y = x²

Δy = -(1/3)

The curve and the triangle are symmetrical about the y-axis, so we can take one half and double it.

We want the height of the triangle, which is the difference between the height of line AB, and the y-intersection of the tangent line.

Tangent line: y = mx + c

m = gradient of line = gradient of curve = dy/dx = 2x

Base of (half) triangle = x

Area of (half) triangle = (1/2) * x * (y - c)

When line AB is at a height of H, then the intersection with the curve is (√H, H) and (-√H, H)

Gradient of the curve and line = 2√H

y-intersection of the tangent line = c = y - mx = H - (2√H)(√H) = H - 2H = -H

Area of (full) triangle = 2 * (1/2) * x * (y - c) = √H * (H - -H) = √H * 2H = 2H√H

A = 2(H)^3/2

dA/dt = dA/dH * dH/dt = 2(3/2)(H)^1/2 * -(1/3) = -H^1/2

At H = 16, dA/dt = -16^1/2 = -4 units² / second

Area is decreasing at a rate of 4 units² / second