title

Implicit differentiation, what's going on here? | Chapter 6, Essence of calculus

description

Implicit differentiation can feel strange, but thought of the right way it makes a lot of sense.
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Timestamps
0:00 - Opening circle example
3:08 - Ladder example
7:43 - Implicit differentiation intuition
12:33 - Derivative of ln(x)
14:23 - Outro
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{'title': "Implicit differentiation, what's going on here? | Chapter 6, Essence of calculus", 'heatmap': [{'end': 386.053, 'start': 363.042, 'weight': 0.862}, {'end': 433.216, 'start': 405.869, 'weight': 0.721}, {'end': 862.544, 'start': 824.571, 'weight': 0.782}], 'summary': 'Covers topics including implicit differentiation, finding slope of tangent line to a circle, solving ladder slipping problem using pythagorean theorem and rates of change, and derivatives of functions with multiple variables and its applications in multivariable calculus.', 'chapters': [{'end': 205.396, 'segs': [{'end': 67.209, 'src': 'embed', 'start': 27.072, 'weight': 0, 'content': [{'end': 33.635, 'text': 'That is, all of the points on the circle are a distance 5 from the origin, as encapsulated by the Pythagorean theorem,', 'start': 27.072, 'duration': 6.563}, {'end': 39.358, 'text': 'where the sum of the squares of the two legs on this triangle equal the square of the hypotenuse 5 squared.', 'start': 33.635, 'duration': 5.723}, {'end': 48.122, 'text': 'And suppose that you want to find the slope of a tangent line to the circle, maybe at the point.', 'start': 40.479, 'duration': 7.643}, {'end': 55.564, 'text': "Now, if you're savvy with geometry, you might already know that this tangent line is perpendicular to the radius touching it at that point.", 'start': 48.122, 'duration': 7.442}, {'end': 62.587, 'text': "But let's say you don't already know that, or maybe you want a technique that generalizes to curves other than just circles.", 'start': 56.325, 'duration': 6.262}, {'end': 67.209, 'text': 'As with other problems about the slopes of tangent lines to curves,', 'start': 63.687, 'duration': 3.522}], 'summary': 'Geometry concept: calculating slope of tangent line to a circle.', 'duration': 40.137, 'max_score': 27.072, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa427072.jpg'}, {'end': 115.204, 'src': 'embed', 'start': 88.422, 'weight': 3, 'content': [{'end': 94.187, 'text': 'But unlike other tangent slope problems in calculus, this curve is not the graph of a function.', 'start': 88.422, 'duration': 5.765}, {'end': 103.454, 'text': "So we can't just take a simple derivative asking about the size of some tiny nudge to the output of a function caused by some tiny nudge to the input.", 'start': 94.727, 'duration': 8.727}, {'end': 107.077, 'text': 'x is not an input, and y is not an output.', 'start': 104.255, 'duration': 2.822}, {'end': 111.661, 'text': "They're both just interdependent values related by some equation.", 'start': 107.718, 'duration': 3.943}, {'end': 115.204, 'text': "This is what's called an implicit curve.", 'start': 112.882, 'duration': 2.322}], 'summary': "Unlike typical tangent slope problems, this curve is not a function graph; it's an implicit curve.", 'duration': 26.782, 'max_score': 88.422, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa488422.jpg'}, {'end': 205.396, 'src': 'embed', 'start': 162.386, 'weight': 4, 'content': [{'end': 171.572, 'text': 'But if you just blindly move forward with what you get, you can rearrange this equation and find an expression for dy divided by dx,', 'start': 162.386, 'duration': 9.186}, {'end': 176.136, 'text': 'which in this case comes out to be negative, x divided by y.', 'start': 171.572, 'duration': 4.564}, {'end': 184.141, 'text': 'So, at the point with coordinates that slope would be negative 3 divided by 4, evidently.', 'start': 176.136, 'duration': 8.005}, {'end': 188.925, 'text': 'This strange process is called implicit differentiation.', 'start': 185.342, 'duration': 3.583}, {'end': 195.79, 'text': "And don't worry, I have an explanation for how you can interpret taking a derivative of an expression with