i have the function y=x^2 and a point B(3;0) i have to find a point in the function that has the minimal distance from the point B(3;0) whic is that point?
Hi i am in 12 grade and my teacher gave me the problem that i represented to you.
To solve this problem i followed this steps
1-from the point (3;0) i build a straight which is paralel with the tangent of the function y=x^2
2-the straight has thaequation y=kx+t and i subtitute the point B(3;0)
3-the tangent of the function has the same k because they are paralel but i cant find the k and in this way i cant find the point in the function which is with the minimal distance from point B(3:0)
I would do it differently, since we don't know to begin with where the point is on the curve, we don't know the slope of the tangent nor of the normal direction.
Method I. I would start by picking a point on the curve, and drawing the line from there to B:
point (x, x^2) --> (3, 0)
The slope of that line is (0 - x^2)/(3 - x) = x^2/(x - 3) = m1
The tangent to the curve is the derivative evaluated at x.
dy/dx = 2x= m2
....<-- edited
The slope of the line and the slope of the curve must be perpendicular to each other.
Perpendicular slopes are negative reciprocals of each other, so the product must be -1.
Set
...m1*m2 = [x^2
/(x - 3)]*[2x] = -1
and solve for x.
Unfortunately, that is 5th order in x, not easy to find solutions!
EDIT - after fixing errors, it is 3rd-order and same as Method II.
Method II. Again pick a point on the curve, and draw the line from there to B:
point (x, x^2) --> (3, 0)
The square of the length of the line is
s^2 = (3 - x)^2 + (x^2)^2
.......= x^4 + x^2 - 6x + 9
Differentiate s^2 with respect to x, and set the derivative to zero.
This will give a cubic equation in x, Look for simple roots .. there is only one real solution
