Posted by Hal on Friday, November 13, 2009 at 8:03am.
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)?
Answer This Question
More Related Questions
- Calculus - Consider line segments which are tangent to a point on the right ...
- calculus - Consider line segments which are tangent to a point on the right half...
- Calculus AB - Could someone please help me with these tangent line problems? 1) ...
- Calculus AB - Sorry but I've got a lot of problems that I don't understand. 1) ...
- calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...
- Calc AB - Suppose that f(x) is an invertible function (that is, has an inverse ...
- Calc. - Find the area of the region bounded by the parabola y=x^2, the tangent ...
- Calculus - at what point is the tangent to f(x)=x^2+4x-1 horizontal? That would ...
- Calculus - Find an equation of the tangent line to the curve at the given point...
- Calculus- Tangent Line - What is the slope of the tangent line to the curve at ...