two variables like this.", 'start': 189.585, 'duration': 6.205}, {'end': 203.595, 'text': "But first I want to set aside this particular problem and show how it's connected to a different type of calculus problem,", 'start': 196.65, 'duration': 6.945}, {'end': 205.396, 'text': 'something called a related rates problem.', 'start': 203.595, 'duration': 1.801}], 'summary': 'Explanation of implicit differentiation & related rates in calculus.', 'duration': 43.01, 'max_score': 162.386, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4162386.jpg'}], 'start': 11.223, 'title': 'Calculus and implicit differentiation', 'summary': 'Discusses finding the slope of a tangent line to a circle with radius 5 at the origin of the xy-plane, offering a general technique for curves beyond circles, and explains the concept of implicit differentiation in calculus resulting in a slope of -3/4, and connects it to related rates problems.', 'chapters': [{'end': 67.209, 'start': 11.223, 'title': 'Calculus: tangent line to a circle', 'summary': 'Discusses finding the slope of a tangent line to a circle with radius 5 at the origin of the xy-plane, using the pythagorean theorem and the concept of perpendicularity, offering a general technique for curves beyond circles.', 'duration': 55.986, 'highlights': ['The tangent line to the circle is perpendicular to the radius touching it at that point, a fundamental concept in geometry.', "The circle's equation x squared plus y squared equals 5 squared defines all points on the circle as a distance 5 from the origin, illustrating the application of the Pythagorean theorem.", 'The chapter presents a technique for finding the slope of tangent lines to curves beyond circles, providing a more general approach to the problem.']}, {'end': 205.396, 'start': 67.209, 'title': 'Implicit differentiation in calculus', 'summary': 'Explains the concept of implicit differentiation in calculus through the process of finding the slope of an implicit curve using the derivative of both x and y components, resulting in a slope of -3/4, and then connects it to related rates problems.', 'duration': 138.187, 'highlights': ['The process of finding dy/dx for curves like this involves taking the derivative of both sides of the equation, resulting in an expression for dy/dx, which in this case comes out to be -x/y.', 'Implicit differentiation is the procedure for finding the derivative of an implicit curve, where x and y are interdependent values related by some equation, unlike the graph of a function.', 'The chapter also connects the concept of implicit differentiation to related rates problems in calculus.']}], 'duration': 194.173, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa411223.jpg', 'highlights': ['The tangent line to the circle is perpendicular to the radius touching it at that point, a fundamental concept in geometry.', "The circle's equation x squared plus y squared equals 5 squared defines all points on the 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the Pythagorean theorem,', 'start': 206.597, 'duration': 9.124}, {'end': 218.663, 'text': 'means that the bottom is 3 meters away from the wall.', 'start': 215.721, 'duration': 2.942}, {'end': 225.706, 'text': "And let's say it's slipping down in such a way that the top of the ladder is dropping at a rate of 1 meter per second.", 'start': 219.683, 'duration': 6.023}, {'end': 233.73, 'text': "The question is in that initial moment, what's the rate at which the bottom of the ladder is moving away from the wall?", 'start': 226.727, 'duration': 7.003}, {'end': 235.917, 'text': "It's interesting, right?", 'start': 234.997, 'duration': 0.92}, {'end': 244.46, 'text': 'That distance from the bottom of the ladder to the wall is 100% determined by the distance from the top of the ladder to the floor.', 'start': 236.538, 'duration': 7.922}], 'summary': 'A 5m ladder is sliding down with top dropping 1m/s, bottom moves 3/5 m/s away from wall.', 'duration': 37.863, 'max_score': 206.597, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4206597.jpg'}, {'end': 299.428, 'src': 'embed', 'start': 266.386, 'weight': 1, 'content': [{'end': 268.607, 