calculus

Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent point to the x-axis. If the tangent point is close to the y-axis, the line segment is long. If the tangent point is far from the y-axis, the line segment is also very long. Which tangent point has the shortest line segment?

(Suppose C is a positive number. What point on the curve has first coordinate equal to C? What is the slope of the tangent line at that point? Find the x-intercept of the resulting line. Compute the distance between the point on the curve and the x-intecept, and find the minimum of the square of that distance (minimizing the square of a positive quantity gets the same answer as minimizing the quantity, and here we get rid of a square root).)

* calculus - Reiny, Friday, November 13, 2009 at 8:39am

following the hints suggested:
let the point be (c,c^2 + 1)
dy/dx = 2x
so at (c,c^2+1) the slope of the tangent is 2c

let the tangent equation be y = mx + b
y = 2cx + b
for our point,
c^2 + 1 = 2c(c) + b
b = 1-c^2

so the tangent equation is
y = 2cx + 1-c^2
at the x-intercept,
0 = 2cx + 1-c^2
x = (c^2 - 1)/(2c)

then using the distance formula

D^2 = (c^2+1)^2 + (c - (c^2 - 1)/(2c))^2

Ok, I will now expand this. At first I thought to differenctiate using quotient rule for the last term, but it looked rather messy

D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)
= c^4 + 2c^2 + 2 + (1/4)c^2 - 1/2 + (1/4)c^-2

2D(dD/dc) = 4c^3 + 4c + c/2 - 1/(2c^3) = 0 for a max/min of D
8c^6 + 9c^4 - 1 = 0

getting really messy....
let a = c^2
solve 8a^3 + 9a^2 - 1 = 0
a=-1 works !!!!!!
(a+1)(8a^2 + a - 1) = 0
if a=-1, c^=-1 ---> no solution
8a^2 + a - 1 = 0
a = (-1 ± √33)/16 = .2965 or a negative
c^2 = .2965
c = .5145

Please, please check my arithmetic and algebra, the method is right!

* calculus - Hal, Saturday, November 14, 2009 at 8:58pm

Thanks so much, but then do you plug in c back into the first point and tangent line and x-intercept to get the x and y coordinates and tangent line equation and x-intercept, respectively?

* calculus - Hal, Saturday, November 14, 2009 at 9:50pm

Can you explain to me how you found out the distance formula. Because how did you find out D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)? I understand the first part with c^4 + 2c^2 + 1, but I don't understand how you got the second part.

* calculus - Hal, Saturday, November 14, 2009 at 10:03pm

And isn't (c^4-2c^2+1)/(4c^2) supposed to be (c^4+2c^2+1)/(4c^2)?

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  1. I recall doing this question yesterday, it was a good one.

    I think you are stuck with expanding
    D^2 = (c^2+1)^2 + (c - (c^2 - 1)/(2c))^2 to get to
    D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)

    let's look at the last part of that.

    recall (a+b)^2 = a^2 + 2ab + b^2, that is,
    we square the first term, then twice the product of the first and last terms, and finally we square the last term.
    So for (c - (c^2 - 1)/(2c))^2
    squaring c gives us c^2, there it is ..
    D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)
    now twice the product of the two ..
    2(c)(-1)(c^2 - 1)/(2c) = -(c^2-1)= -c^2 + 1
    and there it is
    D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)
    and lastly, square the last term
    [-(c^2-1)/(2c)]^2
    = (c^4 - 2c + 1)/(4c^2)and here it is
    D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4-2c^2+1)/(4c^2)

    at the end I found c = .5145

    The question asked for the point of contact so the distance is the shortest.
    Recall we called that point (c,c^2 + 1)

    so the point is (.5145, 1.2647)

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