A ladder 13 feet long is leaning against the side of a building. If the foot of the ladder is pulled away from the building at a constant rate of 8 inches per second, how fast is the area of the triangle formed by the ladder, the building, and the ground changing in feet squared per second at the instant when the top of the ladder is 12 feet above ground? I can't seem to find anything like this in my notes... Thank you for your help in advance
2 answers
I got -119/36 ft^2/sec... not really sure how I got that...
eh? did you not write down your calculations? How can you possibly not know how you arrived at an answer. I think you mean you are not sure whether your answer is right.
At the moment in question, we have a 5-12-13 right triangle.
x^2 + y^2 = 13^2
2x dx/dt + 2y dy/dt = 0
2(5)(2/3) + 2(12) dy/dt = 0
dy/dt = -5/18 ft/s
the area is 1/2 xy, so
da/dt = 1/2 y dx/dt + 1/2 x dy/dt
= (1/2)((12)(2/3) + (5)(-5/18))
= 119/36
Not sure how you got your minus sign. The ladder has just started to fall, so the area of the triangle will be increasing until x and y are equal.
At the moment in question, we have a 5-12-13 right triangle.
x^2 + y^2 = 13^2
2x dx/dt + 2y dy/dt = 0
2(5)(2/3) + 2(12) dy/dt = 0
dy/dt = -5/18 ft/s
the area is 1/2 xy, so
da/dt = 1/2 y dx/dt + 1/2 x dy/dt
= (1/2)((12)(2/3) + (5)(-5/18))
= 119/36
Not sure how you got your minus sign. The ladder has just started to fall, so the area of the triangle will be increasing until x and y are equal.
1 answer