'text': "written as a function of time, because it's changing.", 'start': 266.386, 'duration': 2.221}, {'end': 273.408, 'text': 'Likewise label the distance between the bottom of the ladder and the wall.', 'start': 269.647, 'duration': 3.761}, {'end': 282.974, 'text': 'x. The key equation that relates these terms is the Pythagorean theorem x squared plus y squared equals 5 squared.', 'start': 273.408, 'duration': 9.566}, {'end': 288.421, 'text': "What makes that a powerful equation to use is that it's true at all points of time.", 'start': 283.955, 'duration': 4.466}, {'end': 299.428, 'text': 'Now, one way that you could solve this would be to isolate x, and then you figure out what y has to be based on that 1 meter per second drop rate,', 'start': 290.12, 'duration': 9.308}], 'summary': 'Pythagorean theorem relates distance and time, true at all points.', 'duration': 33.042, 'max_score': 266.386, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4266386.jpg'}, {'end': 386.053, 'src': 'heatmap', 'start': 363.042, 'weight': 0.862, 'content': [{'end': 374.707, 'text': 'But on the other hand, what do you get when you compute this derivative? Well the derivative of x squared is 2 times x times the derivative of x.', 'start': 363.042, 'duration': 11.665}, {'end': 376.888, 'text': "That's the chain rule that I talked about last video.", 'start': 374.707, 'duration': 2.181}, {'end': 386.053, 'text': "2x dx represents the size of a change to x squared caused by some change to x, and then we're dividing out by dt.", 'start': 377.889, 'duration': 8.164}], 'summary': 'Derivative of x squared is 2x times the derivative of x, representing the chain rule.', 'duration': 23.011, 'max_score': 363.042, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4363042.jpg'}, {'end': 433.216, 'src': 'heatmap', 'start': 387.423, 'weight': 3, 'content': [{'end': 395.986, 'text': 'Likewise, the rate at which y of t squared is changing is 2 times y of t times the derivative of y.', 'start': 387.423, 'duration': 8.563}, {'end': 399.087, 'text': 'Evidently, this whole expression must be 0,', 'start': 395.986, 'duration': 3.101}, {'end': 404.449, 'text': "and that's an equivalent way of saying that x squared plus y squared must not change while the ladder moves.", 'start': 399.087, 'duration': 5.362}, {'end': 413.611, 'text': 'At the very start, time t equals 0, the height y is 4 meters, and that distance x is 3 meters.', 'start': 405.869, 'duration': 7.742}, {'end': 423.374, 'text': 'Since the top of the ladder is dropping at a rate of 1 meter per second, that derivative dy dt is negative 1 meters per second.', 'start': 414.671, 'duration': 8.703}, {'end': 433.216, 'text': 'This gives us enough information to isolate the derivative dx dt, and when you work it out, it comes out to be 4 thirds meters per second.', 'start': 424.874, 'duration': 8.342}], 'summary': "The ladder's rate of change in x and y yields dx/dt = 4/3 m/s.", 'duration': 26.188, 'max_score': 387.423, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4387423.jpg'}, {'end': 458.353, 'src': 'embed', 'start': 434.454, 'weight': 4, 'content': [{'end': 441.5, 'text': 'The reason I bring up this ladder problem is that I want you to compare it to the problem of finding the slope of a tangent line to the circle.', 'start': 434.454, 'duration': 7.046}, {'end': 447.044, 'text': 'In both cases, we had the equation x squared plus y squared equals 5 squared,', 'start': 442.38, 'duration': 4.664}, {'end': 451.327, 'text': 'and in both cases we ended up taking the derivative of each side of this expression.', 'start': 447.044, 'duration': 4.283}, {'end': 458.353, 'text': 'But for the ladder question, these expressions were functions of time, so taking the derivative has a clear meaning.', 'start': 452.148, 'duration': 6.205}], 'summary': 'Comparing ladder problem to circle tangent line, both involve taking derivative of x^2 + y^2 = 5^2', 'duration': 23.899, 'max_score': 434.454, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4434454.jpg'}], 'start': 206.597, 'title': 'Ladder slipping problem', 'summary': 'Discusses a scenario where a 5-meter ladder, positioned 4 meters above the