A boat is pulled into a wharf by a rope at a speed of 20m/min. If the rope is attached to a point on the boat 7m vertically below the wharf, at what rate is the rope being drawn in when the boat is 24m from the wharf?
When the boat is x meters from the wharf, the length r of rope is
r^2 = 7^2 + x^2
r dr/dt = x dx/dt
Now plug in your numbers and solve for dr/dt
I find this a bit odd, since I'd think that dr/dt would be constant. Instead, somehow the boat's speed is being kept constant.
Oh, well.
To find the rate at which the rope is being drawn in when the boat is 24m from the wharf, we can use the concept of related rates.
Let's call the distance between the boat and the wharf "x" and the rate at which the boat is moving towards the wharf "dx/dt". We need to find the rate at which the rope is being drawn in, which can be represented as "dy/dt".
We are given that the boat is pulled into the wharf at a speed of 20m/min. This means that dx/dt = 20m/min.
We also know that the rope is attached to a point on the boat 7m vertically below the wharf. This forms a right-angled triangle between the boat, the wharf, and the point where the rope is attached.
Using Pythagoras' theorem, we can find the relationship between x and y (the length of the rope):
x^2 + y^2 = (7m)^2
Differentiating this equation with respect to time t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
We can substitute the given values to find the rate at which the rope is being drawn in:
(2 * 24m * 20m/min) + (2 * y * dy/dt) = 0
48m/min + 2y(dy/dt) = 0
2y(dy/dt) = -48m/min
dy/dt = -48m/min / 2y
To find y, we can use the Pythagorean theorem:
x^2 + y^2 = (7m)^2
24m^2 + y^2 = 49m^2
y^2 = 49m^2 - 24m^2
y^2 = 25m^2
y = 5m
Substituting y = 5m into the equation for dy/dt:
dy/dt = -48m/min / (2 * 5m)
dy/dt = -48m/min / 10m
dy/dt = -4.8m/min
Therefore, the rope is being drawn in at a rate of -4.8m/min when the boat is 24m from the wharf. The negative sign indicates that the rope is being pulled downwards.