equal to 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 graph this function. then the y axis corresponds to r of t the x axis is t now when t is 0 the value of y will be 20. this is the intersection and now the points of interest are 20 and 100 so when t is 20 if you connect 0 in this expression .5 multiplied by 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 multiplied by 100 is 50? plus 20 that will 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 volume of water that is already in the tank does not represent the total amount of water that is in a tank at a hundred, it simply represents the change in the volume of water in those 80 minutes, so now let's calculate the area under the curve using geometry. It is useful to divide this region into a rectangle and a triangle, so to find the area we need to multiply the x values ​​by the y values. The area of ​​a rectangle is multiplied by the width, so the length is the difference between 120 and 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 that if you multiply the minutes by gallons per minute, the unit of minutes will cancel out and you will get the unit of gallons, so 80 times 30 8 times 3 is 24 and then if we add the two zeros that will give us two thousand four hundred now let's focus on the triangle , the area of ​​a triangle is half the base times the height, in this case we could say that the base is associated with the x axis.

Height is associated with the y-axis, so whenever you're dealing with integration, you'll multiply x by y, when you're dealing with differentiation, you'll divide y by x. Now the base of the triangle is still 80. It's 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 it's half the base times the height, so half of 3200 is 1600, therefore, if we add 1600 and 2400, we will get our answer of 4000. and that is the amount of water that accumulated in the tank in those 80 minutes, that is 4,000 gallons of water.

It does not represent the total amount of water in a tank, but represents the change in water volume over the last 80 minutes. So that's basically it for this video, so to review just remember that limits allow you to evaluate a function as x approaches a certain value. Derivatives are functions that allow you to calculate the instantaneous rate of change of a function at any instant and remember the snapshot. The rate of change is equivalent to the slope of the tangent line at any instant of time. Now you can approximate the slope of the tangent line using the slope of the secant line, which is equivalent to an average rate of change between two points.

Finally we have integration. which is a process that allows us to determine how much something accumulates over time and we can find that value by evaluating the definite integral or calculating the area under the curve. These are the three fundamental concepts taught in a typical calculus course, so make sure you understand the basic idea behind derivative limits and integration and that's basically it for this video. For those of you who haven't done so yet, feel free to subscribe to this channel and don't forget to click the notification bell now. I'm going to post some links in the description section of this video to more problems on derivatives of limits and integration so you can practice more with that and I'll also check out my new playlist of calculus videos because it has specific topics in calculus that can help you.

Yes You're taking that course, so that's all I have for this video. Thanks again for watching.