Two poles are 24 metres and 30 metres high and 20 metres apart. A slack
wire joins the tops of the poles and is 32 metres long. A cable car is moving
along the wire at 5 metres per second away from the shorter pole. When the
car is 12 metres horizontally from the shorter pole, it is 15 metres high and the
length of the wire to the shorter pole is 15 metres. At this instant, find:
a) how fast the car is moving horizontally.
b) how fast the car is moving vertically.
To solve this problem, we can use the concepts of similar triangles and the Pythagorean theorem.
Let's denote the distance of the car from the shorter pole as x at time t.
Step 1: Establishing the relationship between x and t
We are given that the car is moving away from the shorter pole at a rate of 5 meters per second. This means that the horizontal distance, x, is increasing at a rate of 5 meters per second.
Therefore, dx/dt = 5 (equation 1).
Step 2: Establishing the relationship between the heights of the car and the shorter pole
According to the problem, when the car is 12 meters horizontally from the shorter pole, it is 15 meters high. Let's denote the height of the car as y at time t.
Therefore, y = 15 (equation 2).
Step 3: Applying the Pythagorean theorem to establish a relationship between x, y, and the lengths of the wire to the poles
We know that the sum of the squares of the sides of a right-angled triangle equals the square of the hypotenuse. Let's consider the right-angled triangle formed by the car, the shorter pole, and the cable wire.
Using the Pythagorean theorem, we can write the following equation:
x^2 + y^2 = (32 - 15)^2
x^2 + y^2 = 289 (equation 3)
Step 4: Differentiating equation 3 implicitly with respect to time
Differentiating equation 3 implicitly with respect to time will help us find the relationship between dx/dt and dy/dt.
Differentiating equation 3 implicitly with respect to t, we get:
2x(dx/dt) + 2y(dy/dt) = 0
Simplifying this equation, we get:
x(dx/dt) + y(dy/dt) = 0 (equation 4).
Step 5: Substituting the given values into equations 1, 2, and 4 to find the required rates
Now, we can substitute the given values into the equations established in steps 1, 2, and 4 to find the desired rates.
From equation 1, dx/dt = 5.
From equation 2, y = 15.
Substituting these values into equation 4, we get:
x(5) + 15(dy/dt) = 0
Rearranging this equation, we have:
15(dy/dt) = -5x
dy/dt = -x/3 (equation 5).
Therefore, we have found the relationship between the vertical rate of change of the car's height and its horizontal distance from the shorter pole.
Step 6: Evaluating the rates at the given instant
At the given instant, when the car is 12 meters horizontally from the shorter pole, we can substitute this value into equation 5.
dy/dt = -x/3
dy/dt = -12/3
dy/dt = -4 meters per second.
Therefore, at this instant:
a) The car is moving horizontally at a speed of 5 meters per second (as given in equation 1).
b) The car is moving vertically at a speed of -4 meters per second (as obtained from equation 5).
To solve this problem, we can use the concepts of similar triangles and the Pythagorean theorem.
Let's label the shorter pole as pole A and the taller pole as pole B. We can create two right triangles - one with legs of heights 15m and 12m (triangle A) and another with legs of heights 24m and 30m (triangle B).
a) To find how fast the car is moving horizontally (dx/dt), we can use the similar triangles. The ratio of the heights in triangle A to triangle B should be the same as the ratio of the horizontal distances. Let's set up the proportion:
(height of triangle A) / (height of triangle B) = (distance from pole A to car) / (distance from pole B to car)
Using the given values: 15m / 24m = 12m / (20m - x), where x is the distance the car has moved horizontally from pole A.
To solve for x, we can cross multiply and rearrange the equation:
15m * (20m - x) = 24m * 12m
300m - 15m * x = 288m
15m * x = 300m - 288m
15m * x = 12m
x = 12m / 15m
x = 4/5
x = 0.8m
Therefore, the car is moving horizontally at a rate of 0.8 meters per second.
b) To find how fast the car is moving vertically (dy/dt), we can use the Pythagorean theorem to find the length of the wire from pole B to the car in terms of x.
Using the given values, the length of the wire from pole A to the car is 15m, and the length of the wire from pole B to the car is 32m. The length of the wire from pole A to pole B is 20m.
Using the Pythagorean theorem, we have:
(length of wire from A to car)^2 + (length of wire from B to car)^2 = (length of wire from A to B)^2
15m^2 + 32m^2 = 20m^2
225m^2 + 1024m^2 = 400m^2
1249m^2 = 400m^2
849m^2 = 0
This equation does not have any real solutions, which means there is an error in the problem statement. Please double-check the given values and try again.