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The exercise with solution is:

classical ladder problem $$$$ The way I tried to solve it is:

$x^2 + y^2 = 15^2$ $$$$ $dx = t\frac{dx}{dt} = 12\frac{1}{4} =3$

$x_{before} = 10$

$x_{after} = x_{before} - dx = 10 - 3 = 7$ $$$$ $y_{before} = \sqrt{15^2 - x^2_{before}} = \sqrt{15^2 - 10^2} = 11.18$

$y_{after} = \sqrt{15^2 - x^2_{after}} = \sqrt{15^2 - 7^2} = 13.27$

$dy = y_{after} - y_{before} = 13.27 - 11.18 = 2.09$ $$$$ $dy/dt = \frac{2.09}{12} = 0.174$

compared to the result of the original example: $y' = \frac{7}{4\sqrt{176}} = 0.134$ $$$$ What is wrong with my method ? or is it just the margin of error of the derivatives (if so then why it is used in this case?)

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The bit that goes $dx=\dots=3$ doesn't make sense. What does $dx$ mean to you? –  Gerry Myerson Mar 19 '13 at 11:51
1  
You have calculated how fast the top moves on average over the $12$ seconds. The question asks for the instantaneous speed, at $12$ seconds. –  Gerry Myerson Mar 19 '13 at 11:54
    
I may have misused the symbols dx/dy/dt, since I was handling them loosely to actually mean any difference, not as they would normally be used in differential calculus. I see how I overlooked solving for instantaneous speed, thanks @GerryMyerson, problem solved. –  John S. Mar 19 '13 at 17:00
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