Help!!! A 25 ft ladder is leaning against a vertical wall. The bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec. Determine how fast the top of the ladder is sliding when the bottom of the ladder is 15 ft from the wall. A) –4 ft/sec B) –2.25 ft/sec C) –13.375 ft/sec D)–12.25 ft/sec E) –0.75 ft/sec

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Help!!! A 25 ft ladder is leaning against a vertical wall. The bottom of the ladder is pulled horizontally away from the wall at 3 ft/sec. Determine how fast the top of the ladder is sliding when the bottom of the ladder is 15 ft from the wall. A) –4 ft/sec B) –2.25 ft/sec C) –13.375 ft/sec D)–12.25 ft/sec E) –0.75 ft/sec

MIT 18.02 Multivariable Calculus, Fall 2007
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Let x be the horizontal distance from the wall and y be the point where the ladder touches the wall. Then \[x^2+y^2=25^2.\] Then the rate at which the top of the ladder is sliding is \[dy/dt.\] To find \[dy/dt\], differentiate the relation \[x^2+y^2=25^2.\] implicitly with respect to \[t.\] We are given that \[dx/dt=3\] and \[x=15\]. From the relation, we can find \[y\] when \[x=15.\] I don't want to give away the answer, but just differentiate implicitly w.r.t. t and solve for dy/dt. Good luck!

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