Question #336367

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


1
Expert's answer
2022-05-03T16:36:19-0400

x2=400y2x^2=400-y^2

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

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

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

when x=4 y=384=86y=\sqrt{384}=8\sqrt{6}

v(y)j=dy/dt=4/(86)=6/12v(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

v2=v(x)2+v(y)2=12+6/144=150/144v^2=v(x)^2+v(y)^2=1^2+6/144=150/144

v=56/12v=5\sqrt6/12

and it makes an angle z=arctan(v(y)/v(x))=arctan(6/12)/1z=arc tan (v(y)/v(x))=arctan(\sqrt6/12)/1


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