vector calculus using Lagrange Multipliers

159 Views Asked by At

$(1)$ Let $c\in R$ be a constant.

Using Lagrange Multipliers, find all the extrema of

$$f(x,y) = x^2 + (y-c)^2$$

subject to the constraint $$y = x^2$$

I'm pretty sure I've found the critical points. There are 3 of them I believe and all are minimums including the point $(0,c)$.

$(2)$ Give a parametrization of the parabola $y=x^2$ by directly substituting the parametrization into $f(x, y)$, decide for each extremum point you found at $(1)$ whether it is a local maximum or a local minimum point. Justify your answer fully.

I let $t=x$, $y=t^2$ then $f(x,y) = t^2 + (t^2 -c)^2$.

I have no idea how to use this to determine the nature of the extrema points. I tried evaluating at the points but don't know what the results point to. Any tips please?

$(3)$ "The shortest distance from the point $(0,1)$ to the parabola $y=x^2$ is $1$." Is this claim true or false? Using your results in $(1)$ and $(2)$, give a full proof of your answer.

I also don't know how to approach this question. Any hints please? Thanks in advance.

EDIT: I think I've figured out $(3)$ but still don't know how to do $(2)$.

1

There are 1 best solutions below

0
On

Hint: substitute the values of t and for (3) use the magnitude to find the distance.