To determine the rate at which the length of a man's shadow increases as he walks away from a lamp post, we can use similar triangles and related rates in calculus.
Let's set up the problem:
- Let \( h \) be the height of the lamp post.
- Let \( x \) be the distance of the man from the base of the lamp post.
- Let \( s \) be the length of the man's shadow.
When the man is \( x \) meters away from the lamp post, the length of his shadow can be described using similar triangles:
The height of the lamp post \( h \) and the man's height \( m \) create two right triangles: one formed by the lamp post and the man's shadow (formed by the line from the top of the lamp to the tip of the shadow), and the other formed by the man and his shadow.
The relationship can be expressed as follows:
\[ \frac{h}{x+s} = \frac{m}{s} \]
Cross multiplying gives:
\[ h s = m (x + s) \]
Rearranging this yields:
\[ h s = mx + ms \]
Now, isolating \( s \):
\[ h s - ms = mx \quad \Rightarrow \quad s(h - m) = mx \] \[ s = \frac{mx}{h - m} \]
Next, we differentiate this equation with respect to time \( t \) to find the rate of change of the shadow length \( \frac{ds}{dt} \):
Assume the man is walking away from the lamp post at a constant speed:
\[ \frac{dx}{dt} = v \]
Differentiating \( s \):
\[ \frac{ds}{dt} = \frac{m}{h - m} \frac{dx}{dt} \]
Substituting \( \frac{dx}{dt} \):
\[ \frac{ds}{dt} = \frac{m}{h - m} v \]
Thus, the lengthening of the man's shadow depends on both his height \( m \), the height of the lamp post \( h \), and the man's walking speed \( v \). Specifically, if you know these quantities, you can calculate the rate of shadow lengthening \( \frac{ds}{dt} \).