A swimming pool is 5 m wide, 10 m long, 1 m deep at the...

Key Concepts

Volume Calculation in Variable Cross-Section

Calculating the volume of an object where the cross-sectional area changes with a variable involves integrating or formulating an explicit relationship between the changing dimension and the varying area. This setup is critical for relating the volume of fluid to the changing level in dynamic systems.

Similar Triangles

Similar triangles provide a geometric method for relating different dimensions in problems where proportions remain constant. In contexts involving non-uniform shapes, such as a sloped pool, similar triangles help express one dimension as a function of another, which is then used in subsequent calculations.

Implicit Differentiation

Implicit differentiation is applied when variables are related through an equation without one being isolated. It enables the differentiation of both sides of a relation with respect to time, revealing how the rate of change of one variable influences the rate of change of another.

Chain Rule

The chain rule is used in calculus to differentiate composite functions, allowing one to relate the time derivatives of interconnected variables. This is essential in related rates problems since the variables are typically linked through functions that depend on time either directly or indirectly.

Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to time given the rate of another related quantity. These problems require establishing a relationship between the variables involved, then differentiating with respect to time to connect the given rate to the unknown rate.

Transcript

00:01 In the given question, the following figure is given to us and we are asked that this is a swimming pool and it is 5 meters wide and 10 meters long and we are told that it is 1 meter deep at the shallow end and it is 3 meters deep at its deepest point right and now there is a cross section that is shown in the figure that we can see over here and we are asked if the pool is being filled at a rate of 0 .1 meter cube per minute, how fast is the water level rising when the depth at the deepest point is 1 meter.

00:43 So the depth at the deepest point meaning this is the deepest point in the pool right.

00:49 So if there is 1 meter of water in this deepest point in the pool, what is the rate at which what the water level is rising that is the height of the water rising in this pool when it is filled with water at the deepest point where it is the height of water currently is 1 meter.

01:19 So we can see that this portion of the swimming pool is having a total height of 2 meters.

01:28 So when it is filled 1 meter at the shallow end at the deepest point it means that water is filled like this right.

01:37 So let's assume that this height over here is 1 meter and now if you were to take this a 3d view of this pool what you would have is you will have the top of the pool will be like this the top of the pool let's elongate this a bit it will be like this and we will have a trapezium like figure on the side right.

02:14 So this will be how the swimming pool would be looking when you look at it as a 3d figure right.

02:26 So this is what you would have.

02:28 To find the volume of this water that has been filled you can see that this point over here this can be taken as the height of the water that has been filled which is 1 meter and we have been already told that the width of this pool is 5 meters and now we know that this distance over here that is this distance we don't know how much that is but we do know that this portion over here that is the bottom side this is 3 meters in length and the height of this this water that has been filled for 1 meter and now is having a shape of a trapezoid on the side it has a height of 1 meter it has a width of 5 meters and the length of the bottom side over here of the trapezium is 3 meters.

03:37 So to find the volume of any object what we generally do is we take the area of one of its faces and multiply it with the height or in this case we can say the width of the object right...

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