Question

FAN + MEDAL
Calculus Help

Answer 1

Circular cylinder = 16 inches
Radius = 4 inches
error = ± 0.05 inches

Answer 2

The formula for a cylinder's Volume is V=πr^2h

Answer 3

what the error says is that you should plug in the error to the Radius and to the height of the cylinder like so

Answer 4

Circular cylinder = 16 ± 0.05 inches
Radious = 4 ± 0.05 inches

Answer 5

From the formula you can see that we are multiplying the Radious (r) with Height(h)

Answer 6

When you multiply quantities with errors in them you ADD the relative errors together

Answer 7

And you get V ± (what you found from above)

Answer 8

Also keep in mind that r is to the power of 2

Answer 9

so r*r*h

Answer 10

add the relative errors, three times

Answer 11

(1 - radius/0.05) + (1 - radius/0.05) + (1 - height/0.05)

Answer 12

By relative error I mean this --> (1 - value/error)

Answer 13

I think they are saying the error is only on the measurement of the radius (though it's not totally clear)

Answer 14

hm

Answer 15

in this case you just have to do (1 - radius/0.05) + (1 - radius/0.05)

Answer 16

wouldn't you use differentials, i.e.
dV= 2pi r h dr
and then replace the differentials with "deltas"
\[ \Delta V = 2 \pi r h \Delta r \]

Answer 17

uhh I dunno about that technique , would be interested to see how it's done

Answer 18

Was doing it that way in physics experiments

Answer 19

The radius can range from 3.95 to 4.05,
so the volume can range from
\(\large\rm v_1=2\pi(3.95)^2h\)
to
\(\large\rm v_2=2\pi(4.05)^2h\)

And then you can measure the difference from the original volume, ya?
That's how I'm thinking about it at least :o

Answer 20

I guess that could also be done that way yea

Answer 21

Ugh. so confuseddd.

Answer 22

I personally would be interested to see @phi solution :D

Answer 23

Since I can't quite grasp the thinking of it

Answer 24

\[ \Delta V = 2 \pi r h \Delta r \\
\pm \Delta V = 2\pi \cdot 4\cdot 16 \cdot \pm 0.05 \\
\pm \Delta V = \pm 6.4\pi\ cu.in. \approx \pm 20.106\ cu. in. \]

Answer 25

If we compute the upper and lower bound on the volume we would get
\[V_{high}= \pi r^2 h = (4.05)^2 \cdot 16 \pi = 262.44\pi \ cu. \in.\]
\[V_{low}= \pi r^2 h = (3.95)^2 \cdot 16 \pi = 249.64\pi \ cu. \in.\]
and the error is (Vhigh-Vlow)/2 = 12.8 pi/2 = 6.4 pi or about \[\pm 20.106 \ cu.\ in.\]

Answer 26

lol it turned your inches into \in,
i hate that :)

Answer 27

in case it was not clear,
\[ \frac{d}{dr}\left(V= \pi r^2 \ h \right)\\
\frac{dV}{dr} = 2 \pi r h \\
dV= 2\pi r \ h \ dr \]

Answer 28

and the idea is if the uncertainty is small, we can use the approximation
\[ \Delta V \approx 2 \pi r \ h\ \Delta r \]

Answer 29

how did you go from

ΔV=2π⋅4⋅16⋅±0.05± to

ΔV=±6.4π cu.in.≈±20.106 cu.in

Answer 30

2*4*16 ?

Answer 31

I put in 4 for r, 16 for h , 0.05 for \( \Delta r\)
into
\[ \Delta V \approx 2 \pi r \ h\ \Delta r \]

Answer 32

the \( \pm\) is probably in the wrong spot, perhaps I should write is as
\[ \Delta V \approx \pm \ 2 \pi \cdot 4 \cdot 16 \cdot 0.05 \ in^3 \]