用Octave求曲率圆的半径
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曲率圆的半径\(R=\frac{1}{K}\)
求\( y=sin(x)\)在\( x=1\)处的曲率圆的半径.
程序代码如下
function [text_result, numeric_result] = func51(x_value)
pkg load symbolic;
x = sym('x');
question = sin(x);
r = 1 / (abs(diff(diff(question, x), x)) / power(1 + power(1 + diff(question, x), 2), 3/2));
result = limit(r, x, x_value);
text_result = ["\n", disp(result)];
numeric_result = eval(result);
endfunction
结果如下
>> [text_result, numeric_result] = func51(1)
warning: passing floating-point values to sym is dangerous, see "help sym"
warning: called from
double_to_sym_heuristic at line 50 column 7
sym at line 384 column 13
power at line 75 column 5
func51 at line 5 column 7
text_result =
3/2
⎛ 2⎞
⎝1 + (cos(1) + 1) ⎠
──────────────────────
sin(1)
numeric_result = 7.3603