Illustrate a mathematical question concept. The image should depict an overhead view of an abstract 3-dimensional landscape, with a sharp incline similar to y = x^2+1. On this landscape, mark an arbitrary point representative of (3,1). Emphasize the distance between the curve and the point through a dotted line. Also imagine the concept of differentiation subtly indicated through an invisible axis bending along the curve, representing the derivative of the original function. Refrain from including any text in the image.

Find the point on the graph of y = x2+1 that is closest to the point (3,1).


d = √[(x-3)^2+(y-1)^2]

d = √[(x-3)^2+(x^2+1-1)^2]

d = √[(x-3)^2+(x^2)^2]

d = [(x-3)^2+x^4]^1/2

d' = 1/2 [(x-3)^2+x^4)^1/2 [2(x-3)+4x^3]

0 = [1(2x-6+4x^2)] / [2√(x-3)^2+x^4]

0 = 2x-6+4x^3

0 = x-3+2x^3

3 = x+2x^3

is this correct so far? i'm not sure what to do from here.

You are correct to the end, but keep it as

2x^3 + x - 3 = 0

try a few numbers,
to get "nice" answers, (the thing would factor) , x has to be ±1,±3, ±3/2

sure enough x = 1 works
so (x-1) is a factor.
Either do long division or use something like synthetic division to show that
2x^3 + x - 3 = 0
(x-1)(2x^2 + 2x + 3) = 0
The remaining quadratic has no real solutions, so

x = 1, then y = 1^2 + 1 = 2

so the closest point is (1, 2)

OR

I use an easier way, based on the fact that at the closest point the tangent would have to be perpendicular to the line from that closest point to the given outside point

Let the closest point be (a,b)
slope of tangent to y = x^2 + 1 = 2x
so at point (a,b) the slope is 2a
and we know that b = a^2 + 1
slope of the line from (a,b) to (3,1)
= (b-1)/(a-3)
= (a^2 + 1 - 1)/(a-3)
= a^2/(a-3)

but the slopes are negative reciprocals:

a^2/(a-3) = -1/(2a)
cross-multiply
2a^3 = -a + 3
2a^3 + a - 3 = 0 ----> or if you want: 2x^2 + x - 3 = 0

Find the point of the curve y=x^2+1 which is closest to the point(3,1)

Find the point of the curve y = x^2+1 that is closest to the point (3,1).

no

Apologies for the confusion. Let's start over.

To find the point on the curve y = x^2 + 1 that is closest to the point (3, 1), we can use the distance formula:

Distance = √[(x - 3)^2 + (y - 1)^2]

Substituting y = x^2 + 1, we have:

Distance = √[(x - 3)^2 + ((x^2 + 1) - 1)^2]
= √[(x - 3)^2 + (x^2)^2]
= √[(x - 3)^2 + x^4]

To find the minimum distance, we can minimize the square of the distance, which is equivalent to minimizing:

f(x) = (x - 3)^2 + x^4

To find the minimum point, we can differentiate with respect to x and set the derivative equal to zero:

f'(x) = 2(x - 3) + 4x^3
0 = 2(x - 3) + 4x^3
0 = 2x - 6 + 4x^3

Simplifying, we have 4x^3 + 2x - 6 = 0.

Unfortunately, this equation does not have a nice closed-form solution, so we may need to use numerical methods or approximate solutions. You can use methods like Newton's method or numerical integration to find the x-coordinate of the point on the curve that is closest to (3, 1).

Your work so far is mostly correct. However, there is a small mistake in simplifying the derivative. Let's go through the steps together:

We want to find the point on the graph of y = x^2 + 1 that is closest to the point (3, 1). To do this, we can minimize the distance between these two points.

The distance between two points (x1, y1) and (x2, y2) is given by the formula:

d = √[(x2 - x1)^2 + (y2 - y1)^2]

In this case, (x1, y1) = (3, 1):

d = √[(x - 3)^2 + (y - 1)^2]

Now, substitute y with the equation of the graph y = x^2 + 1:

d = √[(x - 3)^2 + (x^2 + 1 - 1)^2]

Simplify the equation inside the square root:

d = √[(x - 3)^2 + (x^2)^2]
d = √[(x - 3)^2 + x^4]

To minimize d, we need to find the minimum point on the graph. To do that, we take the derivative of d with respect to x:

d' = 1/2 [(x - 3)^2 + x^4] ^ (-1/2) [2(x - 3) + 4x^3]

Now we need to solve for x when d' equals zero:

0 = [1/2 √((x - 3)^2 + x^4)] [2(x - 3) + 4x^3]

That's where the mistake is. We need to set the entire expression to zero, not just the first term on the right side.

0 = [1/2 √((x - 3)^2 + x^4)] [2(x - 3) + 4x^3]

Now, we can solve for x.

0 = 2(x - 3) + 4x^3

0 = 2x - 6 + 4x^3

Keep going from here to find the value(s) of x that satisfy this equation.

another way: the distance will be perpendicular to the slope at the tangent point.

y=x^2+1
y'=2x
so the slope of the perpendicular line is -1/2
Now, find the line which has slope -1/2 and goes through the point 3,1
y=mx+b
1=-1/2 (3)+b
b=1+3/2=2.5
so now we have a line intersecting a curve,
y=-1/2 x+2.5 and the curve is y=x^2+1
x^2+1=-x/2 + 2.5
NOW SOLVE THAT quadratic, and you have the x point where the line hits the curve, then of course you have the y point.
then distance^2=(x-3)^2+(y-1)^2