calculus
posted by Hal .
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 xaxis. If the tangent point is close to the yaxis, the line segment is long. If the tangent point is far from the yaxis, 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 xintercept of the resulting line. Compute the distance between the point on the curve and the xintecept, 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 = 1c^2
so the tangent equation is
y = 2cx + 1c^2
at the xintercept,
0 = 2cx + 1c^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^42c^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 xintercept to get the x and y coordinates and tangent line equation and xintercept, 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^42c^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^42c^2+1)/(4c^2) supposed to be (c^4+2c^2+1)/(4c^2)?

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^42c^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^42c^2+1)/(4c^2)
now twice the product of the two ..
2(c)(1)(c^2  1)/(2c) = (c^21)= c^2 + 1
and there it is
D^2 = c^4 + 2c^2 + 1 + c^2  c^2 + 1 + (c^42c^2+1)/(4c^2)
and lastly, square the last term
[(c^21)/(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^42c^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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