ground, is slipping down at a rate of 1 meter per second, requiring the determination of the rate at which the bottom of the ladder is moving away from the wall in the initial moment, with the pythagorean theorem and rates of change being key concepts.', 'chapters': [{'end': 288.421, 'start': 206.597, 'title': 'Ladder slipping problem', 'summary': 'Discusses a scenario where a 5-meter ladder, positioned 4 meters above the ground, is slipping down at a rate of 1 meter per second, requiring the determination of the rate at which the bottom of the ladder is moving away from the wall in the initial moment, with the pythagorean theorem and rates of change being key concepts.', 'duration': 81.824, 'highlights': ['The Pythagorean theorem x squared plus y squared equals 5 squared is crucial for relating the distances x and y, and it holds true at all points of time.', "The ladder, 5 meters long with the top starting 4 meters above the ground, implies that the bottom is 3 meters away from the wall, forming the basis for the problem's setup.", "The ladder's top dropping at a rate of 1 meter per second prompts the inquiry into the initial rate at which the bottom of the ladder moves away from the wall, leading to the exploration of rates of change for the values x and y."]}, {'end': 469.563, 'start': 290.12, 'title': 'Derivative of x and y', 'summary': 'Demonstrates a different method for solving a problem involving the derivative of x and y, resulting in a rate of 4 thirds meters per second, and compares it to finding the slope of a tangent line to the circle.', 'duration': 179.443, 'highlights': ['The derivative of x squared and y squared being 0 is an equivalent way of stating that x squared plus y squared must not change while the ladder moves, resulting in the rate of 4 thirds meters per second.', 'Comparing the problem of finding the slope of a tangent line to the circle with the ladder problem emphasizes the clear meaning of taking the derivative of the expressions in the ladder question.']}], 'duration': 262.966, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4206597.jpg', 'highlights': ["The ladder's top dropping at a rate of 1 meter per second prompts the inquiry into the initial rate at which the bottom of the ladder moves away from the wall.", 'The Pythagorean theorem x squared plus y squared equals 5 squared is crucial for relating the distances x and y, and it holds true at all points of time.', "The ladder, 5 meters long with the top starting 4 meters above the ground, implies that the bottom is 3 meters away from the wall, forming the basis for the problem's setup.", 'The derivative of x squared and y squared being 0 is an equivalent way of stating that x squared plus y squared must not change while the ladder moves, resulting in the rate of 4 thirds meters per second.', 'Comparing the problem of finding the slope of a tangent line to the circle with the ladder problem emphasizes the clear meaning of taking the derivative of the expressions in the ladder question.']}, {'end': 899.985, 'segs': [{'end': 515.373, 'src': 'embed', 'start': 469.563, 'weight': 0, 'content': [{'end': 477.146, 'text': 'which causes x and y to change the derivative, just has these tiny nudges dx and dy just floating free,', 'start': 469.563, 'duration': 7.583}, {'end': 479.888, 'text': 'not tied to some other common variable like time.', 'start': 477.146, 'duration': 2.742}, {'end': 482.829, 'text': 'Let me show you a nice way to think about this.', 'start': 481.228, 'duration': 1.601}, {'end': 488.652, 'text': "Let's give this expression x squared plus y squared a name, maybe s.", 'start': 483.389, 'duration': 5.263}, {'end': 491.073, 'text': 's is essentially a function of two variables.', 'start': 488.652, 'duration': 2.421}, {'end': 495.616, 'text': 'It takes every point xy on the plane and associates it with a number.', 'start': 491.794, 'duration': 3.822}, {'end': 500.658, 'text': 'For points on this circle, that number happens to be 25.', 'start': 496.596, 'duration': 4.062}, {'end': 504.342, 'text': 'If you stepped off the circle, away from the center, that value would be bigger.', 'start': 500.658, 'duration': 