Q.229. Rate of change of area between a curve and a line.

Question 229.
A line joining ( 0, - 6 ) and ( 2, 0 ) intersects the  curve, 6 x^4 – 5 x^2 y^2 + 3 x – 2 y + 5 = 0,  in the 4th quadrant. The line is sliding towards the origin parallel to its original position at the rate of 0.5 unit/s. Find the rate of change of area enclosed between the curve and the line after 0.5 second.

Answer 229.
The slope of line through (0, - 6) and (2, 0) is (0+6)/(2-0) = 3.
The equation of the line through (0, - 6) and (2, 0) having slope 3 is
y = 3(x - 2)

Moving closer to the origin at the rate of 0.5 unit/s, it would go closer to the origin by a distance of 0.25 unit in 0. 5 sec.
When this line moves closer to the origin through a distance 0.25 units, its equation is
y = 3x - 6 + (0.25)√(10)

The intersection points  of this line with the curve as given by Wolfram Alpha Link  are
(1.11924, - 1.85172) and (0.432909, - 3.9107)

The length of this line-segment
≈ √[(1.11924 - 0.432909)^2 + (3.9107 - 1.85172)^2]
≈ √(0.471050241561 + 4.2393986404)
≈ 2.1704

Now suppose the line moves through a distance (0.50) dt in time dt,
the rate of change of area
= (0.50) dt * (2.1704) / dt
= 1.0852 sq. units per second.

Link to YA!