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Find the point on the curve `y=x^2` closest to the point `(3,4)`.

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tarfa | Student, Undergraduate | eNoter

Posted June 3, 2012 at 7:13 PM via web

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Find the point on the curve `y=x^2` closest to the point `(3,4)`.

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted June 3, 2012 at 8:11 PM (Answer #1)

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To answer this question, we must first disect it as a normal calculus question and determine for what is being asked in the problem. We need to find a way to minimize the distance between the curve and the point. To do this, we need to find a function that describes the distance between the points on the curve and the point given in the problem. To do this, we need to use the distance formula:` `

`d(x,y) = sqrt((3-x)^2 + (4-y)^2)`

Now, we can substitute `x^2` for `y` in the problem, and then simplify the expression:

`d(x) = sqrt((3-x)^2 + (4-x^2)^2)`

`d(x) = sqrt(9-6x+x^2 + 16 - 8x^2 + x^4)`

`d(x) = sqrt(x^4 - 7x^2 - 6x + 25)`

Unfortunately, this expression does not appear to be a simple squared polynomial; we'll just need to leave it as it is. 

Now, we must minimize the expression. Of course, every time you say "minimize" or "maximize," you should think "derivative." So, we will now take the derivative of the function. Keep in mind, we must use the chain rule!

`d'(x) =1/(2sqrt(x^4-7x^2-6x+25))*(4x^3-14x-6)`

To find any minima or maxima, we must now find where this derivative becomes equivalent to zero:

`d'(x) = (2x^3-7x-3)/sqrt(x^4-7x^2-6x+25) = 0`

This function can only be equal to zero if the numerator is equal to zero, too. So, we will find the zeros for the numerator. Unfortunately, the zeros for the numerator are all irrational, but they can be listed here:

`x = -1.60, -0.46, 2.06`

If you have a graphing calculator, you can intuitively narrow down the point search further. Remember that this is a graph of the derivative! You can tell what the slopes are based on the graph. If the slope is negative, then `d(x)` is decreasing. Likewise, if the slope is positive, `d(x)` is increasing. Therefore, local minima will be found where the zeros of our derivative have negative values for local `x` less than the root of the derivative and positive values for local `x` greater than the root of the derivative. We can reduce our candidates for minimum distance, then, down to the following two `x`-values:

`x = -1.60, 2.06`

Now, to finish the problem, we simply plug these two values back into the original distance function to see which gives us the lesser value:

`d(-1.60) = sqrt((-1.60)^4 -7(-1.60)^2-6(-1.60)+25) = 4.82`

`d(2.06) = sqrt((2.06)^4 - 7(2.06)^2-6(2.06)+25) = 0.97` 

Thus, we have our result: the minimum distance is found at the `x`-value of 2.06.

Per the original question in the problem, we must now find the associated `y`-value to find the closest point. We simply evaluate the given function to find it:

`y = x^2 = 2.06^2 = 4.24`

Thus, we have our closest point: (2.06, 4.24)

To confirm, let's examine a graph of the function, point, and our solution:

It looks like we have the correct answer!

Sources:

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