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Lelechka [254]
4 years ago
9

A 15 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
Mathematics
1 answer:
hoa [83]4 years ago
7 0
<span>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.</span>
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x = - (-4) / 2(1)

x = 4/2 = 2

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