Bisection Method of Solving a Nonlinear Equation â More Examples ...
Bisection Method of Solving a Nonlinear Equation â More Examples ...
Bisection Method of Solving a Nonlinear Equation â More Examples ...
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03.03.2 Chapter 03.03<br />
Solution<br />
From the physics <strong>of</strong> the problem, the dipstick would be wet between h 0 and h 2r ,<br />
where<br />
r radius <strong>of</strong> the tank,<br />
that is<br />
0 h 2r<br />
0 h 2(3)<br />
0 h 6<br />
Let us assume<br />
h <br />
0,<br />
hu<br />
6<br />
Check if the function changes sign between h and h .<br />
3 2<br />
0<br />
0 90 3.8197 3. 8197<br />
f ( h<br />
) f<br />
3 2<br />
f(h u<br />
) f (6) (6) 9(6) 3.8197 104.18<br />
Hence<br />
f h<br />
f<br />
hu<br />
<br />
f 0f<br />
6<br />
3.8197<br />
104.18 0<br />
So there is at least one root between h and h that is between 0 and 6.<br />
<br />
<br />
u<br />
Iteration 1<br />
The estimate <strong>of</strong> the root is<br />
h hu<br />
hm<br />
2<br />
0 6<br />
<br />
2<br />
3<br />
3 2<br />
f hm<br />
f 3 3 9 3 3.1897 50.<br />
180<br />
f h<br />
f<br />
hm<br />
<br />
f 0f<br />
3<br />
3.1897<br />
50.180 0<br />
Hence the root is bracketed between h<br />
and hm<br />
, that is, between 0 and 3. So, the lower and<br />
upper limits <strong>of</strong> the new bracket are<br />
h <br />
0,<br />
hu<br />
3<br />
At this point, the absolute relative approximate error <br />
a<br />
cannot be calculated, as we do not<br />
have a previous approximation.<br />
Iteration 2<br />
The estimate <strong>of</strong> the root is<br />
h hu<br />
hm<br />
2<br />
0 3<br />
<br />
2<br />
1.5<br />
u