Ladder rate-of-change problem

Khan Academy

7 Jun 200810:59

EducationalLearning

32 Likes 10 Comments

TLDRThe video script presents a classic calculus problem involving a ladder sliding down a greased wall. The problem explores the rate of change of the ladder's top as its bottom moves horizontally at a constant speed of 4 meters per second. Utilizing the Pythagorean theorem and principles of implicit differentiation, the video demonstrates how to calculate the rate of descent of the ladder's top, resulting in a solution of -5.33 meters per second. This engaging problem highlights the application of calculus concepts in real-world scenarios.

Takeaways

📌 The ladder rate of change problem is a classic calculus problem used for educational purposes.
🧱 The problem involves a ladder of length 10 meters leaning against a greased wall and ground.
🚪 The bottom of the ladder is moving horizontally to the left at a speed of 4 meters per second.
🔄 The scenario is an example of a rate of change problem, specifically focusing on the motion of the top of the ladder.
📐 The Pythagorean theorem is applied to relate the horizontal (x) and vertical (y) distances from the wall and ground to the ladder's length.
🎓 The implicit function of x and y is formed based on the Pythagorean theorem (x^2 + y^2 = 100).
📈 Rates of change with respect to time (dx/dt and dy/dt) are calculated to find the velocity components.
🔢 At a given moment with x = 8 meters, y is found to be 6 meters using the Pythagorean theorem.
🌀 The derivative is taken of both sides of the equation with respect to time to find the rate of change of y.
📉 The rate of change of the top of the ladder downwards (dy/dt) is found to be -5.33 meters per second.
🔽 The negative sign of dy/dt indicates that the top of the ladder is moving downward as x increases horizontally.

Q & A

What is the ladder rate of change problem?
-The ladder rate of change problem is a classic calculus problem that involves a ladder sliding down a greased wall and floor, and the goal is to find the rate at which the top of the ladder is descending when the bottom of the ladder is moving horizontally at a given speed.
How is the problem visualized in the script?
-The problem is visualized with a diagram where the ground is brown, the wall is yellow, and the ladder is leaning against the wall. The ladder is 10 meters long, and the bottom of the ladder is moving to the left at 4 meters per second. The distance between the wall and the point where the ladder touches the ground is 8 meters.
What is the significance of the greased wall and floor?
-The greased wall and floor are assumed to prevent any friction that might otherwise affect the ladder's motion. This simplifies the problem by allowing the ladder to slide smoothly along the ground and down the wall.
How is the Pythagorean theorem used in this problem?
-The Pythagorean theorem is used to relate the lengths of the ladder, the horizontal distance (x), and the vertical distance (y). It states that x^2 + y^2 = 10^2, which equals 100, since the ladder's length is 10 meters.
What are the variables x and y representing in the problem?
-In the problem, x represents the horizontal distance from the wall to the base of the ladder, and y represents the vertical height of the ladder from the ground.
What is the given rate of change for x with respect to time?
-The given rate of change for x with respect to time (dx/dt) is 4 meters per second to the left.
How is the rate of change of y with respect to time (dy/dt) calculated?
-The rate of change of y with respect to time (dy/dt) is calculated by taking the derivative of both sides of the Pythagorean equation with respect to time and then solving for dy/dt using the given values for x, dx/dt, and y.
What is the value of y when x is 8 meters?
-When x is 8 meters, y can be found by substituting x into the Pythagorean equation: 8^2 + y^2 = 100, which gives y^2 = 100 - 64 = 36, so y is 6 meters.
What is the final calculated rate of change for the top of the ladder?
-The final calculated rate of change for the top of the ladder (dy/dt) is -5.33 meters per second, indicating that the top of the ladder is descending at a rate of 5.33 meters per second.
Why is the rate of change of y negative?
-The rate of change of y is negative because it represents the downward velocity of the top of the ladder. While the x-value is increasing positively, the y-value is decreasing, hence the negative sign.
How does the rate of change problem vary at different times?
-The rate of change problem varies at different times because the position of the ladder and the values of x and y change as time progresses. The rate of change calculated is specific to the given moment when x is 8 meters and the ladder is sliding at 4 meters per second.

Outlines

00:00

📐 Introduction to the Ladder Rate of Change Problem

The paragraph introduces a classic calculus problem involving a ladder sliding down a greased wall. The scenario describes a 10-meter long ladder leaning against a wall with its base sliding horizontally to the left at a speed of 4 meters per second. The challenge is to determine the rate of change of the top of the ladder moving downwards as the base moves horizontally. The problem is set up using the Pythagorean theorem, establishing a relationship between the ladder's length (hypotenuse), the horizontal distance (x), and the vertical height (y).

05:03

🧠 Deriving the Rate of Change for the Ladder

This paragraph delves into the process of finding the rate of change of the ladder's top (dy/dt) with respect to time. It starts by explaining the need to take the derivative of the Pythagorean equation (x^2 + y^2 = 100) with respect to time. The derivative of x^2 with respect to time is 2x(dx/dt), and the derivative of y^2 with respect to time is 2y(dy/dt). Since the ladder's length doesn't change, the derivative of 100 with respect to time is zero. Using the given values of x (8 meters) and dx/dt (4 meters per second), and solving for y from the Pythagorean equation, the paragraph outlines the steps to calculate dy/dt, which represents how fast the top of the ladder is moving down.

