# math

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.)

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1. Almost forgot

(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 π/3, find dw/dt=

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2. Karthik,where is 2 questions??

We draw a diagram with X=distance from wall to base of ladder.
Z=distance along ladder that firefighter has climbed.
H =height of firefighter
W=distance from wall to firefighter
Theta=angle of base of ladder with the ground.

Ladder is sliding away at a constant rate of 3ft/sec------>dx/dt=3
Firefighter is climbing up ladder at rate of 2ft/sec------->dz/dt=2
Cos(theta)=x/29
-sin(theta)d*theta/dt=1/29 dx/dt
B)
Cos(theta)=w/(29/-z)
W=(29-z)cos(theta)
dw/dt=(-1 dz/dt)cos(theta)+(29-z)(-sin(theta) d(theta)/dt)
dw/dt=-cos(theta)dz/dt-(29-z)sin(theta)d(theta)/dt

Now you can solve the remaining portion...

Hmm..I don't know whether it is right or not.OK

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3. I think that the θ is just a way of giving us the slope of the ladder. If the firefighter has climbed a distance z to a height h, then we have

x^2+y^2 = 29^2
w^2+(y-h)^2 = (29-z)^2

when θ=π/3, x = 29/2 and y=29√3/2
w = (29-z)/2 and h = z√3/2
when z=6, then,
w = 23/2 and h=3√3

2x dx/dt + 2y dy/dt = 0
2(29/2)(3) + 2(29√3/2) dy/dt = 0
dy/dt = -√3

See if you can finish it off, using the other equation.

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