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?
Big right triangle, C at base of light pole, A at top, B at shadow tip
Then D at base of man
AC = 4
CD = 10 at start = y(t) = 10 -t
CB = x with velocity dx/dt
in general
x/4 = (x-y)/1.89 similar triangles
4x - 4y = 1.89 x
2.11 x = 4 y
2.11 (dx/dx) = 4(dy/dx)
2.11 (dx/dx*dx/dt) = 4 (dy/dx*dx/dt) chain rule
2.11 dx/dt = 4 dy/dt
dx/dt = (4/2.11)(-1m/s)
The second problem is similar.
To solve these problems, we can use similar triangles and the concept of rates. Let's break down each question step by step:
a) To find the velocity at which the young man's shadow decreases when he is 10m away from the lamppost, we can use the concept of similar triangles. The initial height of the lamppost is 4m, and the distance from the young man to the lamppost is continuously decreasing while he walks towards it.
We can set up a proportion using the similar triangles formed by the height of the lamppost, the height of the man, and the length of the shadow. Let's call the height of the young man's shadow h and the distance from the man to the lamppost x.
Initial height of the lamppost/Length of the shadow = Height of the man/Distance from the man to the lamppost
4/h = 1.89/x
Now, we can differentiate both sides of the equation with respect to time, keeping in mind that the height of the man is constant as he walks towards the lamppost. Therefore, the rate of change of the shadow's length with respect to time is:
0/dt = 0/dt + (-4/x^2) * dx/dt
But we are interested in finding dx/dt when x = 10m, so we rearrange the equation:
0 = (-4/x^2) * dx/dt
Simplifying further, we find:
dx/dt = 0
Therefore, the velocity at which the young man's shadow decreases when he is 10m away from the lamppost is 0 m/s.
b) To find the velocity at which the boat approaches the dock when it is 3m away, we can use a similar approach.
We can set up a proportion using the similar triangles formed by the height of the dock, the height of the girl's hands, and the distance of the boat from the dock. Let's call the height of the boat h and the distance of the boat from the dock x.
Height of the dock/Distance of the boat from the dock = Height of the girl's hands/Height of the boat
h/(3m) = (2.5m - h)/0.3m
Simplifying the equation, we get:
0.3h = 7.5m - 3h
3.3h = 7.5m
h = 7.5m / 3.3
h ≈ 2.27m
Now, let's differentiate both sides of the equation with respect to time, considering that the height of the boat is decreasing as it approaches the dock:
0/dt = (7.5 - 3h) * dx/dt + h * d(3m)/dt
0 = (7.5 - 3(2.27)) * dx/dt - 2.27 * 3 * dt/dt
0 = (7.5 - 6.81) * dx/dt - 6.81 * 1
0.19 * dx/dt = 6.81
dx/dt ≈ 6.81 m/s
Therefore, the velocity at which the boat approaches the dock when it is 3m away is approximately 6.81 m/s.