Question

Use differentials to approximate the maximum error propagated when calculating the volume of a sphere, if radius is 5 +/- .01 meters.

Answer 1

@Nishant_Garg

Answer 2

Ok so first of all we need to formulate our function, volume of a sphere is a function of it's radius. Thus we define
\[V(r)=\frac{4}{3}\pi r^3\]
And the formula for differentials as we have it is
\[dy=\frac{dy}{dx}.dx\]
Or in our case,
\[dV=\frac{dV}{dr}.dr\]
When a change is made in radius
\[\Delta r\] a corresponding change in volume is then
\[\Delta V\]
When these changes are small, not infinitely small but still quite small, we can approximate for the differentials
\[dr \approx \Delta r \space \space \space dV \approx \Delta V\]
\[\Delta V \approx \frac{dV}{dr}.\Delta r\]
Now we are given the error in the radius, which is our delta r, it is the small change that may occur in the radius
\[\Delta r=\pm 0.1\]
Thus now we can easily find delta V or the error expected in volume
\[\Delta V=\pm \frac{dV}{dr} \times 0.1\]
and of course we will put the actual value of r when we compute the derivative
so infact I should say\[\Delta V=\pm f'(5) \times 0.1\]

Answer 3

First find\[\frac{dV}{dr}\]
then put the value of r, which is given as 5 in the derivative, can u do that :)

Answer 4

I though radius is 5 +/- .01 so we will have 2 different radius for this problem right

Answer 5

No, our radius is only one and that is 5 meters, the plus minus .01 is actually the error in the radius, what we are trying to say is the radius may be more or less by 0.01 units, similarly using the plus minus notation for resultant error in volume we are trying to say that the volume may be more or less than the original amount by the value we've calculated

Answer 6

sorry I should've actually written
\[\Delta V=\pm f'(5) \times 0.01\]

Answer 7

hmm okay makes sense now. Alright so dv/dr means the direvitive of the volume/ the radius?

Answer 8

Yep! derivative of the volume function with respect to the radius

Answer 9

4pi r^2/ 5

Answer 10

yep! 4pi r^2, now just plug in r=5

Answer 11

Leave pi as it is, don't use any value for it

Answer 12

cool! so 100pi/5

Answer 13

There's no 5 in the denominator!
\[\frac{d}{dr}(\frac{4}{3}\pi r^3)=\frac{4}{3}\pi \frac{d}{dr}(r^3)=\frac{4\pi}{3}.3 r^2=4\pi r^2\]

Answer 14

so 100pi, now we just plug that into
\[\Delta V=\pm f'(5) \times 0.01\]
in place of f'(5)

Answer 15

plus minus 100pi *.01

Answer 16

yes!

Answer 17

now do u want me to go ahead and substitute the value for pi?

Answer 18

don't put the value of pi yet, rather I'd want to you to simplify that 100 and 0.01

Answer 19

+/- 1

Answer 20

+/- 1 pi

Answer 21

yep!
\[\Delta V=\pm \pi\]
Now if you want you may use 3.1416 or whatever value of pi you want to

Answer 22

or u can leave it at that

Answer 23

I will leave it like that. :)

Answer 24

So if the radius is more or less by 0.01 than the original value, the volume would be more or less by about 3.1416 cubic metres!

Answer 25

yes, tysm makes sense now. I was thinking two radius haha was on the wrong track

Answer 26

It's alright, take care!

Answer 27

I have another question. it's about Newton's method I need serious help with that one. do u know that? I can open that as a new question

Answer 28

hmm, not really familiar with Newton's method, I've heard about it but never encountered in syllabus anywhere so I didn't bother either, maybe ask someone else?It'd be worse if I were to misdirect you accidently!

Answer 29

hm okay let me openanother question than. U might know that one.

Answer 30

sure