(12 points) a 13 foot long ladder leans against a wall and the foot of the ladder is sliding away from the wall at a constant rate of 3 feet/sec. meanwhile, a firefighter is climbing up the ladder at a constant rate of 2 feet/sec. at the instant the firefighter has climbed 4 feet of the ladder, the foot of the ladder is 5 feet from the wall. let h be the vertical height of the firefighter above the ground. find dh/dt at that instant.

  19/13 ft/s ≈ 1.4615 ft/s

Step-by-step explanation:

It may be convenient to choose the time reference as the instant where we want to compute dh/dt. That is, we want to find h'(0).

If w represents the height of the ladder on the wall, we can write two relations: one for w, and one for h, the height of the firefighter.

  w² + x² = 13² . . . . . . where x=5+3t is the distance of the foot of the ladder from the wall

  h = w·(4 +2t)/13 . . . . where 4+2t is the length along the ladder that the firefighter has climbed

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Differentiating with respect to time, we get ...

  2ww' +2xx' = 0

  w' = (-x/w)x' = -3x/w . . . . . solving for w'

and ...

  h' = w'(4 +2t)/13 +w(2/13) . . . . . . . using the product rule

  h' = (-3x/w)(4 +2t)/13 + w(2/13) . . . . filling in w'

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At t=0, x=5, w=12. and we have ...

  h' = (-3·5/12)(4/13) +(12·2)/13 = 19/13 ≈ 1.4615

dh/dt = 19/13 ft/s ≈ 1.4615 ft/s at the given instant

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The attachment shows a solution using a graphing calculator. It gives the same result.

(x, y)→(x − 8, y − 7)

step-by-step explanation:

enjoy!