A 6 ft tall person walks away from a 10 ft lamppost at a constant rate of 5 ft/s. What is the rate (in ft/s) that the tip of the shadow moves away from the pole when the person is 10 ft away from the pole?
I have drawn out the picture, but I am not sure where in fact ot get the equation?s
Does your picture show the similar triangles (the big triangle made by the light post light to the end of the person's shadow... and the little triangle the person makes with his own shadow)? If so then dx/dt = 5 ft/s, and we need ds/dt when x = 10.
To find the rate at which the tip of the shadow moves away from the pole, we can create a right triangle with the person, lamppost, and the tip of the shadow. Let's denote the distance from the person to the lamppost as x, and the length of the shadow as y.
Since the person is 6 ft tall and the lamppost is 10 ft tall, the length of the shadow is proportional to the distance between the person and the lamppost. This means we can write the following proportion:
(Length of the shadow) / (Height of the lamppost) = (Distance between the person and the lamppost) / (Height of the person)
Expressed algebraically:
y / 10 = x / 6
To find the rate at which the tip of the shadow moves away from the pole, we need to differentiate this equation with respect to time. Since x and y are changing with time, we can write:
(d/dt)(y / 10) = (d/dt)(x / 6)
The left side represents the rate at which the length of the shadow is changing with time, and the right side represents the rate at which the distance between the person and the lamppost is changing with time.
To find the rate at which the tip of the shadow moves away from the pole when the person is 10 ft away from the pole (x = 10), we can substitute x = 10 into the equation and solve for dy/dt.
By substituting x = 10 into the equation y / 10 = x / 6, we get:
y / 10 = 10 / 6
Simplifying the equation, we find:
y = 100 / 6 = 50 / 3
Now, we can differentiate the equation:
(d/dt)(y / 10) = (d/dt)(10 / 6)
(dy/dt) / 10 = 0 / 6
dy/dt = 0
Thus, when the person is 10 ft away from the pole, the rate at which the tip of the shadow moves away from the pole is 0 ft/s.
To solve this problem, you can use similar triangles and apply the concept of rates.
Let's label the following lengths:
- Height of the person: 6 ft (h)
- Distance of the person from the lamppost: x ft
- Length of the person's shadow: y ft
As the person walks away from the lamppost, the length of their shadow increases. We want to find how fast the tip of the shadow is moving away from the pole when the person is 10 ft away from the pole.
To set up an equation, consider the similar triangles formed by the person and their shadow:
1. The large triangle formed by the person, the lamppost, and their shadow:
- The height of the person (6 ft) corresponds to the length of their shadow (y ft).
- The distance of the person from the lamppost (x ft) corresponds to the distance of the shadow from the lamppost (x + 10 ft), since the person is 10 ft away from the pole.
2. The small triangle formed by the person, the tip of the shadow, and the lamppost:
- The height of the person (6 ft) corresponds to the length of the shadow (y ft).
- The distance of the person from the lamppost (x ft) corresponds to the distance of the tip of the shadow from the lamppost (10 ft), as the person is 10 ft away from the pole.
Now, let's differentiate both sides of the equation with respect to time (t) to find the rates:
1. Differentiating the large triangle equation:
6 / y = (x + 10) / x
2. Differentiating the small triangle equation:
6 / y = 10 / x
To find the rate at which the tip of the shadow is moving away from the pole, we need to find dy/dt (the derivative of y with respect to time).
1. Equate the two equations to find the relationship between x and y:
(x + 10) / x = 10 / x
x + 10 = 10
2. Solve for x:
x = 10
3. Substitute x = 10 into either of the original equations to find y:
6 / y = (10 + 10) / 10
6 / y = 2
y = 3
Now, substitute x = 10 and y = 3 into one of the original equations and differentiate both sides with respect to time (t):
6 / 3 = 10 / 10 * (dy/dt)
2 = dy/dt
Therefore, the rate at which the tip of the shadow moves away from the pole is 2 ft/s.