a man 2 meters tall walks at a rate of 1.2 meters per second away from a light that is 5 meters above the ground.
(a) when he is 3m from the base of the light, at what rate is the tip of his shadow moving?
(b) when he is 3m from the base of the light, at what rate is the length of his shadow changing?
If the shadow has length s, then when the man is x meters from the spot below the light,
s/2 = (x+s)/5
s = 2/3 x
(b) ds/dt = 2/3 dx/dt = 2/3 * 6/5 = 4/5 m/s
(a) ds/dt + dx/dt = 4/5 + 6/5 = 2 m/s
Ah, shadows and walking tall! Let's tackle these questions with a touch of humor, shall we?
(a) When the man is 3m from the base of the light, the rate at which the tip of his shadow is moving can be calculated using similar triangles. Now, I could go into a long explanation about proportions and ratios, but why not clown around a bit instead? Picture this: the shadow is like a scared little puppy chasing its owner. As the man walks away from the light, the shadow gets longer and the tip frantically tries to catch up. So, the rate at which the tip of his shadow is moving will be 1.2 times the rate at which the man is walking. In other words, it will be 1.2 meters per second. Woof!
(b) Now, let's figure out how fast the length of the shadow is changing when the man is 3m from the base of the light. This time, we need to channel our inner circus acrobat. Imagine the shadow as a flexible rope, and the man is holding one end while walking away. As the man gets farther from the light, he pulls the rope, and the length of the shadow changes. The rate at which the length of his shadow is changing can be calculated using the concept of similar triangles. It turns out that it will also be 1.2 meters per second. So, the length of his shadow will increase by 1.2 meters per second. Quite the shadow show, isn't it?
Remember, these calculations rely on the assumption that the man and the light source are in a flat, open space, with no clowns or complicated obstacles around. Enjoy your mathematical adventures!
To solve this problem, we can use similar triangles and related rates. Let's solve each part of the problem step-by-step:
(a) To find the rate at which the tip of the man's shadow is moving when he is 3m from the base of the light, we need to find the rate at which the distance from the tip of the shadow to the man changes with respect to time.
Let's denote the height of the man as h = 2m, the height of the light as H = 5m, and the distance from the base of the light to the man as x = 3m. We need to find dx/dt, the rate at which x changes with respect to time.
Using similar triangles, we can set up the following proportion:
h/H = x/(x + d)
where d represents the distance from the tip of the shadow to the man, and dx/dt is the rate at which x is changing.
Rearranging this proportion, we get:
hx + hd = Hx
hd = Hx - hx
hd = (H-h)x
d = (H-h)x / h
Now, we can differentiate both sides of the equation with respect to time:
d/dt [d] = d/dt [(H-h)x / h]
ddt = (H-h)/h * dx/dt
Finally, substituting the given values, we have:
ddt = (5-2)/2 * dx/dt
ddt = 3/2 * dx/dt
ddt = 1.5 * dx/dt
Therefore, when the man is 3m from the base of the light, the tip of his shadow is moving at 1.5 times the rate at which the man is moving away from the light.
(b) To find the rate at which the length of the man's shadow is changing when he is 3m from the base of the light, we need to find dL/dt, the rate at which the length of the shadow changes with respect to time.
Using the same similar triangles, we can set up the following proportion:
h/H = L/(L + d)
where L represents the length of the shadow, and d represents the distance from the tip of the shadow to the man.
Rearranging this proportion, we get:
hL + hd = HL
hd = HL - hL
hd = (H-h)L
d = (H-h)L / h
Differentiating both sides of the equation with respect to time, we have:
d/dt [d] = d/dt [(H-h)L / h]
ddt = (H-h)/h * dL/dt
Substituting the given values, we have:
ddt = (5-2)/2 * dL/dt
ddt = 3/2 * dL/dt
ddt = 1.5 * dL/dt
Therefore, when the man is 3m from the base of the light, the length of his shadow is changing at 1.5 times the rate at which the man is moving away from the light.
To find out the rate at which the tip of the man's shadow is moving and the rate at which the length of his shadow is changing, we can use similar triangles and related rates.
Let's consider a right triangle formed by the man, the light, and the tip of his shadow. The height of the triangle is the man's height (2 meters), and the base of the triangle is the distance between the man and the light (3 meters). The length of the shadow is the hypotenuse of this triangle.
(a) To find the rate at which the tip of the man's shadow is moving, we need to find the rate at which the hypotenuse is changing when the base is changing.
Let's call the rate at which the base is changing "x" (in meters per second). We want to find the rate at which the hypotenuse is changing, which we can call "y" (in meters per second).
Using the Pythagorean theorem, we have:
(2)^2 + (3)^2 = (length of shadow)^2
4 + 9 = (length of shadow)^2
13 = (length of shadow)^2
Now, we differentiate both sides of the equation with respect to time t:
d/dt(13) = d/dt((length of shadow)^2)
0 = 2(length of shadow)(d/dt(length of shadow))
Since we know that the length of the shadow is changing with the rate x (which is given as 1.2 meters per second), we can substitute this value into the equation:
0 = 2(length of shadow)(1.2)
Solving for the rate at which the tip of the man's shadow is moving (y), we have:
y = 0 / (2(length of shadow))
y = 0
Therefore, when he is 3m from the base of the light, the tip of his shadow is not moving. It is stationary.
(b) To find the rate at which the length of his shadow is changing, we differentiate both sides of the Pythagorean theorem equation with respect to time t:
d/dt(2)^2 + d/dt(3)^2 = d/dt((length of shadow)^2)
0 + 0 = 2(length of shadow)(d/dt(length of shadow))
Again, substituting the given rate x (which is 1.2 meters per second) into the equation, we have:
0 = 2(length of shadow)(1.2)
Solving for the rate at which the length of his shadow is changing, we have:
d/dt(length of shadow) = 0 / (2(length of shadow))
d/dt(length of shadow) = 0
Therefore, when he is 3m from the base of the light, the length of his shadow is not changing. It remains constant.