3.684}, {'end': 508.927, 'text': 'For other points x, y closer to the origin, that value would be smaller.', 'start': 505.143, 'duration': 3.784}, {'end': 515.373, 'text': 'Now what it means to take a derivative of this expression a derivative of s,', 'start': 509.888, 'duration': 5.485}], 'summary': 'The derivative of the expression x squared plus y squared, named s, is essentially a function of two variables.', 'duration': 45.81, 'max_score': 469.563, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4469563.jpg'}, {'end': 766.428, 'src': 'embed', 'start': 740.188, 'weight': 2, 'content': [{'end': 745.411, 'text': "From there, depending on what problem you're trying to solve, you have something to work with algebraically.", 'start': 740.188, 'duration': 5.223}, {'end': 750.494, 'text': 'And maybe the most common goal is to try to figure out what dy divided by dx is.', 'start': 745.931, 'duration': 4.563}, {'end': 761.665, 'text': 'As a final example here, I want to show you how you can actually use this technique of implicit differentiation to figure out new derivative formulas.', 'start': 753.621, 'duration': 8.044}, {'end': 766.428, 'text': "I've mentioned that the derivative of e to the x is itself.", 'start': 762.606, 'duration': 3.822}], 'summary': 'Using implicit differentiation to find new derivative formulas.', 'duration': 26.24, 'max_score': 740.188, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4740188.jpg'}, {'end': 862.544, 'src': 'heatmap', 'start': 811.918, 'weight': 3, 'content': [{'end': 816.861, 'text': 'Since we know the derivative of e to the y, we can take the derivative of both sides here,', 'start': 811.918, 'duration': 4.943}, {'end': 823.285, 'text': 'effectively asking how a tiny step with components dx dy changes the value of each one of these sides.', 'start': 816.861, 'duration': 6.424}, {'end': 837.853, 'text': 'To ensure that a step stays on the curve, the change to this left side of the equation must equal the change to the right side.', 'start': 824.571, 'duration': 13.282}, {'end': 846.977, 'text': 'Rearranging that means that dy divided by dx, the slope of our graph, equals 1 divided by e, to the y.', 'start': 837.853, 'duration': 9.124}, {'end': 851.739, 'text': "And when we're on the curve, e to the y is, by definition, the same thing as x.", 'start': 846.977, 'duration': 4.762}, {'end': 855.821, 'text': 'So evidently, this slope is 1 divided by x.', 'start': 851.739, 'duration': 4.082}, {'end': 862.544, 'text': 'And of course, an expression for the slope of a graph of a function written in terms of x like this is the derivative of that function.', 'start': 855.821, 'duration': 6.723}], 'summary': 'The derivative of e to the y is 1/x.', 'duration': 25.935, 'max_score': 811.918, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4811918.jpg'}, {'end': 892.919, 'src': 'embed', 'start': 863.304, 'weight': 4, 'content': [{'end': 867.086, 'text': 'So evidently, the derivative of ln of x is 1 divided by x.', 'start': 863.304, 'duration': 3.782}, {'end': 877.885, 'text': 'By the way, all of this is a little sneak peek into multivariable calculus,', 'start': 873.001, 'duration': 4.884}, {'end': 883.751, 'text': 'where you consider functions that have multiple inputs and how they change as you tweak those multiple inputs.', 'start': 877.885, 'duration': 5.866}, {'end': 892.919, 'text': 'The key, as always, is to have a clear image in your head of what tiny nudges are at play and how exactly they depend on each other.', 'start': 884.811, 'duration': 8.108}], 'summary': 'Derivative of ln x is 1/x. multivariable calculus involves functions with multiple inputs and their changes.', 'duration': 29.615, 'max_score': 863.304, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4863304.jpg'}], 'start': 469.563, 'title': 'Derivatives and differentiation', 'summary': 'Discusses the concept of taking the derivative of a function of two variables and uses the example of x squared plus y squared to illustrate its behavior on a circle with a radius of 5. it also explains implicit differentiation, demonstrating how to find the derivative of a function with multiple variables