Solving Calculus Question: x^4 + y^4 = 17, x + 8y = 30

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In summary, the conversation discusses finding the closest points between the equations x^4 + y^4 = 17 and x + 8y = 30. One suggestion is to find dy/dx for both equations and make them equal, then solve simultaneously. Another suggestion is to find a point on the x^4 + y^4 = 17 curve with the same slope as the straight line x + 8y = 30, and then find the distance between that point and the line.
  • #1
orange
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Hey hey.

I have this question I have a hard time figuring out. Here is the assignment as I got it:

What points are the closests between the functions

x^4 + y^4 = 17

and

x + 8y = 30?

I have started a few times but never really got far. I thought the line between the closets points must be a right angle with the plane x + 8y = 30, which might be solved with dot products. That, and that x^4 + y^4 = 17 describes some type of circle, but I can't really see it in front of me.

Anyone have any pointers? :!)
 
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  • #2
Well, one way of doing it would be to find dy/dx for both, and then make the dy/dx for the first equation equal the dy/dx for the second. You should get a function in x and y, then solve simultaniously with either of the equations of the lines. Like I said, not the best way, but it should work. You could try and do it with cross products if you like.

e.g.
[itex] 4x^3 + 4y^3 \frac{dy}{dx} = 0 [/itex] then do the same for the other one and substitute one into the other.
 
  • #3
x^4 + y^4 = 17 is like a square with curved corners (centred on the origin).
x + 8y = 30 isn't a plane, it's a straight line.

What you have to do is find a point (Px, Py) on the x^4 + y^4 = 17 curve where the slope is the same as the slope of the straight line.
If you draw this tangent, it will be equidistant everywhere to the straight line x + 8y = 30.
Now all you have to do is find the distance beteen a point P and a straight line, x + 8y = 30.
 
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