A person 1.8m tall is walking away from a lamppost 6m high at the rate 1.3m/s. At what rate is the end of the person's shadow moving away from the lamppost?

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A person 1.8m tall is walking away from a lamppost 6m high at the rate 1.3m/s. At what rate is the end of the person's shadow moving away from the lamppost?

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Answer 1 · 2016-07-03T17:29:16

$= \frac{13}{7} \frac{m}{s} \approx 1.86 \frac{m}{s}$ $= 13/7 m/s approx 1.86 m/s$

Explanation:

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the point of what follows and the choice of variables in the drawing is this. we know that $. x = 1.3$ $dot x = 1.3$ . we have been asked to find the rate at which the end of the shadow is moving away from the lamppost - that's $. y$ $dot y$ !

so similar triangles tells us that

$\frac{6}{y} = \frac{1.8}{y - x}$ $6/y = 1.8/(y - x)$

$y = \frac{6 x}{4.2} = \frac{10}{7} x$ $y = (6x)/4.2 = 10/7 x$

and now for the little bit of calculus, taking the deriv of each side with respect to time

$\Rightarrow . y = \frac{10}{7} . x$ $implies dot y = 10/7 dot x$
$= \frac{10}{7} \cdot 1.3 = \frac{13}{7} \frac{m}{s} \approx 1.86 \frac{m}{s}$ $= 10/7 * 1.3 = 13/7 m/s approx 1.86 m/s$

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A person 1.8m tall is walking away from a lamppost 6m high at the rate 1.3m/s. At what rate is the end of the person's shadow moving away from the lamppost?

1 Answer

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