A street light is at the top of a 17 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 6 ft/sec along a straight path. How fast is her shadow lengthening when she is 30 ft from the base of the pole?
Well, let's shed some light on this problem! So, we have a woman walking away from a street light with a certain speed, and we want to know how fast her shadow is lengthening.
First, let's assume the woman's shadow is represented by a similar triangle formed by the woman, the street light, and her shadow. We have a right triangle here!
Now, let's call the length of the woman's shadow "s" and the distance between the woman and the base of the pole "x". According to the geometry, we have similar triangles, so we can set up a proportion:
s/6 = (s+17)/x
Cross-multiplying, we get:
s*x = 6*(s+17)
Now, we can differentiate both sides of the equation with respect to time t:
s*dx/dt + x*ds/dt = 6*(ds/dt)
Great! We know that the woman's speed is 6 ft/sec, so ds/dt = 6. Also, we want to find dx/dt, which is the rate at which x is changing when the woman is 30 ft from the base of the pole.
Plugging in the values, we get:
6*(dx/dt) + 30*(ds/dt) = 6*(ds/dt)
6*(dx/dt) = -30
So, dx/dt = -5 ft/sec
Therefore, when the woman is 30 ft from the base of the pole, her shadow is lengthening at a rate of 5 ft/sec in the opposite direction.
I hope that brings a ray of sunshine to this problem!
To find how fast the woman's shadow is lengthening, we can use similar triangles.
Let's denote:
- x as the distance between the woman and the base of the pole.
- y as the length of the shadow.
We can see that the two triangles formed by the woman, the pole, and the shadow are similar.
Using the similar triangles, we can set up the following proportion:
(x + y) / y = (17 + 6) / 6
Simplifying the equation, we get:
(x + y) / y = 23 / 6
Cross-multiplying, we have:
6(x + y) = 23y
Expanding, we get:
6x + 6y = 23y
Rearranging the equation, we have:
6x = 23y - 6y
6x = 17y
Now, let's differentiate both sides of the equation with respect to time (t):
d/dt (6x) = d/dt (17y)
6(dx/dt) = 17(dy/dt)
Since the woman is walking away from the pole, dx/dt = -6 ft/sec.
Substituting the values we know, we have:
-6 = 17(dy/dt)
Solving for dy/dt, we find that the rate at which the shadow lengthens is:
dy/dt = -6 / 17 ft/sec
Therefore, when the woman is 30 ft from the base of the pole, her shadow is lengthening at a rate of -6/17 ft/sec.
To find the rate at which the woman's shadow lengthens, we can use similar triangles and related rates. Let's call the length of the shadow x.
We have two similar triangles here: the triangle formed by the woman, her shadow, and the pole, and the triangle formed by the pole, its shadow, and the ground.
The height of the 17 ft pole corresponds to the 6 ft height of the woman, and the length of the shadow corresponds to the distance between the woman and the base of the pole.
Let h be the distance between the woman and the pole (which is also the distance between the pole and the base of the pole).
By similar triangles, we have the following equation:
17 ft / 6 ft = (h + x) / x
To find the rate at which the woman's shadow lengthens, we need to differentiate this equation implicitly with respect to time t:
d(17 ft / 6 ft) / dt = d((h + x) / x) / dt
Now, let's differentiate each term:
0 = (6 ft * dx/dt - x * 6 ft * dh/dt) / x^2
Simplifying, we have:
0 = 6 ft * dx/dt - 6 ft * dh/dt
Rearranging, we get:
6 ft * dh/dt = 6 ft * dx/dt
Now, let's plug in the given values: dh/dt = 6 ft/sec, h = 30 ft
6 ft * (6 ft/sec) = 6 ft * dx/dt
36 ft/sec = dx/dt
Therefore, the shadow is lengthening at a rate of 36 ft/sec when the woman is 30 ft from the base of the pole.
I hope you made a diagram.
let her distance from the pole be x ft
let the length of her shadow be y ft
by similar triangles,
6/y = 17/(x+y)
simplifying,
11y = 6x
11 dy/dt = 6 dx/dt
but dx/dt = 6
so dy/dt = 6/11 ft/sec
notice the 30 ft is irrelevant.
Also be careful to notice the wording.
You asked, "how fast is her shadow lengthening" and that answer is 6/11
Had you asked, " how fast is her shadow moving", we would have had to add her own velocity to it, namely 6 + 6/11 ft/sec
There is 1 error in the above answer.
11dy/dt = 6 dx/dt is true
dx/dt = 6 is also true
but dy/dt is not 6/11 ft/sec
the 6 must be multiplied by the 6 from dx/dt giving dy/dt = 36/11