A ladder 20 feet long leans against a vertical building.If the bottom of the ladder slides away from the building horizontally at a rate of 3ft/sec,how fast is the ladder sliding down the building when the top of the ladder is 8ft from the ground?

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    $\begingroup$ What have you tried already? What do you already know. How do you think you should solve this problem? Where are you stuck? Unless you add this information to your question, people might downvote your question. $\endgroup$ – Pedro Mar 14 '15 at 14:55

let $x$ be the distance of the ladder from the wall and $y$ its height from the floor. then you have $$x^2 + y^2 = 20^2, \frac{dx}{dt} = 3\,ft/sec$$ you want to find $\frac{dy}{dt}$ when $y = 8\, ft.$

diffrencing you get $$0 = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \to 0 = \sqrt{20^2 - 8^2} \times 3 + 8\frac{dy}{dt} $$

you can solve the equation for $\frac{dy}{dt}.$ you should get a negative answer. reason why.

  • $\begingroup$ You get a negative answer because the ladder is sliding downwards?? $\endgroup$ – Frank Mar 16 '15 at 5:46
  • $\begingroup$ @Frank, that is correct. $\endgroup$ – abel Mar 16 '15 at 9:56
  • $\begingroup$ Thanks for your assistance $\endgroup$ – Frank Mar 17 '15 at 6:39

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