calculus
Determine the points on the parabola $y=x^2 – 25$ that are closest to $(0,3)$
I would like to know how to go about solving this. I have some idea of solving it. I believe you have to use implicit differentiation and the distance formula but I don't know how to set it up. Hints would be appreciated.
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.
Draw a picture.
For either method, we want to minimize $(x-3)^2+y^2$, subject to $y=x^2$.
Substituting $x^2$ for $y$ in $(x-3)^2+y^2$, we find that we want to minimize $f(x)=(x-3)^2+x^4$. Note that $f'(x)=2(x-3)+4x^3$. The critical points of $f(x)$ are where $2(x-3)+4x^3=0$, or equivalently $$2x^3+x-3=0.$$ In general cubic equations are unpleasant to solve. However, the above equation has the obvious root $x=1$. And since $f''(x)=3x^2+1\gt 0$, the function $f'(x)$ is increasing, and therefore can only have one zero. By the geometry, there is at least one point on the parabola at minimum distance from $(3,0)$, so we have found it. The minimum distance is the distance from $(3,0)$ to $(1,1)$, which is $\sqrt{5}$.
We now sketch the Lagrange multipliers approach. The Lagrangian is $$(x-3)^2+y^2-\lambda(y-x^2).$$ Take the partial derivatives, and set them equal to $0$. We get $2(x-3)+2\lambda x=0$ and $2y-\lambda=0$.
So $\lambda=2y$. Substituting in the first equation we get $2(x-3)+4yx=0$. Putting $y=x^2$ we obtain $2(x-3)+4x^3=0$, and we are at an equation we have already dealt with.
Best Answer
Just set up a distance squared function:
$$d(x) = (x-0)^2 + (x^2-25-3)^2 = x^2 + (x^2-28)^2$$
Minimize this with respect to $x$. It is easier to work with the square of the distance rather than the distance itself because you avoid the square roots which, in the end, do not matter when taking a derivative and setting it to zero.