Question

Find a point on the curve y= x^2 that is closest to the point (18, 0).?

Answer 1

1. I'd suggest graphing this parabola and the given point.
2. Draw (by eye) the shortest line segment connecting (18,0) and the graph of y=x^2.
3. Represent the point at which this line intersects the curve by (x,x^2).
4. Write out an expression for the distance between (18,0) and (x, x^2) (use the distance formula).
5. Differentiate this specific distance formula.
6. Set the result = to 0
7. solve for x, noting that x must be (+) because of its location on the graph of y=x^2.
8. If desired, find the second derivative of this distance and find its sign at this x value. That will tell us whether or not our x-value results in a min. (as desired).

Answer 2

Having gone through all but the last step, I obtained the equation x-18+2x^3=0, which is easily solved for x. Then the point on the parabola y=x^2 closest to (18,0) is written in terms of that x: (x, x^2).