Problem

A particle is moving along a hyperbola $ xy = 8. …

01:57

Problem 11 Easy Difficulty

If $ x^2 + y^2 + z^2 = 9, dx/dt = 5, $ and $ dy/dt = 4, $ find $ dz/dt $ when $ (x, y, z) = (2, 2, 1). $

Name: if x2 y2 z2 9 dxdt 5 and dydt 4 find dzdt when x y z 2 2 1
Uploaded: 2021-01-01
Duration: 2 min 13 s
Channel: Numerade
Description: If $ x^2 + y^2 + z^2 = 9, dx/dt = 5, $ and $ dy/dt = 4, $ find $ dz/dt $ when $ (x, y, z) = (2, 2, 1). $

Answer

$\frac{d z}{d t}=-18$

Grace H.

Kayleah T.

Harvey Mudd College

Samuel H.

University of Nottingham

Michael J.

Idaho State University

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 in this problem, the first thing what we want to focus on is just looking at what we have. So we have X squared bus y squared a Z squared People's Nine and ultimately, this problem is helping us to do a better job of understanding related rates problems. So what we have here is, uh, we're gonna want to start by taking the derivative. So when we differentiate with respect to T that is on both sides, we'll end up getting two x times dx DT Class two I times d y d t class choosy. I'm dizzy E t. And we know this is equal to zero because we differentiate the nine. And with all this in mind, we end up seeing that X is equal toe one are actually X is equal to two. But we can factor out the two out of here. That makes it much easier. We know the X is equal to two. Why is equal to two z is equal to one so we can substitute those. We know that DX DT is equal to five d y d t is. You go to form and we're trying to find these e D t. So, since this we're multiplying by one here, we can just simplify that. That this is eight. This is 10. So 10 plus eight is 18. So now we can put this over here. When we subtract it, it'll be negative. 18. So what we've just done is found busy, bt, um, which essentially means we found the derivative of Z with respect to t. Um and we know that to be 18, a negative 18.

Carson M.

California Baptist University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 9

Related Rates