For the following question I am asked to
find the radius for circle of curvature for the function: $-0.18e^{4.88x}$
I found the radius 1.681459915, by using the formula: R =1/ρ and this was correct however
when my answers for finding the centre is incorrect and I am unsure why. It is close though. I have attached a photo of my working out but I will also note it here. I used the forumula:
(x+R(dy/dx), y+R)
and from that I got:
(-0.1902818505, 1.65590175)
the correct answers are:
(x,y) = (-.0608106, -1.69409)
Any explanation as to how these correct answers were found and where I went wrong? am I using the wrong forumla? I have gotten 4 of the same questions wrong and I am stumped.
We have $f(x,y) = y + 0.48 e^{4.88 x} = 0$ and the normal to that function at the point $p=(-0.40,-0.48 e^{4.88 (-0.40)})$ is $\vec v = (0.8784e^{4.88(-0.40)},1)$ then the circle's center is located at
$$ (x_0,y_0)= p - R\frac{\vec v}{\|\vec v\|} = (-0.4,-0.0255582)-1.681459915(0.123765, 0.992312) = (-0.608106, -1.69409) $$
In red $-\frac{\vec v}{\|\vec v\|}$ in blue $f(x,y)=0$ and in dotted black, the osculating circle: the black dot it's center $(x_0,y_0)$ and the selected point in red.