A man who is 1.6 m tall walks away from a lamp which is 4 m above ground at the rate of 30 m/min. How fast is the man's shadow lengthening?

Answer

To solve the problem step-by-step, we can follow these steps:

### Step 1: Understand the Geometry
We have a lamp post that is 4 meters tall and a man who is 1.6 meters tall. The man is walking away from the lamp post, and we need to find out how fast his shadow is lengthening.

### Step 2: Set Up the Variables
Let:
- \( A \) be the height of the lamp (4 m).
- \( B \) be the height of the man (1.6 m).
- \( a \) be the distance from the lamp to the man.
- \( b \) be the length of the man's shadow.

### Step 3: Establish the Relationship
Using similar triangles, we can set up the following relationship based on the heights and distances:
\[ \frac{B}{b} = \frac{A}{a + b} \]
Substituting the known values:
\[ \frac{1.6}{b} = \frac{4}{a + b} \]

### Step 4: Cross Multiply
Cross multiplying gives us:
\[ 1.6(a + b) = 4b \]
Expanding this:
\[ 1.6a + 1.6b = 4b \]
Rearranging gives:
\[ 1.6a = 4b - 1.6b \]
\[ 1.6a = 2.4b \]
Dividing both sides by 0.8:
\[ 2a = 3b \quad \text{(Equation 1)} \]

### Step 5: Differentiate with Respect to Time
Now, differentiate both sides of Equation 1 with respect to time \( t \):
\[ \frac{d}{dt}(2a) = \frac{d}{dt}(3b) \]
This gives:
\[ 2 \frac{da}{dt} = 3 \frac{db}{dt} \]

### Step 6: Substitute Known Values
We know the man walks away from the lamp at a rate of \( \frac{da}{dt} = 30 \) m/min. Substituting this into the equation:
\[ 2(30) = 3 \frac{db}{dt} \]
This simplifies to:
\[ 60 = 3 \frac{db}{dt} \]

### Step 7: Solve for \( \frac{db}{dt} \)
Now, divide both sides by 3:
\[ \frac{db}{dt} = \frac{60}{3} = 20 \text{ m/min} \]

### Conclusion
Thus, the man's shadow is lengthening at a rate of 20 m/min.

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