Let's first set up the problem. At noon, ship A is at position \((-170, 0)\) and ship B is at \((0, 0)\) in a coordinate system where east is the positive x-direction and north is the positive y-direction.
At time \( t \), measured in hours from noon:
- The position of ship A, which is sailing east at 35 km/h, can be represented as: \[ A(t) = (-170 + 35t, 0) \]
- The position of ship B, which is sailing north at 20 km/h, can be represented as: \[ B(t) = (0, 20t) \]
The distance \( d(t) \) between the two ships at time \( t \) is given by the distance formula: \[ d(t) = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} \] Substituting the positions of ships A and B: \[ d(t) = \sqrt{(0 - (-170 + 35t))^2 + (20t - 0)^2} \] This simplifies to: \[ d(t) = \sqrt{(170 - 35t)^2 + (20t)^2} \]
Now, we need to expand the expression inside the square root: \[ d(t) = \sqrt{(170 - 35t)^2 + (20t)^2} = \sqrt{(170 - 35t)^2 + 400t^2} \]
Expanding \((170 - 35t)^2\): \[ (170 - 35t)^2 = 28900 - 11900t + 1225t^2 \]
Thus, we have: \[ d(t) = \sqrt{28900 - 11900t + 1225t^2 + 400t^2} = \sqrt{28900 - 11900t + 1625t^2} \]
Now we need to find the rate of change of distance \( \frac{dd}{dt} \). To do this, we will use implicit differentiation. First, differentiate \( d(t) \): \[ \frac{dd}{dt} = \frac{1}{2\sqrt{28900 - 11900t + 1625t^2}} \cdot ( -11900 + 2 \cdot 1625t) \]
Now let's find \( \frac{dd}{dt} \) at \( t = 4 \) (which is 4 hours after noon): \[ \frac{dd}{dt} = \frac{1}{2\sqrt{28900 - 11900 \cdot 4 + 1625 \cdot 16}} \cdot (-11900 + 2 \cdot 1625 \cdot 4) \]
Calculating \( t = 4 \):
-
Calculate \( 28900 - 11900 \cdot 4 + 1625 \cdot 16 \): \[ 28900 - 47600 + 26000 = 7300 \]
-
Thus: \[ d(4) = \sqrt{7300} \]
-
Now calculate \( -11900 + 2 \cdot 1625 \cdot 4 \): \[ -11900 + 13000 = 1100 \]
-
Therefore, we have: \[ \frac{dd}{dt} = \frac{1100}{2\sqrt{7300}} \] To compute \( 2\sqrt{7300} \): \[ 2\sqrt{7300} = 2 \cdot \sqrt{7300} \approx 2 \cdot 85.44 \approx 170.88 \] So: \[ \frac{dd}{dt} \approx \frac{1100}{170.88} \approx 6.44 \]
Thus, the rate at which the distance between the ships is changing at 4:00 PM is approximately \(\boxed{6.44}\) km/h.