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)?

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)

To find the tangent point with the shortest line segment, we need to minimize the square distance between the point on the curve and the x-intercept of the tangent line.

1. Let the point on the curve be (c, c^2 + 1), where c is a positive number.
2. The slope of the tangent line at that point is given by dy/dx = 2x. So, at the point (c, c^2 + 1), the slope of the tangent is 2c.
3. Let the equation of the tangent line be y = mx + b. Substituting the values, we have y = 2cx + b.
4. For the given point (c, c^2 + 1), we can substitute the coordinates into the equation of the tangent line: c^2 + 1 = 2c(c) + b. Solving for b, we get b = 1 - c^2.
5. So, the equation of the tangent line is y = 2cx + 1 - c^2.
6. To find the x-intercept of the tangent line, we set y = 0: 0 = 2cx + 1 - c^2. Solving for x, we get x = (c^2 - 1)/(2c).
7. Now, we can compute the distance between the point on the curve and the x-intercept using the distance formula: D^2 = (c^2 + 1)^2 + (c - (c^2 - 1)/(2c))^2.
8. Expanding and simplifying, we get D^2 = c^4 + 2c^2 + 1 + c^2 - c^2 + 1 + (c^4 - 2c^2 + 1)/(4c^2).
9. Further simplifying, we have D^2 = c^4 + 2c^2 + 2 + (1/4)c^2 - 1/2 + (1/4)c^-2.
10. To find the minimum of D^2, we can find the critical point by differentiating 2D with respect to c and setting it equal to 0: 2D(dD/dc) = 4c^3 + 4c + c/2 - 1/(2c^3) = 0.
11. Solving this equation, we get the messy equation 8c^6 + 9c^4 - 1 = 0.
12. To simplify, let a = c^2, then we have 8a^3 + 9a^2 - 1 = 0.
13. By factoring, we find that (a+1)(8a^2 + a - 1) = 0. If a = -1, there is no solution.
14. So, we can solve 8a^2 + a - 1 = 0, which gives us a = (-1 ± √33)/16 ≈ 0.2965.
15. Taking the positive value, we have a = c^2 = 0.2965.
16. Taking the square root, we find c ≈ 0.5145.

Therefore, the tangent point that has the shortest line segment is approximately (0.5145, 0.5145^2 + 1) ≈ (0.5145, 1.2641). To find the length of the line segment, you can calculate the distance between this point and the x-intercept using the distance formula.