[Math] Rate Of Change Of Shadow

calculus

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

This is my diagram of the scenario:

Please have a look at my solution and confirm if correct and if not, give me a hint as to what I'm doing wrong.

  • $H =$ height of the shadow on the building

  • $h = $height of the man

  • $X = $distance from building to the man

  • $x = $distance from spotlight to the man

From the diagram $x + X = 12, $ and $h = 2$

$x(t) = 1.6t$, with $t$ in $seconds$.

$\frac{dx}{dt} = 1.6$

By similar triangles: $\frac{h}{H} = \frac{x}{x+X} \Leftrightarrow H = \frac{(x+X)h}{x} = \frac{(12)(2)}{x} = \frac{24}{x}$

$\Rightarrow \frac{dH}{dt} = \frac{d}{dt}(\frac{24}{x}) = -\frac{24}{x^2}\frac{dx}{dt}$

When the man is $4m$ from the building, $x = 12 – 4=8 \Rightarrow \frac{dH}{dt} = -\frac{24}{8^2}(1.6) = -\frac{3}{5}m/s$

Best Answer

From the diagram $x+X=12$, and $h=2$ $x(t)=1.6t$, with t in seconds. $$dxdt=1.6$$ By similar triangles: $$h/H=x/(x+X)\Leftrightarrow H=(x+X)h/x=(12)(2)/x=24/x$$ $$\implies dH/dt=d/dt(24x)=−24/x^2dx/dt$$ When the man is 4m from the building, $x=12−4=8\implies dH/dt=−\frac{24}{8^2}(1.6)=−\frac35m/s$ Length of his shadow on the building is decreasing at a rate $0.6m/s$ when he is 4m from the building.

Yes it's correct.

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