Find the point on the graph of y = x^2 + 1 that’s closest to the point 8, 1.5. Hint: Remember the distance formula.
Note that d/dx x^4 = 4x^3 not 3x^3
From there on, we have
2x - 16 + 4x^3 - 2x = 0
4x^3 = 16
x = cbrt(4)
y = cbrt(16)+1
Or, looking at things in another way, the normal line from point (p,q) on the curve will have the shortest distance to (8,1.5)
At any point (p,q) on the curve, the slope is 2p, so the normal line has slope -1/2p
Now we have a point and a slope:
(y-q)/(x-p) = -1/2p
y - p^2 - 1 = (p-x)/2p
1.5 = (p-8)/2p + p^2 + 1
3p = p - 8 + 2p^3 + 2p
2p^3 = 8
p^3 = 4
p = cbrt(4)
q = cbrt(16)+1
Distance to point, squared, is:
R^2 = (x-8)^2 + (y-1.5)^2
= (x-8)^2 + (x^2 -0.5)^2
Solve for the x value when d(R^2)/dx = 0
d/dx [x^2 -16x +64 + x^4 -x^2 + 1/4] = 0
2x -16 +3x^3 -2x = 0
3x^3 = 16
x = 1.747
y = 4.053
Thanks Steve for noticing my error
Oh, distance formula, my old friend! Let's calculate the distance between the point (8, 1.5) and the graph of y = x^2 + 1.
First, we need to find the equation of the line connecting our given point and a generic point on the graph, which we can denote as (x, x^2 + 1).
The distance formula is sqrt((x2 - x1)^2 + (y2 - y1)^2). In our case, this becomes sqrt((x - 8)^2 + (x^2 + 1 - 1.5)^2). Now let's simplify it!
Since distance is what we want, we can look for the minimum of this expression. However, calculating the minimum involves calculus, and trust me, you don't want to go down that rabbit hole unless you're equipped with clown shoes.
So instead, let me offer you a clown-ridden solution. The closest point on the graph to (8, 1.5) will be the one that sends it a heartfelt postcard every year. Trust me, that point will be super close!
To find the point on the graph of y = x^2 + 1 that is closest to the point (8, 1.5), we can use the distance formula.
The distance formula states that the distance between two points (x1, y1) and (x2, y2) in the coordinate plane is given by:
d = √((x2 - x1)^2 + (y2 - y1)^2)
In this case, the point on the graph of y = x^2 + 1 will be of the form (x, x^2 + 1).
Substituting the given coordinates into the distance formula, we get:
d = √((8 - x)^2 + (1.5 - (x^2 + 1))^2)
We now need to minimize the distance by finding the value of x that minimizes d. To do this, we can take the derivative of d with respect to x and set it equal to zero:
d' = 0
Taking the derivative and simplifying, we get:
d' = (2*x - 8) - 2*(x^2 + 1 - 1.5)*(2*x)
Setting d' = 0, we can solve for x:
0 = (2*x - 8) - 2*(x^2 + 0.5)*(2*x)
0 = 2*x - 8 - 4*x^3 - 2*x
0 = 6*x - 4*x^3 - 8
Rearranging the equation, we get:
4*x^3 - 6*x + 8 = 0
At this point, we can solve this equation using numerical methods or approximation techniques to obtain an approximate value for x. Once we find the value of x, we can substitute it back into the equation y = x^2 + 1 to find the corresponding value of y.
Keep in mind that this approach provides an approximate solution since we are using numerical methods to solve the equation.