Calculus help??

Question

Find the point(s) on the curve, y = x^2 + x + 1 that is closest to the point, (-1, 2). Hint: Use the Pythagorean formula for distance to find a function that relates the x coordinate of a point on the curve with the distance of that point to (-1, 2), and find the x that minimizes that functions. Another Hint: Observe that if you minimize the square of the distance, you also minimize the distance.

Answer 1

Let us consider a generic point on the curve. The coordinates of that point will be (x, x^2+x+1), because y=x^2+x+1

The distance between this point and the point (-1,2) is:

sqrt((x-(-1))^2+(x^2+x+1-2)^2), but the good news is that the (x,y) coordinates where you obtain the minimum of this distance is the same as the one where you obtain the minimum of the squared distance. It is enough to find the point where (x-(-1))^2+(x^2+x+1-2)^2 is minimum. Let us expand and derivate it. Finally, we obtain a nice and simple:

4x^3+6x^2=0 derivative=0
, true for x=0 and x=-3/2. The y coordinate corresponding to these two are: y=1 and y=7/4. The two points on the curve (parabola) where you obtain local extreme points are: (0,1) and (-3/2, 7/4). The derivative comes from negative value, becomes positive in the point (-3/2, 7/4), and get zero back in (0,1) that can be demonstrated it's an inflexion point (second derivative is 0). As long as it comes from negative values to positive, (-3/2, 7/4) is a local minimum point. I don't see in the problem that you nee to find the minimum distance so I stop here. Good luck!

Answer 2

The distance between the y coordinate and 2 is

y - 2

Since y = x² + x + 1 then y - 2 = x² + x - 1

The distance between the x coordinate and -1 is

x - (-1) = x +1

Therefore in terms of x, using pythagoras
we have

Distance² = (x² + x - 1)² + (x +1)²
= (x^4 + 2x³ - x² - 2x + 1) +(x² + 2x + 1) = x^4 + 2x³ + 2

It's minimum occurs where to derivative of x is 0.

The derivative is 4x³ + 6x²

4x³ + 6x² = 0
2x²(2x + 3) = 0

so x = 0 or x = -3/2

Substitution shows that the smallest value for the distance² is 0.3125 = 5/16. The shortest distance is √(5) / 4.

Substituting in (-3/2) for x into the original curve gives:

(-3/2)² + (-3/2) + 1 = 7/4

Therefore the point (-3/2, 7/4) is the closest point of the curve x² + x + 1 to the coordinate (-1, 2)

Answer 3

Let d be the distance from (x, y) to (-1, 2). Then by the distance formula, d² = (x+1)² + (y-2)² = (x+1)² + (x²+x+1-2)². Simplifying this:

(x+1)² + (x²+x+1-2)²
(x+1)² + (x²+x-1)²
x² + 2x + 1 + x⁴ + x² + 1 + 2x³ - 2x² - 2x
x⁴ + 2x³ + 2

The minimum of this function will be at a point where its derivative is equal to zero. So first we find the derivative:

4x³+6x²

Setting this equal to zero, we find:

x²(4x+6) = 0
x=0 ∨ x = -3/2

Substituting these values into the distance function:

x=0 → d² = 2
x=-3/2 → d² = (-3/2)⁴ + 2(-3/2)³ + 2 = 81/16 - 27/4 + 2 = 81/16 - 108/16 + 32/16 = 5/16

Therefore, the point on the curve which is closest to (-1, 2) is at x=-3/2, y = (-3/2)²-3/2+1 = 9/4-6/4+4/4 = 7/4. The distance from (-3/2, 7/4) to (-1, 2) is √5/4