Question

At noon, ship A is 170 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

Answer 1

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 \( t = 4 \):
1. Calculate \( 28900 - 11900 \cdot 4 + 1625 \cdot 16 \):
\[
28900 - 47600 + 26000 = 7300
\]
2. Thus:
\[
d(4) = \sqrt{7300}
\]

3. Now calculate \( -11900 + 2 \cdot 1625 \cdot 4 \):
\[
-11900 + 13000 = 1100
\]

4. 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.