Question

"A 13-foot ladder is leaning against a building when its base begins to slide away from the base of the building. By the time the base is 12 feet from the building, the base is moving at the rate of 6 ft/sec. How fast is the top of the ladder sliding down the wall at that point in time?"

I get that you use a^2 + b^2 = c^2 but when you differentiate it, would you get 2a(da/dt) + 2b(db/dt) = 2c(dc/dt)? Then plug numbers in?

Answer 1

yes

5^2 + 12^2 = 13^2 (you should know that by the way)
so base b at 12 and height a at 5

c is 15 feet forever so dc/dt = 0

2 a (da/dt) + 2 b (db/dt) = 0
ao
5 da/dt = - 12 (6)
da/dt = - 72/5

Answer 2

Thanks!

When it asks to find the rate the area is changing at that point in time, did you get the answer -357/5? Because I did, but it doesn't seem right...

Answer 3

I assume you mean the triangle that the ladder makes with the ground and the wall.

Area = (1/2)ab
d(Area)/dt = (1/2)a db/dt + (1/2)b da/dt
so when a= 5, b=12 , db/dt = 6, da/dt = -72/5

d(Area)/dt = (1/2)(5)(6) +(1/2)(12)(-72/5)
= 15 - 86.4
= -71.4