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 the 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. The 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 antiderivative. 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 look in 2, let me do that again. So let's say it's x squared minus 4 over x minus 2. What is f of 2? Can we evaluate the function at x equals 2?
Well let's find out. So if we plug in 2, 2 squared is 4, 4 minus 4 is 0, 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 2 of x squared minus 4 over x minus 2, we know it's equal to 4. But to show you work, what you need to do is factor x squared minus 4. 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 2, so now we can use direct substitution, so we can replace x with 2, so this becomes 2 plus 2, and this gives us the limit, which is 4. So this tells us that as x gets closer and closer to 2, the y value, or f of x, will approach a value of 4. So even though f of 2 does not exist, the limit as x approaches 2 of the function f of x that exists, and that's equal to 4. 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 you 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 3x 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 4x raised to the 4 minus 1, which is 4x 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 2. So the slope of the tangent line is going to be f prime of 2, 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 1, 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 2. We need to choose two points such that the midpoint of those two points is equal to 2. So we can use 1 and 3 for an example.
We can use 1.9 and 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 1 to 3. 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. y1 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 3 to the 3rd power, and f of 1 is 1 to the 3rd power.
3 to the 3rd is 27, 1 cubed is 1, 3 minus 1 is 2. So we get 26 over 2, which is 13. 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, it's 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 2 over x minus 2. Now f of x, we know it's x cubed. f of 2, if you plug in 2 into that expression, 2 to the 3rd 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 ab plus b squared. So to factor x cubed minus 8, x is a, b is the cube root of 8, which is 2, a squared is x squared, ab 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 2 of x minus 2 times x squared plus 2x plus 4 over x minus 2. So now these two factors will cancel, and now we can substitute 2 with x. So we have the limit as x approaches 2. of x squared plus 2x plus 4, and so it becomes 2 squared plus 2 times 2 plus 4. 2 squared is 4, 2 times 2 is 4. And if you add 4 three times, it's the same as taking 4 and multiplying it by 3, which gives you 12. 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 antiderivative.
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 4x cubed should give us x to the fourth and here's the formula for finding the antiderivative 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 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 4x to the 3 plus 1 divided by 3 plus 1. So that's 4x to the 4th over 4. 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, it's 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 0.01t squared plus 0.5t plus 100. 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 0. If we plug in 0, it's going to give us this value, which is 100. 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 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 a 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?
Are we 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 2t 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 1t raised to the 1 minus 1, which is 1t to the 0. anything raised to the 0 power is 1. So this is going to be 0.5 times 1. Now what about the derivative of a constant, like 100? The derivative of any constant is 0. If you were to use the power rule, this would be 100 and then times 0, because you need to bring this to the front, and then it's going to be t to the 0 minus 1. 100 times 0, the whole thing is 0. So therefore, the derivative function is going to be A prime of T, which is 0.01 times 2, that's 0.02T plus 0.5.
So this is the derivative. It's going to tell us how fast the amount of water is changing in a tank at any time T. So we want to find out how fast it's 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 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 a11 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.71 gallons.
So if we were to average the changes from 9 to 10 and 10 to 11, it will give us.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, which is equal to 0.5t plus 20, where R 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 a tank in those 80 minutes, we can see that A is 20, that's the lower limit, B is 100, R is 0.5T plus 20. So let's go ahead and find the antiderivative of 0.5T and 20. So the antiderivative of t to the first power is going to be t to the second power divided by 2. Just add 1 to the exponent and then divide by that result. Now for 20, you can imagine 20 as being 20t to the 0. So if you add 1 to 0, you'll get 1 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 100. So it's 0.5 times 100 squared divided by 2 plus 20t 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 2 plus 20 times 20. So now let's work out the math.
100 squared is 10,000, and 10,000 divided by 2 is 5,000. Half of 5,000 is 2,500. And then 20 times 100 is 2,000.
Here we have 20 squared, so 20 times 20 is 400. Divided by 2, that's 200. and then half of 200, that's 100. And here we have 20 times 20, which is 400. So now we have 4,500 minus 500, so the net change should be 4,000. So from T equals 20 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 100. 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 100 and 20, 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 2400. Now let's focus on the triangle. The area of a triangle is 1 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 the x by y. When you're dealing with differentiation, you're dividing y by x.
Now, the base of the triangle is still 80, it's still the difference between 100 and 20, 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 4000. And so that's how much water was accumulated in the tank in those 80 minutes. It's 4000 gallons of water. It doesn't represent the total amount of water in a tank, but it represents the change.
the volume of the water in the last 80 minutes. So that's basically it for this video. So to review, just remember limits allow you to evaluate a function when X approaches a certain value. Derivatives are functions that allow you to calculate the instantaneous rate of change of a function at any instant of time. And remember, the instantaneous 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 by calculating the area under the curve. So these are the three fundamental concepts taught in a typical calculus course.
So make sure you understand the basic idea behind limits, derivatives, and integration. And that's basically it for this video. So for those of you who haven't done so already, feel free to subscribe to this channel, and don't forget to click on that notification bell.
Now I'm going to post some links in the description section of this video with more problems on limits, derivatives, and integration so you can get more practice with that. And also check out my new Calc List video playlist because it has specific topics in Calc that can help you if you're taking that course. So that's all I got for this video. Thanks again for watching.