MHB Calculus: Find Closest Point to (3,0) on y=x^2

[zimmertr]

Hello, I've been assigned two calculus problems and have completed both of them. I'm pretty sure the first one is correct but I'm iffy on the second one. I would really appreciate it someone here could check my work on the second problem, and maybe even on the first problem if they have the time.

The problem in question is: "Use calculus to find the point (x,y) on the parabola traced out by y=x2that is closest to the point (3,0)."

Here is a copy of my work: https://www.dropbox.com/s/8jp4ad5qdeo4wtv/Miniproject.pdf?dl=0

 

[mathmari]

We want to minimize the function $L(x,y)=\sqrt{(x-3)^2+y^2} \overset{y=x^2}{\Longrightarrow} L(x,x^2)=L(x)=\sqrt{(x-3)^2+x^4}$.

For simplicity, we can minimize the function $L_{\star}(x)=(x-3)^2+x^4$. (The minimum of $L_{\star}$ will be the same as the minimum of $L$).

To find the minimum of $L_{\star}$ we have to find first the critical points:

$$L_{\star}'(x)=2(x-3)+4x^3 \\ L_{\star}'(x)=0 \Rightarrow 2(x-3)+4x^3=0 4x^3+2x-6=0 \Rightarrow 2(x-1)(2x^2+2x+3)=0 \Rightarrow x=1 \\ L_{\star}(1)=4+1=5>0 \Rightarrow \text{ The point is a minimum. } $$

So, the point of the parabola $y=x^2$ that is closest to the point $(3,0)$ is $(1,1)$.