Question

Find the two points on the parabola y=x^2 - 1 that are closest to the orgin. Using calculs. start with Distance formula? then go from there.

Answer 1

this doesn't make much sense

Answer 2

Make an equation for distance from the origin to a point on your line. Distance is sqrt(dx^2+dy^2)
So you want sqrt((x-0)^2+(y-0)^2)=sqrt(x^2+y^2). Next you want to differentiate this in terms of x. Remember y's will become dy/dx, not just 1:
dD = 1/2 * (x^2+y^2)^-1/2 * (2x+2ydy/dx)
Now we need to find a way to plug in y and dy. y comes from our equation, y=x^2-1
dy is just the derivative of that: dy=2x
Plug those in and you get dD = 1/2 * (x^2+(x^2-1)^2)^-1/2 * (2x+2*(x^2-1)*2x)
Set dD to 0 because maxima and minima of f can only occur at critical points, or when f' is 0. Once you find all of the 0's of f', plug all those x's back into f and you can find which are the lowest

Answer 3

Here's the basic explanation. We want to find the value x where the distance between x^2-1 and the origin (0,0) is a minimum. To find minimum in calculus, set the derivative of whatever you want the minimum of and set it equal to 0 (because that is where it goes from increasing to decreasing or decreasing to increasing often). In our case, it's the distance from (0,0) to (x,y). So take the derivative of that (be careful not to plug in x^2-1 for y until after you differentiate, I believe that's important.) and set it equal to 0. You will have x, y, and dy still in your equation. This is where the y=x^2-1 comes in. Replace y's with x^2-1 and dy's with 2x. Now you have a single variable equation set equal to 0, so just solve that. You'll get 2 points as your potential minimum. Since the question tells you there are 2 minima, you could stop now and say (x1,y1) (x2,y2) (make sure to give y values too! The question asks for the POINTS, not just x-values!). If you didn't know there were 2, you would need to test all your points in the y= equation, and find out which were smallest.