A10-ft ladder is leaning against a house when its base starts to slide away. by the time the base is 6 ft from the house, the base is moving away at the rate of 8 ft/sec. a. what is the rate of change of the height of the top of the ladder? b. at what rate is the area of the triangle formed by the ladder, wall, and ground changing then? a coordinate plane has a horizontal x-axis and a vertical y-axis. a thin rectangle falling from left to right labeled 10 foot ladder starts on the y-axis at y(t) and ends on the x-axis at x(t). an arrow on the y-axis below y(t) is pointing down and an arrow on the x-axis to the right of x(t) is pointing to the right. the angle created between the segment and the x-axis is labeled theta. theta x(t) y(t) 0 1

a. what is the rate of change of the height of the top of the ladder?
Let y be the height
Let x be the base
dx/dt=8 ft/sec, x=6 ft, hypotenuse=10 ft
x²+y²=10²
2x(dx/dt)+2y(dy/dt)=0
solving for dy/dt in terms of x,y and dx/dt:
dy/dt=(-x/y)(dx/dt)
but now x=6 and y=8
dy/dt=(-6/8)(8)=-6 ft/sec

b]b. at what rate is the area of the triangle formed by the ladder, wall, and ground changing then?
Here the rate of change of the area is dA/dt

dA/dt=1/2(x*dy/dt+y*dx/dt)
but
x=6, y=8, dy/dt=-6, dx/dt=8
plugging in our values we get:
dA/dt=1/2(6×(-6)+8(8))
dA/dt=1/2(-36+64)=14 ft²/sec

c. At what rate is the angle between the ladder and the ground changing then?
The relationship that relates the angle with sides x and y of a right angle:

dθ/dt=-1/sinθ1/10dx/dt
sinθ=8/10
dx/dt=8
thus
-10/8×1/10×8
=-1 rad/sec

Ab=dc ac=de ad=bce i hope this !