Question

what is the difference between dv and Δv?
here is my problem: a sphere increases in diameter from 5'' to 5.1''
a)calculate dv
b)calculate Δv
c)calculate the error

Answer 1

dv is used in calculus to represent an infitescimal. An infiniteschimal can be thought of as a change (in this case a change in v) that is almost zero. Actually, it is the closest "number" to zero without being zero. On the other hand, Δv is simply the difference (final volume - initial volume). Does this help?

Answer 2

ok.. so Δv would be .1
but im still confused with dv? how would you calculate that.

Answer 3

I don't think that's what it is asking. I think it is talking about the change in volume of the sphere when the diameter is changed by 0.1 and the infintescimal change in volume at d=5.

Answer 4

im really confused... lol.

Answer 5

Δv = V₂ - V₁
dv is derivative of V ( instantaneous volume rate)

Answer 6

*rate of change

Answer 7

ok.. so does that mean that dv is just 0?

Answer 8

No, you need to take derivative to get dv !

Answer 9

but i thought the derivative of 5 is 0

Answer 10

No, take derivative of V ( Volume of Sphere formula)

Answer 11

oooh ok so dv is \[4\pi r^2\]
and then Δv is just .1?

Answer 12

Oh, you don't know how to find dv, don't you!
-> dv = 4π r² dr

Answer 13

@zepdrix lol.

Answer 14

sandy potato \:D/

Answer 15

♥ help me.

Answer 16

Calculating error? :U Oh crapsicle... I don't remember doing this.. .hmmm.

Answer 17

do you know how to do A or B? i dont understandddd! ):

Answer 18

sec thinking c:

Answer 19

@zepdrix Percentage error: dv/ V

Answer 20

Hmm let's start with ΔV first.

So they told us the change in DIAMETER.
We're going to write those diameters in terms of Radii, and then calculate the 2 volumes that they give us.

Then, as Phyll stated, we'll be able to calculate the AVERAGE change in volume (ΔV) By taking the difference of the volumes!

ΔV=V₂ - V₁

Answer 21

For the first volume, we'll call V₁, it has a diameter of 5" which will give us a radius of 2.5" right? :o

Answer 22

If we plug that into our equation for a sphere, we get that V₁ is,\[\huge V_1=\frac{4}{3}\pi (2.5)^3 \approx 65.4498\]

Still wif me? :3

Answer 23

i get it (:

Answer 24

So V₂ will be the same process.
A diameter of 5.1" will give us a radius of 2.55" I think...
So for V₂, that gives us...\[\huge V_2=\frac{4}{3}\pi(2.55)^3 \approx 69.4559\]

Answer 25

Δv = V₂ - V₁
= (4π/3) ( 2.55 - 2.5)³

Answer 26

Pshhhh look at you getting all fancy with your factoring :D hah

Answer 27

how do you do those cute little subscripts and superscripts in normal text like that?? D:

Answer 28

ZEP ITS CALLED COPY AND PASTE FROM GOOGLE

Answer 29

or that.

Answer 30

Oh goodness :P bunch of noobs :3

Answer 31

Ok ok ok ok so to get ΔV, we simply take the difference of those 2 volumes, what'd you get? :O

Answer 32

@ksandoval Pleaze read our explanation :)

Answer 33

oh wait. nevermind

Answer 34

Nooooo :O
Δd, the difference in diameters is 5.1 - 5 = 0.1
Δr, the difference in radii is 2.55 - 2.5 = 0.05

ΔV, the difference in volumes is 69.4559 - 65.4498 = ?

Answer 35

and calculate with us, would you :)

Answer 36

4.0061 x.o

Answer 37

cool, so the AVERAGE rate of change of volume over whatever the time interval was, is about 4 cubic inches! c: cool.

ΔV, average rate of change
dV, instantaneous rate of change

Ok we got the first part done it looks like! :D

Answer 38

well wait, so is dv just the derivative of the formula for the volume of a sphere?

Answer 39

-.- ok.

Answer 40

lolol simmer down phyll XD that happens sometimes. students need to hear things multiple times before it actually sinks in :D don't get frustrated heh.

Answer 41

well i was just confused because i thought for the error you need to subtract dv from Δv, but dv is a formula..

Answer 42

Hmm I'm a little confused about the dV part myself.
Do they want you to calculate the derivative at 5 or at 5.1?
Maybe @Chlorophyll knows? D:

Answer 43

dv with r = 2.5 and r' = .05

Answer 44

Oh oh oh ok that makes sense :D

Answer 45

-> dv = 4π r² dr

Answer 46

ok. sooo error is .0791?

Answer 47

Oh sorry :O hmm

Answer 48

What was dv? 3.93 or something like that? :O

Answer 49

dV=4π r² dr --> dV=4π (2.5)² (.05) --> dV=3.92699

Answer 50

\[\large \frac{dV}{V}=\frac{3.92699}{V_1}\]I assume we're using V1 for the volume? Like we did with the radius earlier? :o

Answer 51

I'm coming up with 0.06 for the error. hmm

Answer 52

Hmm, I'm not enjoying these problems Sand... -_- do some more fun math lol