Aladder 29 feet long leans against a wall and the foot of the ladder is sliding away at a constant rate of 3 feet/sec. meanwhile, a firefighter is climbing up the ladder at a rate of 2 feet/sec. when the firefighter has climbed up 6 feet of the ladder, the ladder makes an angle of ? /3 with the ground. answer the two related rates questions below. (hint: use two carefully labeled similar right ) if h is the height of the firefighter above the ground, at the instant the angle of the ladder with the ground is ? /3, find dh/dt= feet/sec.(b) if w is the horizontal distance from the firefighter to the wall, at the instant the angle of the ladder with the ground is ? /3, find dw/dt= feet/sec.

dh/dt=1.373 ft/s

dw/dt=2.78 ft/s

Step-by-step explanation:απ

We need to find dh/dt. Looking at the diagram, we'll write h as a function of known parameters:

h=z sin(α) where h is the height of the firefighter from the ground,     and z the distance along the ladder α is the angle of the ladder.

Deriving:

We have given in the exercise but we need

Again, looking at the drawing we can write α in term of x, since we can calculated x and we have

29 *cos(α)=x where x is the distance from the base of the ladder to the wall.

Deriving:

replacing out known values: α=π/3, , z=6,

Now that we have dα/dt:

Now we do a similar process for dw/dt, looking at the drawing:

w=(29-z)*cos(α)

Replacing known values:

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

i will with my problems and all the other things that need to   be on by the way

step-by-step explanation: