A car leaves an intersection traveling west. Its position 5 sec later is 22 ft from the intersection. At the same time, another car leaves the same intersection heading north so that its position 5 sec later is 27 ft from the intersection. If the speed of the cars at that instant of time is 15 ft/sec and 5 ft/sec, respectively, find the rate at which the distance between the two cars is changing. (Round your answer to one decimal place.)
ft/sec
let the distance of the west-bound car from the intersection be x
let the distance of the north-bound car from the intersection be y
and let the distance between them be d
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
d dd/dt = x dx/dt + y dy/dt
when x=22, y=27 , dx/dt = 15 ft/sec , dy/dt = 5 ft/sec
and d^2 = 22^2 + 27^2 = 1213
d = √1213
√1213 dd/dt = 22(15) + 27(5) = 465
dd/dt = 465/√1213 = appr 13.4 ft/s
Well, it seems like these two cars are playing a game of "cat and mouse", or should I say "car and car"? Anyway, let's figure out how fast the distance between them is changing.
The distance between the two cars can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse (in this case, the distance between the cars) is equal to the sum of the squares of the other two sides. In equation form, it's like this: c^2 = a^2 + b^2.
So, let's use this equation to find the distance between the two cars at that instant of time. The car traveling west is 22 ft from the intersection, and the car traveling north is 27 ft from the intersection. Therefore, the side lengths of the right triangle formed by the two cars are 22 ft and 27 ft.
Applying the Pythagorean theorem, we get:
c^2 = 22^2 + 27^2
c^2 = 484 + 729
c^2 = 1213
c ≈ 34.8 ft
Now that we know the distance between the two cars is approximately 34.8 ft, we need to find how fast this distance is changing. To do that, we can take the derivative of the distance equation with respect to time.
d(c^2)/dt = d(22^2 + 27^2)/dt
2c(dc/dt) = 0 + 0 (since both sides are constants)
2c(dc/dt) = 0
Now, we can solve for dc/dt, which represents the rate at which the distance between the cars is changing.
2c(dc/dt) = 0
dc/dt = 0 / 2c
dc/dt = 0 ft/sec
Oh, it seems like the distance between the two cars is not changing at all! Maybe they're enjoying each other's company and decided to maintain a constant distance. So, the rate at which the distance between the two cars is changing is 0 ft/sec.
Well, that's one way to avoid a car accident, I guess!
To find the rate at which the distance between the two cars is changing, we can use the Pythagorean theorem.
Let's denote the position of the car traveling west after 5 seconds as x and the position of the car traveling north after 5 seconds as y.
According to the problem, the car traveling west is 22 ft from the intersection, and the car traveling north is 27 ft from the intersection. Therefore, we have:
x^2 + y^2 = (22)^2 ---(1)
x^2 + y^2 = (27)^2 ---(2)
Differentiating both equations with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0 ---(3)
2x(dx/dt) + 2y(dy/dt) = 0 ---(4)
We know that the speed of the car traveling west at that instant is 15 ft/sec, and the speed of the car traveling north is 5 ft/sec. Therefore, we have:
dx/dt = 15 ---(5)
dy/dt = 5 ---(6)
Substituting equations (5) and (6) into equations (3) and (4), we get:
2x(15) + 2y(5) = 0
2x(15) + 2y(5) = 0
Rearranging these equations, we get:
30x + 10y = 0 ---(7)
30x + 10y = 0 ---(8)
Now, let's solve equations (7) and (8) simultaneously to find the values of x and y:
30x = -10y
x = (-10/30)y
x = (-1/3)y
Substitute this value of x into equation (1):
(-1/3)y^2 + y^2 = (22)^2
(-1/3)y^2 + y^2 = 484
(2/3)y^2 = 484
y^2 = (3/2) * 484
y^2 = 726
Taking the square root of both sides, we get:
y = √726
Now, substitute the value of y into equation (6) to find the value of x:
x = (-1/3)(√726)
Now, we can find the distance between the two cars, which is the hypotenuse of the right triangle formed by x and y:
Distance = √(x^2 + y^2)
Distance = √((-1/3)(√726)^2 + (√726)^2)
Distance = √((1/9)(726) + 726)
Distance = √(726/9 + 726)
Distance = √(726/9 + 726/1)
Distance = √((726 + 6564)/9)
Distance = √(7290/9)
Distance = √810
Finally, to find the rate at which the distance between the two cars is changing, we differentiate the equation for distance with respect to time (t):
d(distance)/dt = d/2√810 * (d/dt √810)
d(distance)/dt = 1/(2√810) * (d/dt √810)
To find d/dt √810, let's use the chain rule:
d(distance)/dt = 1/(2√810) * (1/2√810) * d/dt (810)
d(distance)/dt = 1/(2√810) * (1/2√810) * (d/dt (810))^2
d(distance)/dt = 1/(2√810) * (1/2√810) * 0
d(distance)/dt = 0
Therefore, the rate at which the distance between the two cars is changing is 0 ft/sec.
To find the rate at which the distance between the two cars is changing, we can use the Pythagorean theorem to find the distance between the two cars at any given time.
Let's assume that the position of the first car at time t is (x,0) and the position of the second car is (0,y). We know that the speed of the first car is 15 ft/sec and the speed of the second car is 5 ft/sec.
Using the distance formula, the distance between the two cars at any given time t can be calculated as follows:
Distance = √((x - 0)^2 + (y - 0)^2)
= √(x^2 + y^2)
We are given that the position of the first car 5 sec later is 22 ft from the intersection, which means at time t=5, x = -22.
Similarly, we are given that the position of the second car 5 sec later is 27 ft from the intersection, which means at time t=5, y = 27.
Plugging these values into the distance formula, we get:
Distance = √((-22)^2 + (27)^2)
= √(484 + 729)
= √(1213)
≈ 34.83 ft
Now, we need to find how fast this distance is changing with respect to time. To do this, we differentiate the distance formula with respect to time (t):
d(Distance)/dt = (1/2)*(1/√(x^2 + y^2))*2x*dx/dt + (1/2)*(1/√(x^2 + y^2))*2y*dy/dt
Where dx/dt and dy/dt are the rates at which x and y are changing with respect to time, respectively.
Since dx/dt = -15 ft/sec (negative because the car is moving west) and dy/dt = 5 ft/sec, and at t=5, x = -22 and y = 27, we can plug these values into the equation to find the rate at which the distance between the two cars is changing:
d(Distance)/dt = (1/2)*(1/√((-22)^2 + (27)^2))*2*(-22)*(-15) + (1/2)*(1/√((-22)^2 + (27)^2))*2*27*5
Calculating this expression, we get:
d(Distance)/dt ≈ -1327.4 + 498.6
≈ -828.8 ft/sec
Therefore, the rate at which the distance between the two cars is changing is approximately -828.8 ft/sec (negative because the distance is decreasing).