calculus
Shortest distance of the point $(0,c)$ from the parabola $y=x^2$ ? (Where $0\le c \le5$)
My approach: I wrote the distance formula by taking parametric coordinates as $(t,t^2)$ and then differentiated the equation.I got the extremum as $x=\sqrt (c)$,but that's the wrong answer? I can't figure out my mistake.
Please help!
If you have a curve $y=f(x)$, the length of the curve between points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is given by $$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{(dx)^2 + (df(x))^2} = \int_{x_1}^{x_2}\sqrt{1+\left(\dfrac{df}{dx} \right)^2} dx$$ In your case, $f(x) = x^2$. Hence, $\dfrac{df}{dx} = 2x$. Hence, the length of curve between the points $(x_1,x_1^2)$ and $(x_2,x_2^2)$ is $$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{1+(2x)^2} dx = \int_{x_1}^{x_2} \sqrt{1+4x^2} dx$$ Can you evaluate this integral?
$$\int_{x_1}^{x_2} \sqrt{1+4x^2} dx = \left.\dfrac{2x \sqrt{4x^2+1} + \log(2x+\sqrt{1+4x^2})}4 \right \vert_{x_1}^{x_2}$$
EDIT
Added picture to highlight the different lengths.
It is a projection.
Make a vector with the starting point as the given point, and the ending point as any point one the plane. We here call is $v$. The distance is exactly the projection of $v$ on an unit vector $n$ perpendicular to the plane. And $n$ can be easily calculated according to the plane.
And this is the result.
Best Answer
Let $f(t):=(t-0)^2+(t^2-c)^2$. The necessary condition for the minimum of $f$ is $$ \frac{d}{dt}\left\{(t-0)^2+(t^2-c)^2\right\}=2t+4t(t^2-c)=0, $$
which yields
$$ t_1=0, t_{2,3}=\pm\sqrt{c-1/2}. $$
However, $t_{2,3}$ are real only for $c\ge1/2$. In addition, the second order condition shows that when $c\ge 1/2$, $f(t_1)$ is a local maximum. So, the minimum distance squared is $c^2$ when $c<1/2$ and $c-1/4$, otherwise.