Refer to the diagram below. We have the following points
- A = base of the streetlamp
- B = base of the person
- C = tip of the person's shadow
- D = top of the streetlamp
- E = head of the person
The shadow extends from point B to point C. If we let x be the the horizontal distance from the lamp to the person, then dx/dt represents the speed at which the person is walking away from the lamp. In this case, dx/dt = 4 feet per second.
Let y be the length of the shadow. We can use similar triangles and proportions to help find what y is equal to in terms of x
AD/BE = AC/BC
15/6 = (x+y)/y
15y = 6(x+y)
15y = 6x+6y
15y-6y = 6x
9y = 6x
y = 6x/9
y = 2x/3
y = (2/3)*x
The length of the shadow is 2/3 that of the distance from the person to the lamp.
Now apply the derivative to both sides to compute dy/dt, which represents how fast the shadow is changing.
y = (2/3)x
dy/dt = d/dt[ (2/3)x ]
dy/dt = (2/3)*d/dt[ x ]
dy/dt = (2/3)*dx/dt
dy/dt = (2/3)*4
dy/dt = 2.667
The rate in which the shadow is lengthening is approximately 2.667 ft per second.
