Find the x-coordinate of the point P on the parabola y=1-x^2 (0<x=<1) where the triangle enclosed by the tangent line at p and the coordinate axes has the smallest area.
let the point of contact be P(a,1-a^2)
dy/dx = -2x, so at our point P the slope of the tangent is -2a
equation of tangent:
y - (1-a^2) = -2a(x-a)
y - 1 + a^2 = -2ax +2a^2
2ax + y = a^2 + 1
the base of the triangle is the x-intercept of this line,
the height of the triangle is the y-intercept of this line.
x-intercept: x = (a^2 + 1)/(2a)
y-intercept: y = a^2 + 1
Area of triangle
= (1/2)(a^2+1)^2/(2a)
= (a^4 + 2a^2 + 1)/a
= a^3 + 2a + 1/a
d(Area)/da = 3a^2 + 2 - 1/a^2
= 0 for a max/min of Area
the only real solution for the above is
a = ± 1/√3
so P is (1/√3 , 2/3)
To find the x-coordinate of the point P on the parabola where the triangle enclosed by the tangent line and the coordinate axes has the smallest area, we can make use of calculus and optimization techniques.
Let's start by finding the equation of the tangent line at point P on the parabola. We need to find the derivative of the given function with respect to x:
dy/dx = d(1-x^2)/dx = -2x
Now, we can find the slope of the tangent line by evaluating the derivative at point P:
slope (m) = -2x
At point P, the slope of the tangent line will be equal to the slope of the line from the origin to point P. Let's denote the coordinates of point P as (x, 1-x^2).
Using the slope-intercept form of a line, we can write the equation of the line from the origin to point P as:
y = mx = -2x*x = -2x^2
Now, let's find the x-coordinate of the point where the line y = -2x^2 intersects the parabola y = 1 - x^2. To find this, we can set the two equations equal to each other and solve for x:
1 - x^2 = -2x^2
Simplifying the equation:
3x^2 = 1
Dividing both sides by 3:
x^2 = 1/3
Taking the square root of both sides:
x = ±√(1/3)
Since the given condition is 0 < x ≤ 1, the x-coordinate of the point P where the triangle has the smallest area is:
x = √(1/3)
Therefore, the x-coordinate of the point P is √(1/3).
To find the x-coordinate of the point P on the parabola where the triangle enclosed by the tangent line at P and the coordinate axes has the smallest area, we can follow these steps:
Step 1: Find the equation of the tangent line at P.
Step 2: Find the points of intersection of the tangent line with the coordinate axes.
Step 3: Calculate the area of the triangle enclosed by the tangent line and the coordinate axes.
Step 4: Repeat steps 1-3 for different x-values in the interval (0, 1) and determine the x-coordinate that gives the smallest triangle area.
Let's go through these steps one by one:
Step 1: Find the equation of the tangent line at P.
To find the equation of the tangent line at P, we need to find the derivative of the parabola y = 1 - x^2 and then evaluate it at x = P.
The derivative of y = 1 - x^2 is dy/dx = -2x.
Evaluate this derivative at x = P to get the slope of the tangent line at P.
Step 2: Find the points of intersection of the tangent line with the coordinate axes.
To find the points of intersection, we set the y-coordinate of the tangent line to zero to find the x-intercept and set the x-coordinate of the tangent line to zero to find the y-intercept.
Step 3: Calculate the area of the triangle enclosed by the tangent line and the coordinate axes.
The area of a triangle can be calculated using the formula: Area = 0.5 * base * height.
In this case, the base is the distance between the x-intercept and the y-intercept, and the height is the y-coordinate of the point P.
Step 4: Repeat steps 1-3 for different x-values in the interval (0, 1) and determine the x-coordinate that gives the smallest triangle area.
Calculate the triangle area for different x-values in the interval (0, 1), and determine the x-coordinate that corresponds to the smallest area.
By following these steps, you can find the x-coordinate of the point P on the parabola where the triangle enclosed by the tangent line at P and the coordinate axes has the smallest area.