Certainly! Let's approach this problem by using the concept of similar triangles and related rates.
1. Identify the variables and known values:
- Height of the man: [tex]\( h_m = 6 \)[/tex] feet
- Height of the lamp post: [tex]\( h_l = 18 \)[/tex] feet
- Rate at which the man is moving away from the lamp post: [tex]\( \frac{dD}{dt} = 5 \)[/tex] ft/sec
Let [tex]\( L \)[/tex] be the length of the shadow and [tex]\( D \)[/tex] be the distance of the man from the lamp post.
2. Set up the relationship using similar triangles:
When the man stands some distance from the lamp post, he and his shadow form one right triangle, while the lamp post and the length from the top of the lamp post to the tip of the shadow form a larger right triangle.
From the similar triangles, we have:
[tex]\[
\frac{h_l}{L + D} = \frac{h_m}{L}
\][/tex]
3. Express the relationship algebraically:
Since [tex]\( h_l = 18 \)[/tex] ft and [tex]\( h_m = 6 \)[/tex] ft, plug these values into the equation:
[tex]\[
\frac{18}{L + D} = \frac{6}{L}
\][/tex]
4. Cross-multiply to solve for [tex]\( L \)[/tex] and [tex]\( D \)[/tex]:
[tex]\[
18L = 6(L + D)
\][/tex]
Simplify by expanding and combining like terms:
[tex]\[
18L = 6L + 6D
\][/tex]
Subtract [tex]\( 6L \)[/tex] from both sides:
[tex]\[
12L = 6D
\][/tex]
Divide by 6:
[tex]\[
2L = D
\][/tex]
Therefore, we have:
[tex]\[
L = \frac{D}{2}
\][/tex]
5. Differentiate the relationship with respect to time [tex]\( t \)[/tex]:
We need to find the rate at which the shadow lengthens, [tex]\(\frac{dL}{dt}\)[/tex]. Differentiate both sides of [tex]\( L = \frac{D}{2} \)[/tex] with respect to [tex]\( t \)[/tex]:
[tex]\[
\frac{dL}{dt} = \frac{1}{2} \frac{dD}{dt}
\][/tex]
6. Substitute the known rate [tex]\( \frac{dD}{dt} \)[/tex]:
Given [tex]\( \frac{dD}{dt} = 5 \)[/tex] ft/sec:
[tex]\[
\frac{dL}{dt} = \frac{1}{2} \cdot 5
\][/tex]
Simplify the expression:
[tex]\[
\frac{dL}{dt} = 2.5 \text{ ft/sec}
\][/tex]
Therefore, the rate at which the man's shadow is lengthening is [tex]\( 2.5 \)[/tex] ft/sec.
