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Boyle's Law states that when a sample of gas is c…

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Problem 36 Hard Difficulty

A faucet is filling a hemispherical basin of diameter $60 \mathrm{~cm}$ with water at a rate of $2 \mathrm{~L} / \mathrm{min}$. Find the rate at which the water is rising in the basin when it is half full. [Use the following facts: $1 \mathrm{~L}$ is $1000 \mathrm{~cm}^{3}$. The volume of the portion of a sphere with radius $r$ from the bottom to a height $h$ is $V=\pi\left(r h^{2}-\frac{1}{3} h^{3}\right),$ as we will show in Chapter $6 .$


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Amrita Bhasin

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 9

Related Rates

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Derivatives

Differentiation

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Lectures

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04:40

Derivatives - Intro

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.

Video Thumbnail

44:57

Differentiation Rules - Overview

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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Watch More Solved Questions in Chapter 3

Problem 1
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Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50

Video Transcript

hemispherical basin. So hemispherical means half a sphere and its diameter. 60 centimeters that's being filled at a rate of two liters per minute. Find the rate the water is rising when the basin is half full. Okay, so first we need to draw a picture. So let's just draw side view half a circle and then put some water in there at the bottom. Okay. What we know so far is that the radius of the bowl is 37 30 centimeters, that's all. Okay, the water is going in at a rate of two liters per minute. So what that is doing is changing the volume of the water. So that's D V D t. Let's make that one blue D v D T rate of change of the volume with respected time. Two leaders per minute Find the right. The water is rising. So we're trying to find let's call it H here from the bottom of the basin to the top of the water. So find a d h d t when the basin is half full. So when it's half full, then h will be 15. So when H equals 15 centimeters okay, so we know the volume of a sphere is four thirds pi r cubed. But we don't have a whole sphere here. We're only feeling got a part of it filled up. So we need a special formula which they gave us right here. The volume of a portion of the sphere is pi. Times are h squared, minus one third h cute. So that's what we're gonna take the derivative of. Oh, and that are right. There is not the radius of the water but the radius of the sphere. Okay, First, we got to change this DVD t into Centimeters Cube. So we also are given this little bit of information. One leader is 1000 centimeters cube, so D V D t 2000 centimeters huge per minute. All right, so v is pi. Times are h squared minus one third h cube. So we're gonna take the drift, Give with respect to time So derivative of B Do you really see plays? A constant just comes along Oh, I forgot to put the radius and its 30 derivative of age square to age times the derivative of age D h d t minus one third time's derivative of H cube three h squared D h d t. All right now we gotta do is put the numbers in and simplify. We'll be all done 2000 equals pi times 60 times 15 d h d t minus 15 squared d h d t. So let's divide the pi over here. 2000 Divide by pi equals. So we have 60 times 15 minus 15 times. 15, um 60 is four times 15. I'm just doing it this way. So I don't have to get my calculator out, so I'm gonna factor of the 15 times 15 out I get four minus one there. So 15 times 15, that's to 25 times three 6 75. Yeah. All right. The H d t. So 2000 over 6. 75 pi equals d h d t. Okay, so DVT was in centimeters cubed. We divided by centimeter. Centimeters. So Senator used to keep per minute. So this is in centimeters per minute. Okay, We can reduce the 2000 and the 6 75 by Well, let's just do five. So 401 35 80. And I'm doing five again. I have 27 Okay, So D h d t 80/27 pi centimeters per minutes.

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Top Calculus 1 / AB Educators
Heather Zimmers

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Calculus 1 / AB Courses

Lectures

Video Thumbnail

04:40

Derivatives - Intro

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.

Video Thumbnail

44:57

Differentiation Rules - Overview

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

Join Course
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