John, another great simplification. I agree with some of the commenters that a connected video showing a real world word problem would help cement the concept for students. Physics always has many places to use Calculus. Thanks for listening.
@ Robert Arguello Many things, in all the sciences; in medicine, insurance, economics, physics, chemistry, biology, metallurgy, optics, social science, mathematics… and the list goes and goes, you name it! For instance, in physics, the equation for falling bodies, that is, the distance function for falling bodies is, s(t) = 1/2gt^2. It tells us how many feet a body falls in t seconds, or how many meters it falls in t seconds if we’re using the metric system. And g is the gravitational constant, the pull of gravity on all bodies, it’s constant, always the same, and the same on all bodies, and is roughly 32 feet per second per second (32ft/s^2), or 32 feet per second squared (32ft/s^2), same thing. So, let’s say you’re at some height and you dropped a ball from your hand and wanted to know how far it fell right at 6 seconds, that is, exactly at 6 seconds later. You’d simply substitute 6s (6 seconds) into the equation for t (time), and substitute 32ft/s^2 for g (the acceleration due to gravity) (32 feet per second per second, that is, 32 feet per second squared, like so: s(6s) = 1/2(32ft/s^s)(6s)^2 = 648 feet! The ball would fall 648 feet after exactly 6 seconds. But, and more particularly to your question (“what is the derivative useful for”), we could or may even also need to ask, “how fast was the ball falling at exactly 6 seconds?” Well, we would need the velocity function in order to answer that question! That is, we’d need to find the derivative of the distance function for falling bodies in order to answer that question. So we’d simply take the derivative of our above equation, that is, of our above function. And the derivative of the above equation, or the above function, that is, our above distance function is the velocity function. And it is v(t) = gt! Now, all we have to do in order to find out exactly how fast the ball was falling at exactly 6 seconds we can now simply substitute 6s (6 seconds) for t into our derivative of the distance function which gives us our velocity function, and we also substitute into that equation 32ft/s^2 for g (gravity) like so, v(6s) = (32ft/s^2)(6s) = 192ft/s = 192 feet per second! So… at exactly 6 seconds after you dropped your ball it was falling at a speed of 192 feet per second! Neat, eh! …and that’s super fast! But Notice how the units of distance and time, given in feet and seconds, cancel out in both procedures to give only feet in the distance function, and to give both feet and seconds in the velocity function! The distance function gives answers in distance only, given in feet for this example. the other units cancel out, simply because it’s concerned only with distance traveled and not at all with time expired; and the velocity function gives answers in both units, that is, in both feet and seconds because it’s concerned with the rate of speed which involves and can only be given in terms of both time expired and distance traveled. So it’s crucial that when substituting values for the variables g and t, as in this example, and all such examples, the units of time t, seconds s, and the units of distance, feet or meters, etc. must be included in the equation for it to be not only correct, but for it to also be definite and specific and to make sense. When we’re talking speed, for instance, it would be senseless to say, he was going 192 feet when what we really mean is, he was going 192 feet per second! And you’ll always know that when your answer doesn’t have the proper units or has misplaced units or has no units at all when it should, or when it shouldn’t, depending on the equation we’re working with, then you’ve made a mistake somewhere in the calculation with regards to units and or their substitution or lack thereof into the equation! And the derivative of the velocity function, called the second derivative of the distance function, the derivative of a derivative, naturally, is acceleration, or, the acceleration function, in this case, and it is, a(t) = g! It’s g simply because it’s the gravitational constant, the pull of gravity, the acceleration of gravity, or, the of acceleration due to gravity, which is always constant and is 32ft/s^2! And it’s derivative is 0, …simply because it’s always “constant,” and the derivative of a constant, as stated above, is always 0! But that’s just one of the things a derivative can find and do, and that’s just one of the uses of the derivative and it’s usefulness. And that’s just one instance of its almost endless power and usefulness in real and even imaginary life! In your study of calculus you’ll discover that