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  • A man, 2m tall, walks at the rate of m/s towards a street light which is m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is m from the base of the light?

Q: A man, 2m tall, walks at the rate of 1\frac{2}{3}m/s towards a street light which is 5\frac{1}{3} m above the ground. At what rate is the tip of his shadow moving? At what rate is the length of the shadow changing when he is 3\frac{1}{3} m from the base of the light?

Answers (1)

Given: a 2m tall man walks at the rate of 1\frac{2}{3}m/s towards a 5\frac{1}{3} m tall street light.

To find: calculate the rate of movement of the tip of the shadow and also the rate of change in the length of the shadow when he is 3\frac{1}{3} m from the base of the light.

Explanation:

Here the street light is AB = 5\frac{1}{3}

And man is DC = 2m

Let BC = x m and CE = y m

The rate of the man’s walk towards the streetlight is 1\frac{2}{3}m/s, and as the man is moving towards the street light, the entity carries a negative charge

Hence, \frac{\mathrm{dx}}{\mathrm{dt}}=-1 \frac{2}{3} \mathrm{~m} / \mathrm{s}....(i)

Now consider ΔABE and ΔDCE

∠DEC = ∠AEB (same angle)

∠DCE = ∠ABD = 90°

Hence by AA similarity,

ΔABE≅ΔDCE

Hence by CPCT,

\frac{AB}{DC}=\frac{BE}{CE}

\\ \begin{aligned} &\text { After substituting the values mentioned in the figure, we get }\\ &\Rightarrow \frac{\left(5 \frac{1}{3}\right)}{2}=\frac{x+y}{y}\\ &\Rightarrow\left(\frac{16}{3}\right) \mathrm{y}=2(\mathrm{x}+\mathrm{y})\\ &\Rightarrow 16 y=6(x+y)\\ &\Rightarrow 16 y=6 x+6 y\\ &\Rightarrow 16 y-6 y=6 x\\ &\Rightarrow 10 y=6 x\\ &\Rightarrow \mathrm{y}=\frac{6}{10} \mathrm{x}\\ &\Rightarrow \mathrm{y}=\frac{3}{5} \mathrm{x} \end{aligned}

Apply the first derivative with respect to t,

\\ \Rightarrow \frac{d y}{d t}=\frac{d\left(\frac{3}{5} x\right)}{d t}$\\ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{3 \mathrm{~d}(\mathrm{x})}{5}
After substituting the value in the above equation from equation (i), we get

\\\Rightarrow \frac{d y}{d t}=\frac{3}{5}\left(-1 \frac{2}{3}\right)$\\ $\Rightarrow \frac{d y}{d t}=\frac{3}{5}\left(-\frac{5}{3}\right)$\\ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=-1 \mathrm{~m} / \mathrm{s}

So, the rate of movement of tip of the shadow is -1m/s, i.e., the length of the shadow is decreasing at the rate of 1m/s.

Let BE = z

So from fig,

z = x+y

Let’s apply the first derivative with respect to t on the above equation.

\\ \begin{aligned} &\frac{\mathrm{dz}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{x})}{\mathrm{dt}}+\frac{\mathrm{dy}}{\mathrm{dt}}\\ &\text { Substituting oft eh corresponding values result in, }\\ &\frac{d z}{d t}=-1 \frac{2}{3}-1\\ &\Rightarrow \frac{d z}{d t}=-\frac{5}{3}-1\\ &\Rightarrow \frac{\mathrm{dz}}{\mathrm{dt}}=\frac{-5-3}{3}\\ &\Rightarrow \frac{\mathrm{dz}}{\mathrm{dt}}=\frac{-8}{3}\\ &\Rightarrow \frac{\mathrm{dz}}{\mathrm{dt}}=-2 \frac{2}{3} \mathrm{~m} / \mathrm{s} \end{aligned}

So, the rate of tip of the shadow moving towards the light source is of 2 \frac{2}{3}  m/s.

 

Posted by

infoexpert22

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