A man 1.7m tall walks at the rate of 2 meters per minute towards a street light that is 10m above the ground a.) At what rate is the tip of his shadow moving? b.) At what rate is the length of the shadow changing?

1 Answer
Apr 27, 2018

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Let at an instant the position of man #NM# is #x# m from the light post when the tip of his shadow #T# is #y# m from the post and length of his shadow is #s# m.
Given that man is walking towards the light post at the rate #2m"/"s#, we have #(dx)/(dt)=2m"/min"#

Here height of the man #MN=1.7m# and

height of the light #LO=10#m.

By simple geometry #Delta TMNand Delta TLO# are similar

So by the property of similar triangles

#(TM)/(TS)=1.7/10#

#=>s/(s+x)=0.17#

#=>s=0.17(s+x)#

#=>83s=17x#

Differentiating w. r. to #t# we get

#83xx(ds)/(dt)=17xx(dx)/(dt)=17xx2m"/min"#

#(ds)/(dt)=34/83m"/min"#

(b) So the length of the shadow changing at the rate

#(ds)/(dt)=34/83m"/min"#

Now #y=s+x#

Differentiating this w. r. to #t# we get

#(dy)/(dt)=(ds)/(dt)+(dx)/(dt)#

#(dy)/(dt)=34/83m"/min"+2m"/min"=200/83m"/min"#

(a) Hence the length of the shadow changing at the rate

#(dy)/(dt)==200/83m"/min"#