Hey there. I have a quick question. Ive done questions like these before so I don't know whats stoping me. Anways the question is this:
Find a point on the graph of
y = x^2
with x>0 that lies closest to (0, 3). What is the x-coordinate of such a point?
So I got 3^2 + y^2 = x^2 and solved for x but after that I'm lost. Normally that's what we did, but our other equations was normally like xy=8 so now I'm kinda lost. Any help would be apperciated. Thanks
The distance d from apoint with coordinates (x,y) to the point (0,3) is given by:
d^2 = x^2 + (y-3)^2
You have to minimize this while satisfying the constraint that he point (x,y) be on te parabola. So the contraint is:
y = x^2
This means that:
d^2 = y + (y-3)^2
You have to minimize this function.
Okay, so would you mind checking this? I've lost so many points already I don't wanna enter any more wrong answers.
If we have d^2 = y + (y-3)^2 then d =(y + (y-3)^2) ^1/2 (aka the root is what im getting at) So then if I take the derivative and then make it equal zero I get y = 3, so then x would equal the root of 3?
Hmm well I guess I went wrong somwheres because I put that it and it says its not right.
Sorry again. I missed something when I did the derivative and while adding something together I forgot something out. So now I have that y=5/2 and x=sqrt5/2
This is correct. Note that you can just differentiate d^2 instead of d. The square of a positive function is minimal (maximal) if and only if the function itself is minimal (maximal).
To find the x-coordinate of the point on the graph of y=x^2 that is closest to (0,3), we need to minimize the distance function d=√(x^2 + (y-3)^2) while satisfying the constraint y=x^2.
First, let's rewrite the distance function as d^2 = x^2 + (y-3)^2. We square both sides to get rid of the square root.
Next, substitute the constraint y=x^2 into the distance function: d^2 = x^2 + ((x^2)-3)^2.
To minimize d^2, we can differentiate it with respect to x, set the derivative equal to zero, and solve for x.
Taking the derivative of d^2 with respect to x, we get: d^2' = 2x + 2(x^2-3)(2x) = 2x + 4x(x^2-3) = 2x + 4x^3 - 24x.
Setting d^2' equal to zero and solving for x:
2x + 4x^3 - 24x = 0
2x(2x^2 - 12 + 1) = 0
2x(2x^2 - 11) = 0
x = 0 or x = sqrt(11/2)
Since x > 0 is given in the question, we choose x = sqrt(11/2).
Now that we have the value of x, we can substitute it back into the constraint y = x^2 to find y.
y = (sqrt(11/2))^2
y = 11/2
So the point on the graph of y = x^2 that is closest to (0,3) has coordinates (sqrt(11/2), 11/2). The x-coordinate of this point is sqrt(11/2).