A ladder 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 π/3 with the ground. Answer the two related rates questions below. (Hint: Use two carefully labeled similar right triangles.)

(a) If h is the height of the firefighter above the ground, at the instant the angle of the ladder with the ground is π/3, find dh/dt=

If w is the horizontal distance from the firefighter to the wall, at the instant the angle of the ladder with the ground is π/3, find dw/dt=

Question:

Does the firefighter clim up 2 ft/sec along the length of the ladder , or 2 ft/sec vertically ?

I think we can safely assume that the firefighter's progress is measured along the ladder.

A) (23*sqrt3)/29 ***when the ladder is 29 feet long

B) got stuck on this one still

To solve these related rates questions, we will use similar right triangles. Let's label the variables and set up the relationship between them.

Let h be the height of the firefighter above the ground, and w be the horizontal distance from the firefighter to the wall. The ladder is 29 feet long, and the foot of the ladder is sliding away from the wall at a rate of 3 feet/sec. The firefighter is climbing up the ladder at a rate of 2 feet/sec.

At a given instant, when the angle of the ladder with the ground is π/3, we can create two similar right triangles:

Triangle 1: This triangle represents the entire ladder, with vertical height h and horizontal distance w.
Triangle 2: This triangle represents the portion of the ladder above the height the firefighter has climbed, with vertical height (29 - h) and horizontal distance (w - 6).

Since the two triangles are similar, we can use their corresponding sides to set up a proportion:

(29 - h)/(w - 6) = h/w

We can simplify this equation by cross-multiplying:

w(29 - h) = h(w - 6)

Expanding both sides:

29w - wh = hw - 6h

Moving all terms to one side:

29w - 6h = hw - wh

Now, let's differentiate both sides of the equation with respect to time (t) to find the related rates:

d(29w)/dt - d(6h)/dt = d(hw)/dt - d(wh)/dt

Since d(29w)/dt represents the rate at which the horizontal distance from the wall is changing (dw/dt), and d(6h)/dt represents the rate at which the height is changing as the firefighter climbs (dh/dt), we can substitute these values into the equation:

29(dw/dt) - 6(dh/dt) = (hw)(dh/dt) - (wh)(dw/dt)

Given that dw/dt = 3 ft/sec and dh/dt = 2 ft/sec, and the angle θ is π/3, we can substitute these values and solve for dh/dt:

29(3) - 6(2) = (h * w)(2) - (w * h)(3)

87 - 12 = 2hw - 3hw

75 = -hw

Dividing both sides by -w:

75/w = -h

Therefore, dh/dt = -75/w ft/sec

For part (a) of the question, we have already found dh/dt by solving the related rates equation.

Now, to find dw/dt at the instant the angle of the ladder with the ground is π/3, we can substitute the known values into the related rates equation:

29(dw/dt) - 6(2) = (h * w)(2) - (w * h)(3)

Simplifying:

29(dw/dt) - 12 = 2hw - 3hw

29(dw/dt) = hw

dw/dt = hw / 29

Substituting dh/dt = -75/w, we have:

dw/dt = (-h * 75) / w

Since we are asked to find dw/dt at the instant the angle of the ladder with the ground is π/3, we can further substitute the known values of h, w, and the angle:

dw/dt = (-(29/tan(π/3)) * 75) / w

Simplifying:

dw/dt = (-75 * √3) / w

Therefore, dw/dt = (-75 * √3) / w ft/sec

To summarize:
(a) dh/dt = -75/w ft/sec
(b) dw/dt = (-75 * √3) / w ft/sec

a) dh/dt= 1.373695468