Question:
Find the nearest point to point $(3,0)$ and lie on the curve of $y = x^3- 2x^2 + 3$
Honestly, it's not the question what i'm dealing with. I have another question, but the answer is predictable if i'm using GeoGebra. Some sites give examples about linear function only that actually very easy to solve using distance formula or similarity of angles property.
And i just need to change the original question and extend it what if the function is not linear?, so "maybe" there is an actual or general way to solve this kind of the problem.
But, the problem is getting hard if we have complicated function (trigonometric or exponential) and we are not allowed to use tools, for example during exams.
So, how to solve this problem? I need your guide and some advice. Thanks
I thought I had a good idea, but my solution turned out to be messy and require a calculator (to get the roots of a $6$th degree polynomial). I will post it anyway, in the hopes that you or someone else can see how to tweak it to not need a calculator.
My idea was to use vectors, and see when the dot product of two vectors was $0$.
Let $P=(3,0)$, and suppose the nearest point on the curve is $Q$. The coordinates of $Q$ will be $(t,t^3-2t^2+3)$ for some $t\in\Bbb{R}$. So the strategy is to take vector $\overrightarrow{PQ}$, and a tangent vector to the curve at $Q$, and make sure their dot product is $0$.
We have $\overrightarrow{PQ}=\binom{t-3}{t^3-2t^2+3}$, which can be saved for later.
Hopefully I'm not glossing over too much of my work, but next I figured out the tangent line to the curve at the point $Q$. It is (in point-slope form),
$y-(t^3-2t^2+3)=(3t^2-4t)\cdot(x-t)$.
Now I needed a second point on that line, so I just let $x=0$ to get the point $(0,-2t^3+2t^2+3)$, which I will call $R$. We can then write down $\overrightarrow{RQ}=\binom{t}{3t^3-4t^2}$.
We need to get $\overrightarrow{PQ}\bullet\overrightarrow{RQ}=0$.
Unfortunately, that means solving $3t^6-10t^5+8t^4+9t^3-11t^2-3t=0$.
The good news is this provides the correct answer, which is approximately $Q=(1.509, 1.882)$, albeit after using a calculator.
The fact that the answer is approximate does not make me confident that a simpler solution that can be done entirely by hand exists, but maybe I have missed a key insight. I hope my work puts you or someone else onto the right track.