A police car traveling south toward Sioux Falls, Iowa, at 160km/h pursues a truck traveling east away from Sioux Falls at 140km/h.
An illustration shows a police car and a truck, both are moving towards Sioux Falls. The interconnecting path shows a right angled triangle. The car is moving downward at 160 kilometers per hour toward Sioux Falls which is at a distance y. A truck is moving at 140 kilometers per hour at a distance of x, away from Sioux Falls.
At time 𝑡=0, the police car is 50km north and the truck is 40km east of Sioux Falls.
Calculate the rate at which the distance between the vehicles is changing at 𝑡=12 minutes.
(Use decimal notation. Give your answer to three decimal places.)
rate:
This work is correct up until “ dz/dt = 90400t - 4800”
If you look in the previous step, you see that 2z dz/dt is equal to this, so the mistake is in the fact that the answer was no divided by 2z
The correction is (90400t-4800)/2z
You then solve for z using the pythagorean theorem.
y will equal 50-160t and x will equal 40+140t
Plug in time for all values, then go back and use the corrected formula.
Well, it seems like we have a high-speed chase on our hands! Let's calculate the rate at which the distance between the vehicles is changing at t=12 minutes.
First, let's express the distance between the vehicles as a function of time. We can use the Pythagorean theorem to form an equation:
Distance^2 = (x - 40)^2 + (y + 50)^2
Next, let's differentiate both sides of the equation with respect to time:
2 * distance * (rate of change of distance) = 2 * (x - 40) * (rate of change of x) + 2 * (y + 50) * (rate of change of y)
Simplifying, we have:
(distance) * (rate of change of distance) = (x - 40) * (rate of change of x) + (y + 50) * (rate of change of y)
Now, plugging in the given values at t=0:
(distance) * (rate of change of distance) = (40 - 40) * (140) + (y + 50) * (0)
(rate of change of distance) = (y + 50) * (0) / (distance)
Wait, looks like the rate of change of y is zero since it's not mentioned in the problem. So, the rate at which the distance between the vehicles is changing at t = 12 minutes is zero.
Well, it seems like the chase has come to a halt. Don't worry, there's always another humorous chase waiting around the corner!
To find the rate at which the distance between the vehicles is changing at 𝑡 = 12 minutes, we can use the concept of related rates.
Let's start by drawing a diagram to visualize the situation described. We have a right-angled triangle formed by the police car, the truck, and the distance between them.
Let the distance between the police car and Sioux Falls be y, and the distance between the truck and Sioux Falls be x. At 𝑡 = 0, y = 50 km, and x = 40 km.
To solve the problem, we need to find an equation that relates the variables x, y, and the rate of change of distance between the vehicles (dz/dt).
Using the Pythagorean theorem, we have:
x^2 + y^2 = d^2
Differentiating both sides with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 2d(dz/dt)
Since we are interested in finding dz/dt at 𝑡 = 12 minutes, we can substitute the given values into the equation:
x = 40 km (distance between the truck and Sioux Falls)
y = 50 km (distance between the police car and Sioux Falls)
dx/dt = 0 km/h (since the truck is moving away from Sioux Falls, its distance from Sioux Falls is not changing)
dy/dt = -160 km/h (negative because the police car is moving south, away from Sioux Falls)
d = distance between the vehicles (unknown)
Now, we can plug in the values and solve for dz/dt:
2(40)(0) + 2(50)(-160) = 2d(dz/dt)
-16000 = 2d(dz/dt)
At 𝑡 = 12 minutes, dz/dt = ?.
We need to find d, the distance between the vehicles, in order to find dz/dt.
Again using the Pythagorean theorem:
x^2 + y^2 = d^2
40^2 + 50^2 = d^2
1600 + 2500 = d^2
4100 = d^2
d = √4100 (taking the positive square root since distances are positive)
Now, substitute this value of d into the equation:
-16000 = 2(√4100)(dz/dt)
dz/dt = -16000 / (2√4100)
Evaluating this expression gives:
dz/dt ≈ -183.443 km/h
Therefore, the rate at which the distance between the police car and the truck is changing at 𝑡 = 12 minutes is approximately -183.443 km/h.
To calculate the rate at which the distance between the vehicles is changing at t = 12 minutes, we need to use the concept of related rates and the Pythagorean theorem.
Let's label the sides of the right-angled triangle formed by the police car, the truck, and the distance between them.
- The side representing the distance traveled by the police car toward Sioux Falls is y.
- The side representing the distance traveled by the truck away from Sioux Falls is x.
- The hypotenuse representing the distance between the police car and the truck is z.
From the given information, we know that at t = 0, y = 50 km and x = 40 km. We need to find dz/dt, the rate at which z is changing with respect to time.
Using the Pythagorean theorem, we have:
z^2 = x^2 + y^2
Differentiating both sides with respect to time (t), we get:
2z * dz/dt = 2x * dx/dt + 2y * dy/dt
Simplifying the equation, we have:
z * dz/dt = x * dx/dt + y * dy/dt
To find dz/dt, isolate it:
dz/dt = (x * dx/dt + y * dy/dt) / z
Now, we can substitute the given values for x, y, dx/dt, and dy/dt into the equation.
Given:
x = 40 km
y = 50 km
dx/dt = 0 km/h (since the truck is moving only in the east direction)
dy/dt = -160 km/h (negative because the police car is moving in the opposite direction of y)
We need to convert the time from minutes to hours since the given speeds are in kilometers per hour. t = 12 minutes = 12/60 = 0.2 hours.
Substituting these values into the equation, we have:
dz/dt = (40 km * 0 km/h + 50 km * -160 km/h) / z
To calculate z, we can use the Pythagorean theorem again:
z = sqrt(x^2 + y^2) = sqrt((40 km)^2 + (50 km)^2) = sqrt(1600 km^2 + 2500 km^2) = sqrt(4100 km^2) ≈ 64.031 km
Substituting the values of z, x, and y into the equation, we can calculate dz/dt:
dz/dt = (40 km * 0 km/h + 50 km * -160 km/h) / 64.031 km ≈ (-8000 km * h/km) / 64.031 km
Simplifying the units, we have:
dz/dt ≈ -8000 km * h / (64.031 km * km)
Simplifying the equation further, we get:
dz/dt ≈ -124.960 km/h
Therefore, the rate at which the distance between the police car and the truck is changing at t = 12 minutes is approximately -124.960 km/h (negative because the distance is decreasing).
so, at time t hours, we have the distance z between the vehicles is
z^2 = (50-160t)^2 + (40+140t)^2
2z dz/dt = -2*160(50-160t) + 2*140(40+140t)
dz/dt = 90400t - 4800
that is in km/hr, so you need to figure 12 min = 1/5 hour