two cars start at the same place and at the same time. One car trvels west at a constant speed of 50 miles per hour and a second car travels south at a constant speed of 60 miles per hour. Approcimately how fast is the distance between them changing one-half hour later

The direction of movement of both cars forms a right angle triangle. The distance travelled due south and west east by both cars represents the legs of the triangle. Their distance apart after t hours represents the hypotenuse of the right angle triangle.

Let x represent the length the shorter leg(west) of the right angle triangle.

Let y represent the length the longer leg(south) of the right angle triangle.

Let z represent the hypotenuse.

Applying Pythagoras theorem

Hypotenuse² = opposite side² + adjacent side²

Therefore

z² = x² + y²

To determine the rate at which the distances are changing, we would differentiate with respect to t. It becomes

2zdz/dt = 2xdx/dt + 2ydy/dt- - - -- - -1

One car travels south at 60 mph and the other car travels west at 50 mi/h. It means that

dx/dt = 50

dy/dt = 60

Distance = speed × time

Since t = 0.5 hour, then

x = 50 × 0.5 = 25 miles

y = 60 × 0.5 = 30 miles

z² = 25² + 30² = 625 + 900

z = √1525

z = 39.05 miles

Substituting these values into equation 1, it becomes

2 × 39.05 × dz/dt = 2 × 25 × 50 + 2 × 30 × 60

78.1dz/dt = 2500 + 3600

78.1dz/dt = 6100

dz/dt = 6100/78.1

dz/dt = 78.1 mph

Sam1s Sam1s · Answer 2 · 2021-12-01T13:03:02+01:00

Given Information :

The direction of movement of both cars forms a right angle triangle.
The distance traveled due south and west east by both cars represents the legs of the triangle.
Their distance apart after t hours represents the hypotenuse of the right angle triangle.

Let x represent the length of shorter leg(west) of the right angle triangle.
Let y represent the length of longer leg(south) of the right angle triangle.
Let z represent the hypotenuse.

Apply Pythagoras theorem :

Hypotenuse² = opposite side² + adjacent side²

z² = x² + y²

2zdz/dt = 2xdx/dt + 2ydy/dt - 1

One car travels south at 60 mph

The other car travels west at 50 mi/h.

dx/dt = 50

dy/dt = 60

Distance = Speed × Time

Since t = 0.5 hour, then

x = 50 × 0.5 = 25 miles

y = 60 × 0.5 = 30 miles

z² = 25² + 30² = 625 + 900

z = √1525

z = 39.05 miles

Substituting these values into equation 1, it becomes

2 × 39.05 × dz/dt = 2 × 25 × 50 + 2 × 30 × 60

78.1dz/dt = 2500 + 3600

78.1dz/dt = 6100

dz/dt = 6100/78.1

dz/dt = 78.1 mph

The distance between them changing one-half hour later is 78.1 mph.

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