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The ladder forms a right triangle against the wall. Length of the ladder (17ft) is the hypotenuse of such a triangle. Second side is given as 6ft.
Let y be the vertical length and x be the horizontal length of the legs of a right triangle that GT has described above. Then the following is true:\[x^2+y^2=17^2 \]Solve the above for y\[y=\sqrt{289-x^2} \]take the Total Derivative of the result\[dy=-\frac{x dx}{\sqrt{289-x^2}} \]and solve for dx\[dx=-\frac{\sqrt{289-x^2} dy}{x} \]Replace x and dy with 6 and -4 respectively and simplify the result.\[dx=\frac{2 \sqrt{253}}{3}=10.604 \]
Thank you, robtobey! That is the correct answer.
Your welcome.
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