A line in the first quadrant is tangent to the graph of y=1/x^2. How can we minimize the area between the line and the axes?
To minimize the area between the line and the axes, we need to find the line that is tangent to the graph of y = 1/x^2 at a point in the first quadrant.
First, let's find the derivative of the function y = 1/x^2. Taking the derivative will give us the slope of the tangent line at any point.
dy/dx = -2/x^3
Next, we need to find the point of tangency. Let's denote the x-coordinate of this point as 'a'. At the point of tangency, the slope of the tangent line should be equal to the derivative of the function at that point.
So, we set the derivative equal to the slope, which is equal to the y-coordinate (since the tangent line is vertical):
-2/a^3 = 1/a
Solving this equation will give us the value of 'a', which represents the x-coordinate of the point of tangency.
Multiplying both sides by 'a^3', we get:
-2 = a^2
Taking the square root of both sides gives us:
a = √2 (positive square root since it lies in the first quadrant)
So, the x-coordinate of the point of tangency is √2.
Now, we can determine the corresponding y-coordinate by substituting the value of 'a' into the equation y = 1/x^2:
y = 1/(√2)^2
y = 1/2
Therefore, the point of tangency is (√2, 1/2).
Now that we have the slope of the tangent line and a point on the line, we can find the equation of the tangent line using the point-slope form:
y - y₁ = m(x - x₁), where (x₁, y₁) is the point of tangency and 'm' is the slope.
Plugging in the values, we get:
y - 1/2 = -2/√2 (x - √2)
Simplifying, we obtain:
y = -2/√2 x + 3/2
To minimize the area between the line and the axes, we need to consider the region bounded by the tangent line, the x-axis, and the y-axis. This area is equal to the integral of the function y = -2/√2 x + 3/2 from 0 to √2.
∫[0,√2] (-2/√2 x + 3/2) dx
Evaluating the integral will give us the minimum area.