How do I round off the answer for the depth to within 2mm?

% Values of constants

syms h

r = 0.3;

V = 0.35;

L = 3;

eqn = (0.5*pi*r^2-r^2*asin(h/r)-h*sqrt(r^2-h^2))*L == V;

answer = vpasolve(eqn, h)

depth = r - answer % depth of water in the trough

answer =

0.041305887729811791004451374702525

depth =

0.25869411227018820899554862529748

[KSSV]

What units they are in now?

[Hamza Aboumaray]

metres

[KSSV]

double(depth)*1000

[Hamza Aboumaray]

Thanks! Just want to ask one question, what it does mean when they ask to get the answer to within 2 mm? Does that mean to 3 decimal places since 2 mm is 0.002 metres?

Answer 1

[Thiago Henrique Gomes Lobato]

If you want to round to a specific unit (m, mm, etc), you can do this by just adding an extra parameter to the round function with the number of commas you want to round. If, however, you want to round in steps of 2 mm , it gets a little bit trickier although also not so difficult. The idea is to verify how many groups of 2 mm you have, round it and then add to the round version of your data without rounding mm. Something like this:

a = 2.355123; % m 
Nonzeromm = (a-round(a,2))*1000; % Find how many mm there is in the value
Nof2mm    = round(Nonzeromm/2);  % Count how many multiple of 2 are there
aRounded = round(a,2)+Nof2mm/1000*2 % Calculate the round without mm and then ad the ones you finded before
aRounded =

    2.3560

[Hamza Aboumaray]

When they say get the answer to within 2 mm does that mean round it to 3 decimal places since 2 mm is 0.002 metres?

[Thiago Henrique Gomes Lobato]

This means that you have a precision of 2 mm. So, for example, depth = 0.283 would not be valid, since this would mean that your precision was 1 mm. It is indeed rare to have precisions of non-unitary dimensions (2 mm instead 1 mm or even 0.5 mm), but it can happen. You must then round the number within intervals of 0.002 m, which is what is in my answer.

[Hamza Aboumaray]

I copied your code but I get an error saying: Unrecognized function or variable 'Non'. Can you please fix this?

[Thiago Henrique Gomes Lobato]

There's no "Non" variable in my code, you probably didn't copied entirely or made some extra modification. Mixing both codes the exact code you should have is this one:

syms h
r = 0.3;
V = 0.35;
L = 3;
eqn = (0.5*pi*r^2-r^2*asin(h/r)-h*sqrt(r^2-h^2))*L == V;
answer = vpasolve(eqn, h)
depth = double(r - answer) % depth of water in the trough

Nonzeromm = (depth-round(depth,2))*1000; % Find how many mm there is in the value
Nof2mm    = round(Nonzeromm/2);  % Count how many multiple of 2 are there
depthRounded = round(depth,2)+Nof2mm/1000*2 % Calculate the round without mm and then ad the ones you finded before

answer =
 
0.041305887729811791004451374702525
 

depth =

   0.258694112270188


depthRounded =

   0.258000000000000

[Hamza Aboumaray]

Thank you very much! Very much appreciated!