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?

Question

Evelyn Agard · Accepted Answer

The length of the shadow is changing at #=2.69(ft)/sec#

Let the distance of the person from the bottom of the light post be #=x ft#

And the length of his shadow is #=y ft#

Form the similar triangles

#(x+y)/(20)=y/7#

#7(x+y)=20y#

#7x+7y=20y#

#13y=7x#

#y=7/13x#

Differentiating wrt #t#

#dy/dt=7/13dx/dt#

We know that #dx/dt=5(ft)/sec#

Therefore,

#dy/dt=7/13*5=35/13=2.69(ft)/sec#

Christopher Agresta · Answer

undefined To find the rate at which the length of the person's shadow is changing when the person is 16 ft from the lamppost, we can use related rates and similar triangles. Let $ x $ be the distance from the person to the lamppost and $ y $ be the length of the person's shadow. Since the person's height is 7 ft and the lamppost's height is 20 ft, the length of the lamppost's shadow is 20 ft. By similar triangles, we have $ \frac{x}{y} = \frac{20}{7} $. Differentiating both sides with respect to time, we get $ \frac{dx}{dt} = -5 $ ft/sec and $ \frac{dy}{dt} = ? $. Solving for $ \frac{dy}{dt} $ when $ x = 16 $, we find $ \frac{dy}{dt} = -\frac{175}{14} $ ft/sec. Therefore, the rate at which the length of the person's shadow is changing when the person is 16 ft from the lamppost is $ -\frac{175}{14} $ ft/sec.