and its applications in multivariable calculus.', 'chapters': [{'end': 515.373, 'start': 469.563, 'title': 'Derivative of a function', 'summary': 'Discusses the concept of taking the derivative of a function of two variables and uses the example of x squared plus y squared to illustrate its behavior on a circle with a radius of 5.', 'duration': 45.81, 'highlights': ['The expression x squared plus y squared, denoted as s, represents a function of two variables, associating every point xy on the plane with a number.', 'For points on the circle with a radius of 5, the value of s is 25, and it increases when moving away from the center, and decreases for points closer to the origin.']}, {'end': 899.985, 'start': 515.373, 'title': 'Implicit differentiation', 'summary': "Explains implicit differentiation, demonstrating how to find the derivative of a function with multiple variables and how it's used to derive new derivative formulas, with a focus on the key points of the technique and its applications in multivariable calculus.", 'duration': 384.612, 'highlights': ['Implicit differentiation is used to find the derivative of a function with multiple variables, allowing for the derivation of new derivative formulas.', 'The technique involves taking the derivative of both sides of an equation to understand how the value of each side changes during a tiny step with components dx and dy.', 'It emphasizes the importance of having a clear understanding of how tiny nudges in multiple inputs affect each other and provides a sneak peek into multivariable calculus.']}], 'duration': 430.422, 'thumbnail': 'https://coursnap.oss-ap-southeast-1.aliyuncs.com/video-capture/qb40J4N1fa4/pics/qb40J4N1fa4469563.jpg', 'highlights': ['The expression x squared plus y squared, denoted as s, represents a function of two variables, associating every point xy on the plane with a number.', 'For points on the circle with a radius of 5, the value of s is 25, and it increases when moving away from the center, and decreases for points closer to the origin.', 'Implicit differentiation is used to find the derivative of a function with multiple variables, allowing for the derivation of new derivative formulas.', 'The technique involves taking the derivative of both sides of an equation to understand how the value of each side changes during a tiny step with components dx and dy.', 'It emphasizes the importance of having a clear understanding of how tiny nudges in multiple inputs affect each other and provides a sneak peek into multivariable calculus.']}], 'highlights': ['Implicit differentiation is used to find the derivative of a function with multiple variables, allowing for the derivation of new derivative formulas.', 'The chapter also connects the concept of implicit differentiation to related rates problems in calculus.', 'The process of finding dy/dx for curves like this involves taking the derivative of both sides of the equation, resulting in an expression for dy/dx, which in this case comes out to be -x/y.', 'The expression x squared plus y squared, denoted as s, represents a function of two variables, associating every point xy on the plane with a number.', 'The tangent line to the circle is perpendicular to the radius touching it at that point, a fundamental concept in geometry.', "The circle's equation x squared plus y squared equals 5 squared defines all points on the circle as a distance 5 from the origin, illustrating the application of the Pythagorean theorem.", "The ladder's top dropping at a rate of 1 meter per second prompts the inquiry into the initial rate at which the bottom of the ladder moves away from the wall.", 'The Pythagorean theorem x squared plus y squared equals 5 squared is crucial for relating the distances x and y, and it holds true at all points of time.', 'The chapter presents a technique for finding the slope of tangent lines to curves beyond circles, providing a more general approach to the problem.', 'The derivative of x squared and y squared being 0 is an equivalent way of stating that x squared plus y squared must not change while the ladder moves, resulting in the rate of 4 thirds meters per second.']}