Point on the curve closest to (18,1)

Faraz Ahmed
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Question:
Question: Find the point on the curve y=x^2 +1 that is closest to the point (18,1).

Please see the image and that’s where I’m stucked- after taking the first derivate. Please solve it further step by step completely. It’d mean a lot.
 

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Which value does the derivative have to reach at the minimum?

Afterwards you'll have to try different values for u to find the solution (there is a formula but that is more complicated than testing different values).
 
Faraz Ahmed said:
Question:
Question: Find the point on the curve y=x^2 +1 that is closest to the point (18,1).

Please see the image and that’s where I’m stucked- after taking the first derivate. Please solve it further step by step completely. It’d mean a lot.

It is against PF rules for us to do complete solutions; we are allowed to offer hints, but not more.

Here is a hint: minimizing the squared-distance will give the same solution as minimizing the distance itself----can you see why? --- and the squared-distance problem is easier.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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