A 10​-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