Question

A 13 feet ladder is leaning against a wall. Then tom begins pulling the foot of the ladder away from the wall at a 0.5 ft/s. How fast is the top of the ladder sliding down when the foot of the ladder is 5ft from the wall?

Answer 1

Pythagorean theorem: \(x^2+y^2=L^2\)

the length of the ladder never changes so when you take the derivative you have
\[2xx'+2yy'=0\]
or
\[yy'=-xx'\]
where y = 12, x = 5 and x' = 0.5

Answer 2

a^2+b^2=c^2
a^2+5^2=13^2
a^2+25=169
a^2=169-25
a^2=144
\[\sqrt{a^2}=\sqrt{144}\]
a=12

Answer 3

Refer to the Mathematica solution attached.

Answer 4

@robtobey2 typo in the pdf. should be \(y\rightarrow\sqrt{13^2-5^2}\)

Answer 5

\[y^2=13^2-x^2=169-x^2,y=\sqrt{169-x^2}\]
\[\frac{ dy }{ dt }=\frac{ 1 }{ 2\sqrt{169-x^2} }*-2x \frac{ dx }{ dt }\]
when x=5 ft,
\[\frac{ dx }{ dt }=0.5=\frac{ 1 }{ 2 }ft/s\]
\[\frac{ dy }{ dt }=\frac{ -5*\frac{ 1 }{ 2 } }{ \sqrt{169-5^2} }=-\frac{ 5 }{ 2*12 }=\frac{- 5 }{ 24 }ft/s\]
negative sign shows it is upper end coming near the foot of wall as lower end is going away from the wall.