calculus texts and teachers will give endless real life examples of calculus’s power and usefulness! But the main purpose of videos like this one is mainly and almost strictly to show you what calculus is and how to use it. We learn what fractions are and do fractions before we use them to bake cakes, before we use them to measure the amount of ingredients to be used in making our cakes! I know I was long, so… my apologies. But just a side note, people new to calculus always ask, and understandably so, “why is the derivative of a constant 0?” Because a derivative calculates rate of change. And a constant, or constant function is just that, constant! It’s rate of change is constant at 0 and never changes and therefore has a 0 rate of change, that is, it’s derivative is 0! The rate of change of a constant, or of a constant function, therefore, is always the same, so it’s rate of change, that is, it’s derivative is always 0. For example, the function y = 2, say, is a “constant” function, or 2 is a constant, same thing, because it is a horizontal line that has a slope of 0, or a “derivative” of 0, same thing, that never changes and is therefore called a constant function, or a constant. That’s why the derivative of a constant, or the derivative of a constant function, in this example of y = 2, say, is 0. The derivative (which is just the slope at every point along the function’s graph) is 0 because the slope (the derivative) of that equation, by the point-slope equation, or the point-slope formula for slope (m), [m = (y2 - y1)/(x2 - x1)], gives m = 0 because (y2 - y1) always equals 0 for every (x2 - x1) along the x-axis simply because y = 2 throughout the entirety of the length of the line y = 2! The derivative is so powerful that it also calculates that truth! Hope I didn’t confuse things, there, by giving more info than you perhaps need or even asked for in answer to your particular question.
John, another great simplification. I agree with some of the commenters that a connected video showing a real world word problem would help cement the concept for students. Physics always has many places to use Calculus. Thanks for listening.
Nice to know that a mathematician found another way of solving this problem.
Watching from Jamaica 🇯🇲🇯🇲🇯🇲
Good video on finding the integral, but it would be great to add how this could be a real-world problem, even in some simple way.
@ Jerry Miller
This isn’t finding the integral, he’s finding the derivative.
Still, a word problem would be nice
I have a math degree. I don’t recall ever having a real world problem for a polynomial.
you are better than my maths teacher love it😍
John, you are doing a great job, when you can, work a word problem in that exemplifies the concept you are teaching
Good explanation. Thanks!
Wow.. what a coincidence. I just did a video on basic differentiation
This is so great! Thank you so much!
Thank you for this video, it was very helpful!
Awesome video!!
Thanks!
I am in 7th grade and trying to learn calculus and I understand this.
What good is finding the derivative? What is it useful for?
@ Robert Arguello
Many things, in all the sciences; in medicine, insurance, economics, physics, chemistry, biology, metallurgy, optics, social science, mathematics… and the list goes and goes, you name it!
For instance, in physics, the equation for falling bodies, that is, the distance function for falling bodies is, s(t) = 1/2gt^2. It tells us how many feet a body falls in t seconds, or how many meters it falls in t seconds if we’re using the metric system. And g is the gravitational constant, the pull of gravity on all bodies, it’s constant, always the same, and the same on all bodies, and is roughly 32 feet per second per second (32ft/s^2), or 32 feet per second squared (32ft/s^2), same thing. So, let’s say you’re at some height and you dropped a ball from your hand and wanted to know how far it fell right at 6 seconds, that is, exactly at 6 seconds later. You’d simply substitute 6s (6 seconds) into the equation for t (time), and substitute 32ft/s^2 for g (the acceleration due to gravity) (32 feet per second per second, that is, 32 feet per second squared, like so: s(6s) = 1/2(32ft/s^s)(6s)^2 = 648 feet! The ball would fall 648 feet after exactly 6 seconds. But, and more particularly to your question (“what is the derivative useful for”), we could or may even also need to ask, “how fast was the ball falling at exactly 6 seconds?” Well, we would need the velocity function in order to answer that question! That is, we’d need to find the derivative of the distance function for falling bodies in order to answer that question. So we’d simply take the derivative of our above equation, that is, of our above function. And the derivative of the above equation, or the above function, that is, our above distance