(It's a related rate question)
A man 6ft tall walks away from a lamp post (15ft) at 5ft/sec. How fast is his shadow lengthening?
---
I have the picture, and the constants, the man and the lamp post. I have 5ft/sec as dw/dt, and I know I'm looking for ds/dt, the rate of the shadow.
Problem is, I don't know the actual equation to figure this out. Help please?
Thanks!
The lamp post makes a right triangle.
let x be the distance from the lamppost to the man (dx/dt=5; given) Let L be the length of is shadow, measured from the man feet to the end of the shadow.
(dL/dt is the rate of the shadow lengthens.
Using similar triangle
6/15=L/(L+x) or
6L+6x=15L
6x=9 L
x=3/2 L
dx/dt= 3/2 dL/dt
well, then dL/dt=2/3 dx/dt= 2/3 5=10/3 ft/sec
Now if you were to ask at what rate the shadow tip was moving , that is dx/dt+dL/dt
let x be the rate of shadow.
x / (x + 5) = 6 / 15
15x = 6(x+5) = 6x + 30
9x = 18
x = 30/9 = 10/3
To solve this related rates problem, we need to establish a relationship between the variables involved. In this case, we can use similar triangles to find a relationship between the height of the man, the length of his shadow, and the distance from the lamp post.
Let's label the height of the man as 'h' (6ft), the length of his shadow as 's', and the distance from the lamp post as 'd' (15ft).
Using similar triangles, we have the following equation:
h / s = (h + d) / d
We want to find ds/dt, the rate at which the length of the shadow is changing. To do this, we need to differentiate both sides of the equation with respect to time, t.
Differentiating the equation implicitly, we get:
dh / dt * (1/s) = (dh / dt + dd / dt) * (1 / d)
Since we're given that the man is walking away from the lamp post at a rate of 5ft/sec (dw/dt = -5ft/sec), we can substitute these values into the equation.
-5 * (1 / s) = (-5 + dd / dt) * (1 / d)
Now we can solve for dd/dt, the rate at which the man is moving away from the lamp post.
dd / dt = -5 * d (1 / s)
Since we're interested in finding ds/dt, we can use the Pythagorean theorem to relate s and d:
s^2 + d^2 = h^2
Differentiating both sides of this equation with respect to t, we get:
2s * (ds / dt) + 2d * (dd / dt) = 2h * (dh / dt)
Substituting the values we found earlier for dh/dt and dd/dt, we have:
2s * (ds / dt) + 2d * (-5 * d / s) = 2h * (-5 / s)
Now we can solve for ds/dt, the rate at which the length of the shadow is changing.
2s * (ds / dt) = -10h - 10d^2 / s
ds / dt = (-10h - 10d^2 / s) / (2s)
Plugging in the values for h (6ft) and d (15ft), we can calculate ds/dt.