10:04

📉 Calculating the Downward Velocity of the Ladder's Top

The final paragraph concludes the problem by calculating the downward velocity (dy/dt) of the ladder's top as -5.33 meters per second. The calculation is based on the previous paragraph's setup, where 2x(dx/dt) + 2y(dy/dt) equals zero, and the known values are substituted in. The result is a negative value, reflecting that while the x-value increases positively, the y-value decreases negatively, indicating a downward motion. The paragraph also emphasizes the importance of considering the position in time and the shape of the right triangle formed by the ladder, the wall, and the ground to fully understand the dynamics of the problem.

Mindmap

Keywords

💡ladder rate of change problem

The ladder rate of change problem is a classic calculus problem that explores the concept of rates of change in a physical scenario. In the video, it involves a ladder sliding down a greased wall and ground, with the goal of finding the rate at which the top of the ladder is descending. This problem is used to illustrate the application of derivatives and the Pythagorean theorem in real-world situations, particularly in calculating instantaneous rates of change.

💡ground

In the context of the video, the ground represents the base on which the ladder rests. It is a key component of the problem as the ladder's bottom end slides along the ground. The ground is integral to understanding the horizontal movement of the ladder, which is described as moving to the left at a rate of 4 meters per second. This movement is crucial in calculating the rate of change of the ladder's top end.

💡wall

The wall is the vertical surface against which the ladder is leaning in the problem scenario. It is important because the ladder's length and the angle it forms with the ground and the wall are used to determine the ladder's position and the rate of change of its top end. The wall's presence and the greased condition of both the wall and the ground contribute to the ladder's sliding motion, which is central to the rate of change problem.

💡ladder

The ladder is the primary object of study in the video. It is described as being 10 meters long and is sliding down a greased wall and ground. The problem focuses on the motion of the ladder's top end as its bottom end moves horizontally. The ladder's length is a fixed constant in the Pythagorean equation used to relate the horizontal and vertical distances, and its motion is central to understanding the rate of change being calculated.

💡rate of change

Rate of change is a fundamental concept in calculus that refers to the speed at which a quantity changes with respect to another quantity. In the video, it is used to describe the velocity at which the top of the ladder is moving downwards as the bottom of the ladder slides to the left. The calculation of the rate of change is the main objective of the problem and is achieved through the use of derivatives and the Pythagorean theorem.

💡Pythagorean theorem

The Pythagorean theorem is a mathematical principle that states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In the video, the theorem is used to relate the lengths of the ladder (the hypotenuse) to the horizontal and vertical distances from the wall and the ground, allowing for the calculation of the rate of change of the ladder's top end.

💡implicit differentiation

Implicit differentiation is a technique used in calculus to differentiate equations that do not explicitly state the relationship between the dependent and independent variables. In the video, this method is employed to find the derivative of the equation derived from the Pythagorean theorem, which is an implicit function of x and y, to solve for the rate of change of the ladder's top end with respect to time.

💡chain rule

The chain rule is a fundamental principle in calculus that is used to find the derivative of a composite function. In the video, the chain rule is applied to the derivative of the squared term in the Pythagorean equation with respect to time, which involves taking the derivative of the function with respect to x and then multiplying by the derivative of x with respect to time.

💡horizontal displacement

Horizontal displacement refers to the change in position of an object along the horizontal axis. In the video, the horizontal displacement is represented by the variable x, which describes the leftward movement of the ladder's bottom end at a rate of 4 meters per second. This displacement is a key factor in determining the rate at which the top of the ladder descends.

💡vertical velocity

Vertical velocity is the speed at which an object moves vertically, either upwards or downwards. In the context of the video, the vertical velocity is the rate at which the top of the ladder is descending, which is the ultimate quantity being calculated. The negative value obtained for the vertical velocity indicates that the top of the ladder is moving downwards.

💡meters per second

Meters per second is a unit of speed, indicating the distance an object travels in one second. In the video, this unit is used to express both the rate at which the ladder's bottom end is moving horizontally (4 meters per second) and the calculated rate at which the top of the ladder is descending (negative 5.33 meters per second). These values provide a clear picture of the ladder's motion in terms of speed and direction.

Highlights

Introduction to the ladder rate of change problem as a classic calculus problem.

Visualization of the problem with a ground, wall, and ladder setup.

Description of the ladder's length as 10 meters and the ground's greasy condition causing the ladder's bottom to slide.

The bottom of the ladder moving to the left at 4 meters per second.

Use of the Pythagorean theorem to relate the distances x and y with the ladder's length.

Implicit function of x and y based on the right-angled triangle formed by the ladder, wall, and ground.

Derivation of the relationship between x and y with respect to time using the chain rule.

Given condition of the left point of the ladder moving at 4 meters per second and the distance being 8 meters.

Calculation of the rate of change of the top of the ladder moving downwards (dy/dt).

Use of the derivative to relate the rate of change of x (dx/dt) to the rate of change of y (dy/dt).

Substitution of known values (x=8, dx/dt=4) into the derived equation to solve for dy/dt.

Solution for dy/dt resulting in a value of -5.33 meters per second.

Explanation of the negative sign in dy/dt indicating the downward direction of the y-value's change.

Discussion on how the x and y velocities are related and their impact on the ladder's movement.

Mention of the ladder problem's dependence on the position in time and the right triangle's configuration.

Conclusion summarizing the ladder rate of change problem as a fundamental calculus question.

Ladder rate-of-change problem

Takeaways

Q & A

What is the ladder rate of change problem?

How is the problem visualized in the script?

What is the significance of the greased wall and floor?

How is the Pythagorean theorem used in this problem?

What are the variables x and y representing in the problem?

What is the given rate of change for x with respect to time?

How is the rate of change of y with respect to time (dy/dt) calculated?

What is the value of y when x is 8 meters?

What is the final calculated rate of change for the top of the ladder?

Why is the rate of change of y negative?

How does the rate of change problem vary at different times?