function is the velocity function. And it is v(t) = gt! Now, all we have to do in order to find out exactly how fast the ball was falling at exactly 6 seconds we can now simply substitute 6s (6 seconds) for t into our derivative of the distance function which gives us our velocity function, and we also substitute into that equation 32ft/s^2 for g (gravity) like so, v(6s) = (32ft/s^2)(6s) = 192ft/s = 192 feet per second! So… at exactly 6 seconds after you dropped your ball it was falling at a speed of 192 feet per second! Neat, eh! …and that’s super fast! But Notice how the units of distance and time, given in feet and seconds, cancel out in both procedures to give only feet in the distance function, and to give both feet and seconds in the velocity function! The distance function gives answers in distance only, given in feet for this example. the other units cancel out, simply because it’s concerned only with distance traveled and not at all with time expired; and the velocity function gives answers in both units, that is, in both feet and seconds because it’s concerned with the rate of speed which involves and can only be given in terms of both time expired and distance traveled. So it’s crucial that when substituting values for the variables g and t, as in this example, and all such examples, the units of time t, seconds s, and the units of distance, feet or meters, etc. must be included in the equation for it to be not only correct, but for it to also be definite and specific and to make sense. When we’re talking speed, for instance, it would be senseless to say, he was going 192 feet when what we really mean is, he was going 192 feet per second! And you’ll always know that when your answer doesn’t have the proper units or has misplaced units or has no units at all when it should, or when it shouldn’t, depending on the equation we’re working with, then you’ve made a mistake somewhere in the calculation with regards to units and or their substitution or lack thereof into the equation! And the derivative of the velocity function, called the second derivative of the distance function, the derivative of a derivative, naturally, is acceleration, or, the acceleration function, in this case, and it is, a(t) = g! It’s g simply because it’s the gravitational constant, the pull of gravity, the acceleration of gravity, or, the of acceleration due to gravity, which is always constant and is 32ft/s^2! And it’s derivative is 0, …simply because it’s always “constant,” and the derivative of a constant, as stated above, is always 0!
But that’s just one of the things a derivative can find and do, and that’s just one of the uses of the derivative and it’s usefulness. And that’s just one instance of its almost endless power and usefulness in real and even imaginary life! In your study of calculus you’ll discover that calculus texts and teachers will give endless real life examples of calculus’s power and usefulness! But the main purpose of videos like this one is mainly and almost strictly to show you what calculus is and how to use it. We learn what fractions are and do fractions before we use them to bake cakes, before we use them to measure the amount of ingredients to be used in making our cakes!
I know I was long, so… my apologies. But just a side note, people new to calculus always ask, and understandably so, “why is the derivative of a constant 0?” Because a derivative calculates rate of change. And a constant, or constant function is just that, constant! It’s rate of change is constant at 0 and never changes and therefore has a 0 rate of change, that is, it’s derivative is 0! The rate of change of a constant, or of a constant function, therefore, is always the same, so it’s rate of change, that is, it’s derivative is always 0. For example, the function y = 2, say, is a “constant” function, or 2 is a constant, same thing, because it is a horizontal line that has a slope of 0, or a “derivative” of 0, same thing, that never changes and is therefore called a constant function, or a constant. That’s why the derivative of a constant, or the derivative of a constant function, in this example of y = 2, say, is 0. The derivative (which is just the slope at every point along the function’s graph) is 0 because the slope (the derivative) of that equation, by the point-slope equation, or the point-slope formula for slope (m), [m = (y2 - y1)/(x2 - x1)], gives m = 0 because (y2 - y1) always equals 0 for every (x2 - x1) along the x-axis simply because y = 2 throughout the entirety of the length of the line y = 2! The derivative is so powerful that it also calculates that truth!
Hope I didn’t confuse things, there, by giving more info than you perhaps need or even asked for in answer to your particular question.
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