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Question
A person 2m tall walks away from a streetlight at the rate of 1m/s.
If the light is 6m above ground level, how fast is the person’s shadow lengthening?
Hello Vsikandari,
)
I attached a drawing to illustrate the exercise.
v is the velocity of travelling, x is the length of the shadow .
From trigonometry we know that :
x/(x+vt)=2/6 . In our case v=1m/s , so
x/(x+t)=(1/3)
x=(1/3)x+(1/3)t
(2/3)x=(1/3)t
2x=t
x(t)=2t
Therefore,
x'(t)=2 m/sec .
Alon.
Answers by Expert:
Alon Mandes
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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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