Answer to Question #336367 in Calculus for John

Question

Related Rates: Problem Solving

Direction: Solve each word problem involving a related rates. Make a sketch of your work

if necessary, and then apply differentiation.

A ladder 20 feet long is leaning against an embankment inclined 60 degrees to the

horizontal. If the bottom of the ladder is being moved horizontally toward the

embankment at 1 foot per second, how fast is the top of the ladder moving when

the bottom is 4 feet from the embankment?

PLEASE ANSWER MY QUESTION QUICKLY!!

DEADLINE : 05/03/2022 11: 00 PM

Answer 1

$x^2=400-y^2$

2x dx/dt=-2y dy/dt

$v(x)i=dx/dt=-(y/x) dy/dt=1$

$dy/dy=-(x/y)$

when x=4 $y=\sqrt{384}=8\sqrt{6}$

$v(y)j=|dy/dt|=4/(8\sqrt6)=\sqrt6/12$

At 4 m from the base the ladder is going towards the wall at v[x] i = 1 i m/s and is climbing up the the wall at v[y] j = [ ( sq rt 6/12] j m/s

The resultant velocity is given by

$v^2=v(x)^2+v(y)^2=1^2+6/144=150/144$

$v=5\sqrt6/12$

and it makes an angle $z=arc tan (v(y)/v(x))=arctan(\sqrt6/12)/1$