You've dropped a lucky penny into the middle of a hemispherical water basin of radius 3m. There is a light on the upper edge of the basin. The penny falls vertically along the axis of symmetry of the basin, and when it is 1m from the bottom of the basin it is falling at 2m/s. How fast is the shadow of the penny moving along the surface of the basin at this instant?
To find the speed at which the shadow of the penny is moving along the surface of the basin, we can use related rates.
Let's denote the position of the penny along the vertical axis as y(t), where y is the distance from the bottom of the basin and t is time.
First, we can find the relationship between the position of the penny and its speed along the axis. We know that when the penny is 1m from the bottom (y = 1m), it is falling at a speed of 2m/s. So, we have:
dy/dt = 2 m/s
Next, we need to relate the position of the penny along the axis to its position on the surface of the basin.
We can use similar triangles to find this relationship. Let's denote the distance of the shadow from the vertical axis as x(t). Since the light source is on the upper edge of the basin, the triangle formed by the vertical axis, the surface of the basin, and the shadow of the penny is similar to the triangle formed by the vertical axis, the bottom of the basin, and the penny itself.
Using the ratio of corresponding sides, we have:
x(t) / y(t) = radius of the basin / height of the basin
Given that the radius of the basin is 3m, and the height of the basin is also 3m (since it is a hemisphere), we have:
x(t) / y(t) = 3 / 3
x(t) / y(t) = 1
Now, we need to differentiate both sides of the equation with respect to time t to find the relationship between the rates of change of x and y.
d/dt [x(t) / y(t)] = d/dt [1]
Using the quotient rule, we have:
(dx/dt * y(t) - x(t) * dy/dt) / [y(t)]^2 = 0
dx/dt * y(t) - x(t) * dy/dt = 0
Now, we can substitute in the values we know. At the given instant, when the penny is 1m from the bottom (y = 1m), it is falling at a speed of 2m/s (dy/dt = 2m/s).
Substituting these values, we have:
dx/dt * 1 - x(t) * 2 = 0
Simplifying the equation, we have:
dx/dt = 2x(t)
Now, we can find the value of x(t). When the penny is 1m from the bottom (y = 1m), we can calculate the corresponding x value using the similar triangles relationship:
x = y * (radius of the basin / height of the basin) = 1 * (3 / 3) = 1m
Now, we can plug in the value of x(t) = 1m into the derived equation:
dx/dt = 2 * 1m = 2 m/s
Therefore, the shadow of the penny is moving along the surface of the basin at a speed of 2 m/s at the instant when the penny is 1m from the bottom of the basin.