Help with Calculus homework please
Not sure if it's the same with other schools, but it's in the Mean Value Theorem and Curve Sketching chapter. The question is:
Find the point on the hyperbola y^2 - x^2 = 4 that are closest to the point (2,0).
And here's a screenshot of what my teacher posted for the answer:
http://i943.photobucket.com/albums/ad275/BlackyoshiCP/calc_zps3e2d30ac.png
What I don't understand is where did that big square root thing come from in the 4th line down, and why does it disappear in the line below that? o.o
May 3, 2013
2 Comments • Newest first
Find the point on the hyperbola y^2 - x^2 = 4 that are closest to the point (2,0).
(x,y) and (2,0) y^2 - x^2 = 4
d = sqr([x-2]^2 + [y-0]^2)
d = sqr([x-2]^2 + [y]^2) || y = +-sqr(x^2 + 4)
d = sqr([x-2]^2 + [sqr(x^2 + 4)]^2)
d = sqr(x^2 - 4x + 4 + x^2 + 4)
d = sqr(2x^2 - 4x + 8) || closest when the value of 2x^2 - 4x + 8 is smallest.
f(x) = 2x^2 - 4x + 8
f'(x) = 4x - 4
f'(x) = 4x - 4 = 0
x = 1
y = +-sqr(x^2 + 4) || x = 1
y = +-sqr(1^2 + 4)
y = +-sqr(5)
(1 , sqr(5)) or (1, -sqr(5))
It's the formula for the linear distance between any two points.
S/he just squared both terms and simplified for the line after. If you differentiate that equation you can use optimization to find the minimum distance for any fixed point (2,0) and a point in the hyperbola. That's what your teacher did.
The (x - 2) term comes from the difference in x-coordinates and the sq.rt(x^2 + 4) term comes from the difference in y-coordinates of those two points. Note that there's no minus sign because in this case y_2 is 0.
The reason that the big radical disappears from the fourth line is due to the fact that removing it simplifies calculations (don't have to differentiate a radical) and that if you think about it, removing the radical won't really change the answer you're looking for.
When you take dl/dx and find the absolute minimum, the point will be the minimum regardless or whether or not you take the square root of it.
Hope that helps.