Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2+7 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?)
answer to C question at the end: If 1st coordinate (x-value) is C, the 2nd coordinate (y-value) is C^2+7.
Now, for the other question:
At (x,y), the slope of the tangent line is
dy/dx = 2x
So, now we have a point (a,a^2+7) and a slope (2a).
The equation of the tangent line where x=a is thus
y-(a^2+7) = 2a (x-a)
y = 2ax - 2a^2 + a^2+7
y = 2ax - a^2 + 7
This line intersects the x-axis where
0 = 2ax - a^2 + 7
x = (a^2-7)/2a
So, the length of the line segment from (a,a^2+7) to (0,(a^2-7)/2a) is
d^2 = (a - (a^2-7)/2a)^2 + (a^2+7)^2
or
d = 1/2a (a^2 + 7) sqrt(4a^2 + 1)
dd/da = (8a^4 + a^2 - 7)/[2a^2 sqrt(4a^2+1)]
dd/da = 0 when x = ±√(7/8) = ±0.935
d = 8.929
To find the tangent point that has the shortest line segment, we need to first find the equation of the tangent line at any given point on the curve y = x^2 + 7.
The equation of a tangent line is given by the point-slope form: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.
In this case, we need to find the slope of the tangent line at a given point (x, y) on the curve y = x^2 + 7. The slope of the tangent line can be calculated as the derivative of the curve at that point.
Taking the derivative of y = x^2 + 7, we get dy/dx = 2x, which represents the slope of the tangent line at any point (x, y) on the curve.
Now, we can substitute the values of x and y into the equation y - y1 = m(x - x1) to find the equation of the tangent line at any given point (x, y) on the curve.
Let's assume the coordinate of the tangent point is (C, C^2 + 7) since C is a positive number. The equation of the tangent line is then y - (C^2 + 7) = 2C(x - C).
Next, we need to find the x-intercept of the tangent line, which is the point where the line intersects the x-axis. To find this point, we set y = 0 in the equation of the tangent line and solve for x.
0 - (C^2 + 7) = 2C(x - C)
Simplifying the equation, we get -C^2 - 7 = 2Cx - 2C^2
Rearranging, -2C^2 - 7 = 2Cx - C^2
Combining like terms, we have -C^2 - 7 = (2C - 1)x
Solving for x, we get x = (-C^2 - 7) / (2C - 1)
From this equation, we can see that the x-coordinate of the point of tangency is dependent on the value of C, and it will change as C varies. Therefore, we need to minimize the length of the line segment by finding the minimum value of x.
To find the minimum value of x, we take the derivative of the equation x = (-C^2 - 7) / (2C - 1) with respect to C, and set it to 0.
d/dC [x] = d/dC [(-C^2 - 7) / (2C - 1)] = 0
Simplifying the equation, we get (4C^2 - 2C + 7) / (2C - 1)^2 = 0
Since we are looking for a positive value of C, we can disregard the denominator (2C - 1)^2 as it is always positive. Therefore, we need the numerator to be equal to 0.
4C^2 - 2C + 7 = 0
Solving this quadratic equation, we can use the quadratic formula:
C = (-(-2) ± √((-2)^2 - 4(4)(7))) / (2(4))
Simplifying further, we get C = (2 ± √(-60)) / 8
Since the square root of a negative number is not defined in real numbers, there are no real solutions for C in this equation. Consequently, we can conclude that there is no tangent point on the curve y = x^2 + 7 that has the shortest line segment.
Therefore, all tangent points on the curve have line segments of equal length.