a) A young man measuring 1.89 m. walks at a velocity of 1m/s towards a lamppost which its light is at 4m from the ground. At what velocity does the young man's shadow decrease when he's 10 m away from the lamppost?
b) A young girl, standing at the edge of a dock, pulls a boat towards herself with a cable attached to the front of a boat at 0.3m from sea level. If the girl pulls the cable at a velocity of 1.6 m/s, her hands being at 2.5m from sea level, at what velocity does the boat approach the dock when it's 3m away?
a) To find the velocity at which the young man's shadow decreases when he's 10m away from the lamppost, we can use similar triangles.
Let's assume the height of the young man's shadow is h meters.
Using the similar triangles formed by the young man, his shadow, and the lamppost, we can set up the following equation:
h / (1.89 - 4) = (h - 4) / 10
Simplifying the equation, we get:
h / (-2.11) = (h - 4) / 10
10h = -2.11h + 8.44
12.11h = 8.44
h = 8.44 / 12.11
h ≈ 0.6959 m
Now, we can differentiate the equation with respect to time to find the rate of change of the shadow's height with respect to time (velocity of the shadow).
d(h) / dt = (0.6959 - 4) / 10 * (dh/dt)
Substituting the values we know, we have:
d(h) / dt = (-3.3041) / 10 * (dh/dt)
We are looking for the velocity of the shadow, so let's find dh/dt when the young man is 10m away from the lamppost.
When the young man is 10m away, the shadow's height is h = 0.6959m
Let's find dh/dt:
dh/dt = (0.6959 - 4) / 10
dh/dt = -3.3041 / 10
Now we can substitute this value back into our equation:
d(h) / dt = (-3.3041 / 10) * (dh/dt)
d(h) / dt = -0.33041 m/s
Therefore, the velocity at which the young man's shadow decreases when he's 10m away from the lamppost is approximately -0.33041 m/s.
b) To find the velocity at which the boat approaches the dock when it's 3m away, we can use a similar approach.
Let's assume the height of the boat is h meters.
Using the similar triangles formed by the girl, the boat, and the dock, we can set up the following equation:
h / (-2.2) = (h - 0.3) / 3
Simplifying the equation:
h / (-2.2) = (h - 0.3) / 3
3h = -2.2h + 0.9
5.2h = 0.9
h = 0.9 / 5.2
h ≈ 0.1731 m
Now, we can differentiate the equation with respect to time to find the rate of change of the boat's height with respect to time (velocity of the boat).
d(h) / dt = (0.1731 - 0.3) / 3 * (dh/dt)
Substituting the values we know, we have:
d(h) / dt = (-0.1269) / 3 * (dh/dt)
We are looking for the velocity of the boat, so let's find dh/dt when the boat is 3m away.
When the boat is 3m away, the boat's height is h = 0.1731m
Let's find dh/dt:
dh/dt = (0.1731 - 0.3) / 3
dh/dt = -0.1269 / 3
Now we can substitute this value back into our equation:
d(h) / dt = (-0.1269 / 3) * (dh/dt)
d(h) / dt = -0.0423 m/s
Therefore, the velocity at which the boat approaches the dock when it's 3m away is approximately -0.0423 m/s.
a) To determine the velocity at which the young man's shadow decreases when he is 10m away from the lamppost, we can use similar triangles.
Let's denote:
- x as the distance between the young man and the lamppost.
- h as the height of the young man's shadow.
- y as the distance between the young man's shadow and the lamppost.
We have the following similar triangles:
1) The triangle formed by the young man, his shadow, and the lamppost.
2) The triangle formed by the young man, the lamppost, and the height of the lamppost.
Using these triangles, we can derive the following equation:
(h + y) / x = h / 4
Simplifying the equation, we get:
h + y = (4h) / x
We can rearrange the equation to solve for y:
y = (4h) / x - h
Now, to find the velocity at which the young man's shadow decreases, we need to differentiate y with respect to x, and then substitute the appropriate values:
dy/dt = (d/dx)((4h) / x - h)
Differentiating, we get:
dy/dt = -4h / x^2
Now we can substitute the given values:
- h = 1.89m (the young man's height)
- x = 10m (the distance between the young man and the lamppost)
Substituting these values, we have:
dy/dt = -4(1.89) / (10)^2
dy/dt = -0.0756 m/s
Therefore, the velocity at which the young man's shadow decreases when he is 10m away from the lamppost is -0.0756 m/s.
b) To determine the velocity at which the boat approaches the dock when it is 3m away, we can again use similar triangles.
Let's denote:
- x as the distance between the boat and the dock.
- h as the height of the boat.
- y as the distance between the boat and the dock along the vertical axis.
Using similar triangles, we have the following equation:
(y + h) / x = (2.5 + h) / 3
Simplifying, we get:
(y + h) = (2.5 + h)(x / 3)
Again, we need to differentiate y with respect to x to find the velocity at which the boat approaches the dock:
dy/dt = (d/dx)((2.5 + h)(x / 3) - h)
Differentiating, we get:
dy/dt = (2.5 + h) / 3
Now we can substitute the given values:
- h = 0.3m (distance of the cable from sea level)
- x = 3m (distance between the boat and the dock)
Substituting these values, we have:
dy/dt = (2.5 + 0.3) / 3
dy/dt = 0.93 / 3
dy/dt = 0.31 m/s
Therefore, the velocity at which the boat approaches the dock when it is 3m away is 0.31 m/s.