A 13 ft ladder is leaning against a house when its base begins to slide away. By the time the base is 3 ft from the house, the base is moving at the rate of 2 ft/sec. Answer the following:

At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?

(dh/dt = 12/(2(sqrt160) )

Question

A 13 ft ladder is leaning against a house when its base begins to slide away. By the time the base is 3 ft from the house, the base is moving at the rate of 2 ft/sec. Answer the following: 

At what rate is the area of the triangle formed by the ladder, wall, and ground changing then?

(dh/dt = 12/(2(sqrt160) )

anonymous · Answer

Is da/dt = (1/2)*(db/dt)*(h)+(1/2)*(b)*(dh/dt) the correct differentiation of area of triangle or did I screw up...