Question

1. a spotlight on the ground shines on a wall 12 m away. if a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Explanation:

Step1: Set up the proportion

Let $x$ be the distance of the man from the spotlight and $y$ be the length of the shadow on the wall. Using similar - triangles, we have $\frac{2}{y}=\frac{x}{12}$, which can be rewritten as $xy = 24$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of $xy = 24$ with respect to $t$ using the product rule $(uv)^\prime=u^\prime v + uv^\prime$, where $u = x$ and $v = y$. We get $x\frac{dy}{dt}+y\frac{dx}{dt}=0$.

Step3: Find the values of $x$ and $y$

The man is $4$ m from the building, so $x=12 - 4=8$ m. From $xy = 24$, when $x = 8$, then $y = 3$ m. The man's speed is $\frac{dx}{dt}=1.6$ m/s.

Step4: Solve for $\frac{dy}{dt}$

Substitute $x = 8$, $y = 3$, and $\frac{dx}{dt}=1.6$ into $x\frac{dy}{dt}+y\frac{dx}{dt}=0$. We have $8\frac{dy}{dt}+3\times1.6 = 0$. Then $8\frac{dy}{dt}=- 4.8$, and $\frac{dy}{dt}=-0.6$ m/s. The negative sign indicates that the length of the shadow is decreasing.

Answer:

$0.6$ m/s