Problem

A swimming pool is 20 $\mathrm{ft}$ wide, 40 $\ma…

01:28

Question

Answered step-by-step

Problem 27 Medium Difficulty

A water trough is $ 10 m $ long and a cross-section has the shape of an isosceles trapezoid that is $ 30 cm $ wide at the bottom, $ 80 cm $ wide at the top, and has height $ 50 cm, $ If the trough is being filled with water at the rate $ 0.2 m^3/min, $ how fast is the water level rising when the water is $ 30 cm $ deep?

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Carson Merrill

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Watch More Solved Questions in Chapter 3

Problem 1

Problem 2

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

So what we see here is we have a water trough and our ultimate goal is to find the volume. Um Oh, are ultimately to find how fast the water level is rising. So we want to, um, refer to these values with different equations so we can take the derivative of them. So we know that the volume is going to be equal to the area of the cross section times the length of the trough. So if we have our trough right here has depth like this. And if this is the cross section right here because this goes all the way through and then we have the cross section times the length that will give us the volume. So we know the length of the trough is given as a constant of 10 m so we can differentiate both sides. Um, with respect to t, we get TV. D t is equal to l times. D a d t. So l is 10. So we could just write that is 10 times g eight et. Then we want to find the area of the cross section, which we know is a It's a nice sauces trapezoid. So if we look for the area of I saw Seles trapezoid. What we end up getting is the length of the upper base, plus the lower base divided by two times the height. So we know that the length of the upper base at any time is two K plus 30. Um and that's going to be also we know that the using triangles within our trapezoid. So if this is our trapezoid right here using this diagram where we have these small triangles with links K, this is 30 30 30. This is gonna be 25 25 thing we have right here the height and then right here is going to be 50. So based on all of this, we know that 25 over K is equal to 50 over H. So knowing that we see that H is equal to two K, um, and therefore, if we have two k 30 that's the same thing. A two K plus 30. We could rewrite that as a judge, plus 30. Then we know that the lower bases length 30. So we add that, and then that's gonna be over to times the height. So another way we could write this now is, um, going to be HCGH squared over to plus 30 h and this is helpful because we now have our area which we can take the derivative of and end up getting that d a d t is equal Thio h plus 30 times d h d t. Thankfully, though we consult itude this expression in for the previous equation because our goal is to find d h d t. So we have the DVD t is equal thio l times h plus 30 times d h d t. We know the values that we've been given our the volume So 0.2, uh, meters cubits per minute And then we also have, um since we're dealing with centimeters the way that we're gonna one right, this is 0.2 times 10 to the sixth centimeters, cubes per minute. And then we know that since the length is 10, the height is 30. So we have a length of 10 right here and a height of 30. So based on that, we now have that two times 10 to the fifth is equal to six times 10 to the fourth D h DT. We divide these and we end up getting that D h d t is going to give us, um, 13th centimeters per minute, so that will be our final result. And that's the rate at which the water level is increasing.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem

A swimming pool is 20 $\mathrm{ft}$ wide, 40 $\ma…

This textbook answer is only visible when subscribed! Please subscribe to view the answer

More Answers

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 1 / AB Educators

Grace He

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

Calculus 1 / AB Courses

Recommended Videos

Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

James Stewart

Calculus: Early Transcendentals