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Water is flowing into a conical reservoir 50 feet high with a top radius of 10 feet, at the rate of 16 cubic feet per second. At what rate is the depth of the water in the reservoir changing when the water is 10 feet deep? (The volume of a cone is $\frac{1}{3} \pi r^{2} h,$ where $r$ is the radius and $h$ its height.)

$$4 / \pi \approx 1.2732 \mathrm{ft} / \mathrm{sec}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Missouri State University

Harvey Mudd College

University of Nottingham

Idaho State University

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

04:36

A conical water tank with …

05:37

For the following exercise…

04:32

A conical tank (with verte…

It's always a good idea to start of related rates. Problems with a picture. So what we're told is we have a conical reservoir. So here's our cone. Okay. What do we know about this cone? We know that it's 50 ft high and has a top radius of 10 ft. These air unchanging numbers so I can put these right in my picture 50 ft high, 10 ft radius at the top. And I have water at a certain height. Okay, now, this height, though, this is changing. So the height here, I'm gonna call H and the radius. I'm going to call our because I'm putting water in so the overall dimensions don't change, but the water dimensions dio. So I get constants for the overall dimensions. Variables for the varying numbers. So let's write down what we know. We know that water is flowing in at a rate of 16 cubic feet per second. Well, that's a change in volume, So D v d t. It's increasing. So it's a positive 16 cubic feet per second. What else do we know? Well, I know I'm looking for I want the rate that the depth of the water is changing. Well, the depth of the water is H so D h d t is my unknown. That's what I'm trying to find. And I want to find this at the point that the water is 10 ft deep. I'm not putting that on my picture, because that's a snapshot, Not a constant. Okay. And we are told, are function in case you weren't sure of it. We're told that the volume of a cone is one third pi r squared H. Okay, so we have some things we know. We know what we're trying to find. We've been given a formula to use. Let's make sure we can were set to use this formula, though, before we proceed. As you can see, I have three variables V r and H. I have DVD t I have d h d t. What I don't have is d r d t. I don't know what it is. I'm not trying to find it. So I really don't wanna have a are in my formula because that's going to give me a d r d t. If I try to take the derivative. So what we're going to dio is we want to take it out of our formula. Can we rewrite that are in terms of one of our other variables? Okay. And we can That's why we're given the overall dimensions of our reservoir. Because the big triangle that I'm doing in red here and the small triangle with the water those are similar triangles. So we can say that the overall radius toe height ratio is the same as the waters radius toe height ratio, okay, or the radius to the height is 1/5. If I cross multiply, I get h equals five are now. I really want to put I want to get rid of the r. I wanna leave H because that's the one I know. So I'm gonna say are is h over five. So let's plug that into our formula that we were given. Volume equals one third pie h over five squared times H. And I'm just going to combine some things here that five in the denominator five squared that becomes we could write. That is pi over 75. Put all of our constants together. H cute. Okay, now, two variables, two rates. We are ready to solve this now we take our derivative, we get DVD T bring down the three. That's pi over 25 h squared times d h d t Plug in Everything that we know DVD T is 16 h at our particular point is 10. So this is 100 and D H d t is are unknown. I can simplify a bit 25 goes into 104 times and then we divide 16 Divided by pi times four Which means D H d t equals four over pie.

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