Question

A man who is 6 feet tall is walking at a rate of 4 feet/second, directly away from a light that is located 10 feet above the ground.

How long is the man's shadow t seconds after he starts walking? The answer will be in terms of t.

Answer 1

2.5t

Explanation:

Let the man be

x

feet away from the street light, and his shadow length be

y

, Let the tip of the shadow be

l

feet away from the light such that

l

=

x

+

y

and we are given that

d

x

d

t

=

2

By similar triangles:

y

6

=

x

+

y

30

∴

30

y

=

6

(

x

+

y

)

∴

30

y

=

6

x

+

6

y

∴

30

y

−

6

y

=

6

x

∴

24

y

=

6

x

∴

y

=

1

4

x

Now,

l

=

x

+

y

⇒

l

=

x

+

1

4

x

∴

l

=

5

4

x

Differentiating wrt

x

gives;

d

l

d

x

=

5

4

And by the chain rule we have:

d

l

d

t

=

d

l

d

x

d

x

d

t

∴

d

l

d

t

=

5

4

×

2

∴

d

l

d

t

=

2.5

feet/sec