Imagine the quadratic curve with equation
y=y(x)=ax^2+bx+c
On the concave side of this curve there is a circle of radius R. The circle is as close the the extremum of the quadratic as possible without resulting in the curves crossing each other. Write a function which takes as inputs a,b,c, and R and returns the coordinates of the center of the circle.
For example, if
a=1; b=0; c=10; R=pi;
then the function returns
T = circ_puzz(a,b,c,R) T = 0 20.1196044010894
This can be visualized as follows:
P = @(x) a*x.^2 + b*x + c; % Quadratic C = @(x) real(-sqrt(R^2-(x-T(1)).^2) + T(2)); % Lower half circle x = linspace(-R,R,10000); % Range of plotted data. plot(x,C(x),'r',x,-C(x)+2*T(2),'r',x,P(x),T(1),T(2),'*k') ylim([0,30]) axis equal
Is there any special in Test 3?
don't know why the assertions fail.
The solution is correct.
If you look at the example posted in the problem description, you will see that the return argument should be a two element row vector.
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