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
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.
6 answers
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
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).
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).
3 answers