A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with right triangles. At what rate is the length of the person's shadow changing when the person is 16 ft from the lamppost?

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Answer 1 · 2018-01-19T18:05:19

The length of the shadow is changing at $= 2.69 \frac{f t}{sec}$ $=2.69(ft)/sec$

Let the distance of the person from the bottom of the light post be $= x f t$ $=x ft$

And the length of his shadow is $= y f t$ $=y ft$

Form the similar triangles

$\frac{x + y}{20} = \frac{y}{7}$ $(x+y)/(20)=y/7$

$7 (x + y) = 20 y$ $7(x+y)=20y$

$7 x + 7 y = 20 y$ $7x+7y=20y$

$13 y = 7 x$ $13y=7x$

$y = \frac{7}{13} x$ $y=7/13x$

Differentiating wrt $t$ $t$

$\frac{d y}{d t} = \frac{7}{13} \frac{d x}{d t}$ $dy/dt=7/13dx/dt$

We know that $\frac{d x}{d t} = 5 \frac{f t}{sec}$ $dx/dt=5(ft)/sec$

Therefore,

$\frac{d y}{d t} = \frac{7}{13} \cdot 5 = \frac{35}{13} = 2.69 \frac{f t}{sec}$ $dy/dt=7/13*5=35/13=2.69(ft)/sec$