A15 foot ladder is leaning against a wall. if the top slips down the wall at a rate of 4 ft/s, how fast will the foot be moving away from the wall when the top is 13 feet above the ground

From the pythagorean theorem, we know that x^2 and y^2 = 15^2, where x and y, where x is the base moving from the wall and y is the distance of the ladder sliding down. This means that y = sqrt(225 - x^2). Plugging in x = 13 gives us y = 2sqrt(14). Differentiating both sides of the original equations gives us dy/dt = 13 x 4 / 2sqrt(14) = 52/2sqrt(14) = 26/sqrt(14) ft/s.

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Idon’t know but you’ll figure it out