find the x-coordinate of the point P on the parabola y=1-x^2 for 0<x<1 where the triangle that is enclosed by the tangent line at P and the coordinate axes has the smallest area.
let the point of contact in the first quadrant be P(a,b) or P(a,1-a^2)
slope of the tangent is -2x (the derivative)
so at P the slope is -2a
equation of the tangent is
(y - (1-a^2))/(x-a) = -2a
which simplifies to
2ax + y = a^2 + 1
the base of the triangle will be the x-intercept of the tangent
the height of the triangle will be the y-intercept of the tangent.
x-intercept : let y = 0, then x = (a^2+1)/(2a)
y-intercept : let x = 0, then y = a^2 + 1
so the Area = (a^2 + 1)^2 /(4a)
use the quotient rule to find the derivative of Area, set that equal to zero and solve.
let me know what you got.
To find the x-coordinate of the point P on the parabola where the enclosed triangle has the smallest area, we can use calculus and optimization techniques.
Let's start by finding the equation of the tangent line at point P on the parabola y = 1 - x^2. To do this, we need to find the derivative of the function y = 1 - x^2 with respect to x.
dy/dx = -2x
The slope of the tangent line at point P is equal to the derivative at that point. So, when the tangent line is a minimum, the derivative will be equal to zero. Therefore, we need to find the x-coordinate where the derivative of y is zero.
Setting -2x = 0, we get:
-2x = 0
x = 0
The x-coordinate of the point P where the tangent line is a minimum is x = 0.
Now, we need to verify that this point gives us the smallest area for the enclosed triangle.
The area of a triangle can be calculated as A = (1/2) * base * height. In this case, the base of the triangle is 2x (since it's enclosed by the x-axis), and the height is y (since it's enclosed by the tangent line). So, the area can be expressed as:
A = (1/2) * 2x * y
Substituting y = 1 - x^2, we get:
A = (1/2) * 2x * (1 - x^2)
A = x - x^3
Now, we need to find the minimum value of A by finding the critical points. To do this, we differentiate A with respect to x:
dA/dx = 1 - 3x^2
Setting dA/dx = 0, we get:
1 - 3x^2 = 0
3x^2 = 1
x^2 = 1/3
x = ±sqrt(1/3)
Since we are looking for a value of x between 0 and 1, the only valid solution is x = sqrt(1/3).
Therefore, the x-coordinate of the point P on the parabola y = 1 - x^2 for 0 < x < 1 where the triangle enclosed by the tangent line at P and the coordinate axes has the smallest area is x = sqrt(1/3).