a road that runs perpendicular to a highway leads to a farmhouse located one mile off the highway an automobile travels down the highway past the road leading to the farmhouse at a speed of 60 mph how fast is the distance between the farmhouse an automobile increasing when the car is 3 miles past the intersection of the highway and the road to the farmhouse

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At a time of t hours, let the distance of the car past the intersection be x miles.
let the distance between the car and the farmhouse be d
I get a simple right-angled triangle with
sides 1, x and hypotenuse d

d^2 = 1^2 + x^2
2d dd/dt = 2x dx/dt
or dd/dt = x dx/dt /d

when x = 3
d^2 = 1 + 9 = 10
d =√10 , and we are also given that dx/dt = 60

dd/dt = 3(60)/√10 = 180/√10 = appr 56.9 mph

if the car is x miles past the road, the distance is

s^2 = 1+x^2
when x=3, s=√10

2s ds/dt = 2x dx/dt
ds/dt = x/s dx/dt
when x=3, then,

ds/dt = 3/√10 * 60 = 18√10 = 56.9 mph

To find the rate at which the distance between the farmhouse and the automobile is changing, we can use the Pythagorean theorem. Let's define the situation as follows:

Let x be the distance from the farmhouse to the intersection of the highway and the road.
Let y be the distance the automobile has traveled past the intersection.

According to the Pythagorean theorem, the distance between the farmhouse and the automobile at any given time is given by:

d = sqrt(x^2 + y^2)

Taking the derivative of both sides with respect to time (t), we get:

dd/dt = (1/2) * (1/sqrt(x^2 + y^2)) * (2x(dx/dt) + 2y(dy/dt))

Since dx/dt is always 0 (as the farmhouse is one mile off the highway and doesn't move), the equation simplifies to:

dd/dt = (x(dx/dt) + y(dy/dt)) / sqrt(x^2 + y^2)

Given that the automobile travels down the highway at a speed of 60 mph (which is equivalent to 60 miles/hr), dy/dt = 60 mph.

We need to find dx/dt at the given point when the car is 3 miles past the intersection. Using the Pythagorean theorem again:

x^2 + y^2 = (1^2 + 3^2) = 10

Hence, x = sqrt(10).

Substituting the values into the derivative equation:

dd/dt = (sqrt(10) * dx/dt + 3 * 60) / sqrt(10)

Since dx/dt = 0 (the farmhouse does not move), the equation further simplifies to:

dd/dt = (3 * 60) / sqrt(10)

Calculating the value, we get:

dd/dt ≈ 180 / 3.16

dd/dt ≈ 57.02 mph

Therefore, when the car is 3 miles past the intersection, the distance between the farmhouse and the automobile is increasing at a rate of approximately 57.02 mph.

To find the rate at which the distance between the farmhouse and the automobile is increasing, we need to use the concept of related rates.

Let's break down the problem and assign variables:
- Let x be the distance the car has traveled down the highway from the intersection.
- Let y be the distance between the farmhouse and the car.

From the problem, we know the following:
- The car is traveling at a speed of 60 mph, so dx/dt (the rate at which x is changing) is 60 mph.
- The distance between the farmhouse and the highway is constant at 1 mile, so the rate at which y is changing (dy/dt) is not given but what we are trying to find.
- We're given that the car is 3 miles past the intersection, so x = 3 miles.

Now, we can form a right-angled triangle with sides x and y, where the hypotenuse represents the distance the car has traveled from the intersection to the farmhouse.

Using the Pythagorean Theorem, we have:
x^2 + y^2 = (distance from intersection to farmhouse)^2
x^2 + y^2 = 1^2
x^2 + y^2 = 1

Differentiating both sides with respect to t (time):
2x(dx/dt) + 2y(dy/dt) = 0

Plugging in the given values:
2(3)(60) + 2y(dy/dt) = 0
360 + 2y(dy/dt) = 0
2y(dy/dt) = -360

Now, we can solve for dy/dt (the rate at which y is changing):
dy/dt = -360 / (2y)
dy/dt = -180 / y

To find the rate at which the distance between the farmhouse and the automobile is increasing when the car is 3 miles past the intersection, we need to substitute x = 3 into the equation:
dy/dt = -180 / y
dy/dt = -180 / sqrt(1^2 - 3^2)
dy/dt = -180 / sqrt(1 - 9)
dy/dt = -180 / sqrt(-8)

Therefore, the distance between the farmhouse and the automobile is increasing at a rate of -180 / sqrt(-8) miles per hour when the car is 3 miles past the intersection. Note that the negative sign indicates that the distance is decreasing. However, the square root of a negative number is undefined in the real number system, so there seems to be an issue with the problem setup.