A girl starts at a point A and runs east at the rate of 10 ft/sec. One
minute later, another girl starts at A and runs north at a rate of 8 ft/sec.
At what rate is the distance between them changing 1 minute after the
second girl starts ?
the distance z between the girls at time t seconds (t > 60) is
z^2 = (10t)^2 + (8(t-60))^2
2z dz/dt = 2(10t) + 64*2(t-60) = 148t-7680 = 4(37t-1920)
so now just figure z(120) and use that to find dz/dt
480√29 dz/dt = 10080
dz/dt = 21/√29 = 3.9 ft/s
At a time of t seconds,
let the distance covered by the east-bound girl be x ft
let the distance covered by the north-bounder by y ft
let the distance between them be d ft
given: dx/dt = 10 ft/s, dy/dt = 8 ft/s
find dd/dt when t = 60
d^2 = x^2 + y^2
2d dd/dt = 2x dx/dt + 2y dy/dt
dd/dt = (x(10) + y(8) )/d
when t = 60, x = 600, y = 480, d = 120√41
dd/dt = ( 600(10) + 480(8) )/(120√41) = ...... ft/s
Well, if two girls are running away from you, the distance between them is definitely increasing! But let's find out how fast it's increasing.
We can use the Pythagorean theorem to find the relationship between the distance and the rates at which the girls are running. Let's call the horizontal distance x and the vertical distance y. Therefore, according to the Pythagorean theorem, x² + y² = distance².
Now, since the first girl is running east at a rate of 10 ft/sec, she would have covered a distance of x = 10 ft in 1 minute. Similarly, the second girl is running north at a rate of 8 ft/sec, so her distance covered in 1 minute would be y = 8 ft.
Let's differentiate both sides of the equation x² + y² = distance² with respect to time (t):
2x(dx/dt) + 2y(dy/dt) = 2distance(d(distance)/dt)
Now, plugging in the values we know:
2(10 ft)(10 ft/sec) + 2(8 ft)(8 ft/sec) = 2distance(d(distance)/dt)
Simplifying:
20 ft/sec * 10 ft + 16 ft/sec * 8 ft = 2distance(d(distance)/dt)
200 ft/sec + 128 ft/sec = 2distance(d(distance)/dt)
328 ft/sec = 2distance(d(distance)/dt)
Therefore, the rate at which the distance between the two girls is changing 1 minute after the second girl starts is 328 ft/sec. But just remember, this is relative to the Clown Bot's super speed!
To find the rate at which the distance between the two girls is changing, we can use the concept of the Pythagorean theorem.
Let's assume that after 1 minute, the first girl has traveled a distance of "d" feet east from point A. Since she is running at 10 ft/sec, her distance traveled after 1 minute is 10 ft/sec * (1 minute * 60 seconds) = 600 feet.
The second girl starts at point A and runs north at a rate of 8 ft/sec for 1 minute. Therefore, she has traveled a distance of 8 ft/sec * (1 minute * 60 seconds) = 480 feet.
Now, let's consider a right triangle formed by connecting the positions of the two girls and the distance between them. The sides of this triangle are:
1) The distance traveled by the first girl (d) in the east direction, which is 600 feet.
2) The distance traveled by the second girl in the north direction, which is 480 feet.
3) The distance between the two girls, which we'll call "x".
According to the Pythagorean theorem, the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. In this case, we have:
d^2 + x^2 = 480^2
To find the rate at which the distance between the two girls is changing, we need to differentiate this equation with respect to time. But first, let's solve the equation for x.
x^2 = 480^2 - d^2
x^2 = 230,400 - d^2
x = sqrt(230400 - d^2)
Now, let's differentiate both sides of the equation with respect to time.
d/dt(x) = d/dt(sqrt(230400 - d^2))
To differentiate the right side of the equation, we apply the chain rule.
d/dt(x) = (1/2)(230400 - d^2)^(-1/2)(-2d)(d/dt(d))
d/dt(x) = -d(230400 - d^2)^(-1/2)
Plugging in the known values, we have d = 600.
d/dt(x) = -600(230400 - (600)^2)^(-1/2)
Simplifying further,
d/dt(x) = -600(230400 - 360000)^(-1/2)
d/dt(x) = -600(-129600)^(-1/2)
Since we want to find the rate at which the distance between the two girls is changing 1 minute after the second girl starts, we need to evaluate this expression at t = 1 minute or 60 seconds.
d/dt(x) = -600(-129600)^(-1/2)
d/dt(x) = -600/(-12,9600)^(1/2)
d/dt(x) = 600/(12,9600)^(1/2)
d/dt(x) = 600/360
d/dt(x) = 5/3
Therefore, the rate at which the distance between the two girls is changing 1 minute after the second girl starts is 5/3 feet per second.
To find the rate at which the distance between the two girls is changing, we can use the concept of related rates.
Let's consider the point A as the origin of a coordinate system, with the first girl running east (x-direction) and the second girl running north (y-direction).
After the second girl starts running, the distance between them can be represented by the hypotenuse of a right-angled triangle. We can use the Pythagorean theorem to express this relationship:
d^2 = x^2 + y^2
To find the rate of change of distance (dd/dt), we'll differentiate both sides of the equation with respect to time (t):
2d * (dd/dt) = 2x * (dx/dt) + 2y * (dy/dt)
Let's find the values for x, dx/dt, y, and dy/dt.
The first girl starts running east at a rate of 10 ft/sec. Since she runs for one minute, she covers a distance of x = (10 ft/sec) * (60 sec) = 600 ft. So, x = 600 ft.
The second girl starts running north at a rate of 8 ft/sec. Since she also runs for one minute, she covers a distance of y = (8 ft/sec) * (60 sec) = 480 ft. So, y = 480 ft.
Differentiating both x and y with respect to time gives us dx/dt = 10 ft/sec and dy/dt = 8 ft/sec, as these rates remain constant.
Now, we can substitute the known values into the derivative equation and solve for dd/dt:
2 * d * (dd/dt) = 2 * x * (dx/dt) + 2 * y * (dy/dt)
2 * d * (dd/dt) = 2 * 600 ft * (10 ft/sec) + 2 * 480 ft * (8 ft/sec)
2 * d * (dd/dt) = 12,000 ft^2/sec + 7,680 ft^2/sec
2 * d * (dd/dt) = 19,680 ft^2/sec
Since we are interested in the rate of change of the distance after 1 minute, we can substitute d = sqrt(x^2 + y^2) = sqrt((600 ft)^2 + (480 ft)^2) into the equation:
2 * sqrt((600 ft)^2 + (480 ft)^2) * (dd/dt) = 19,680 ft^2/sec
Now, we can solve for dd/dt:
sqrt((600 ft)^2 + (480 ft)^2) * (dd/dt) = 9,840 ft^2/sec
dd/dt = 9,840 ft^2/sec / sqrt((600 ft)^2 + (480 ft)^2)
Calculating the value, we find:
dd/dt ≈ 12.19 ft/sec
Therefore, the distance between the two girls is changing at a rate of approximately 12.19 ft/sec, 1 minute after the second girl starts running.