Calculus HELP!!!

Question

Find the point on the parabola x^2 =2y that is closest to the point (4,1)

Answer 1

x^2 =2y

y = (1/2)x²

So, any random point on the graph of x^2 =2y can be written as (x, (1/2)x²)

The distance between that random point (x, (1/2)x²) and the point (4, 1) can be calculated using the distance formula:

d = √((x₂ - x₁)² + (y₂ - y₁)²) and we can let (x₁, y₁) = (4, 1) and (x₂, y₂) = (x, (1/2)x²). Plugging those in we get:

d = √((x - 4)² + ((1/2)x² - 1)²) and then simplifying:

d = √(x² - 8x + 16 + (1/4)x⁴ - x² + 1)

d = √((1/4)x⁴ - 8x + 17) or, writing with an exponent to replace the radical:

d = ((1/4)x⁴ - 8x + 17)^1/2

To minimize this distance, take the derivative and set it equal to zero:

d' = (1/2)((1/4)x⁴ - 8x + 17)^-1/2•(x³ - 8)

d' = (x³ - 8)/[2((1/4)x⁴ - 8x + 17)^1/2]

(x³ - 8)/[2((1/4)x⁴ - 8x + 17)^1/2] = 0

This only can equal zero if the numerator equals zero:

x³ - 8 = 0

x³ = 8

x = 2

When x = 2, the y-value of the point on the parabola is 2 so the point is (2, 2)

Answer 2

desmos.com/calculator/4gvqadsmge

At the above graph you can enable the graphs of D(x) and D'(x) to see that the derivative graph crosses the x-axis at x = 2.

Answer 3

the point (2,2) is the point on the parabola x^2 =2y that is closest to the point (4,1)

2y = x^2

y = .5x^2

dy/dx = x = slope of the parabola = 2 at the point (2,2)

-1/x = slope of the perpendicular line= the negative inverse = -1/2 = -.5

line with that slope m through (4,1) is

y - 1 = (m)(x-4)

m = (y-1/x-4) = (2-1)/(2-4)=-1/2

shortest distance from parabola to (4,1) is

d where d^2 = 1 + 4 = 5

shortest distance = sqr5 = about 2.236 units

it might help to graph the parabola and visually look at what point on it seems closest to (4,1)

(2,2) looks about right