Vertical component of velocity

Calculus Level 3

A 10 m long stick is resting against the wall in the picture. Someone is pulling the stick's lower end horizontally to the left with a velocity of 4 cm/s. Find the velocity (in cm/s ) of the upper end of the stick when the lower end is 6 m away from the wall.


The answer is 3.

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Srinivasa Satti by Srinivasa Satti

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1 solution

Relevant wiki: Related Rates of Change - Basic

By pythagorean theorem,

1 0 2 = y 2 + x 2 10^2=y^2+x^2

100 = y 2 + x 2 100=y^2+x^2

Differentiate both sides with respect to t t .

100 = 2 y d y d t + 2 x d x d t 100 = 2y \dfrac{dy}{dt}+2x \dfrac{dx}{dt}

y d y d t = − x d x d t y \dfrac{dy}{dt} = -x \dfrac{dx}{dt}

d y d t = − x y d x d t \dfrac{dy}{dt} = \dfrac{-x}{y} \dfrac{dx}{dt}

Solve for y y when x = 6 x=6 . By pythagorean theorem, we have

1 0 2 = y 2 + 6 2 10^2=y^2+6^2

y 2 = 64 y^2=64

y = 8 y=8

Substitute:

d y d t = − x y d x d t = − 6 8 ( 4 ) = − 3 \dfrac{dy}{dt} = \dfrac{-x}{y} \dfrac{dx}{dt}=\dfrac{-6}{8}(4)=\boxed{-3}

The negative sign indicates that y y is decreasing as t t increases.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

The question should really ask for the "speed" or "magnitude of velocity" because I put in negative three at first and got confused that it was wrong.

Tristan Goodman - 2 years, 1 month ago

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