A road perpendicular to a highway leads to a farmhouse located 6 mile away. An automobile traveling on the highway passes through this intersection at a speed of 75mph.
How fast is the distance between the automobile and the farmhouse increasing when the automobile is 10
miles past the intersection of the highway and the road?
To solve this problem, we can use the concept of related rates. We are given that the automobile is traveling at a speed of 75 mph on the highway and it is approaching the farmhouse. Let's define the following variables:
- Let "x" be the distance between the automobile and the intersection of the highway and road.
- Let "y" be the distance between the automobile and the farmhouse.
We want to find dy/dt, the rate at which the distance between the automobile and the farmhouse is changing when x is equal to 10 miles.
From the problem, we know that x is decreasing at a rate of 75 mph because the automobile is moving away from the intersection.
First, let's draw a diagram to visualize the situation:
A ---------------- B ------------- C
|<---------- 6 miles ----------->|
- A is the farmhouse
- B is the intersection of the highway and the road
- C is the automobile
At any specific time, let's say when the automobile is at point C, we can form a right-angled triangle ABC:
|\
| \
y | \ x
| \
|____\
Using the Pythagorean theorem, we have the equation:
x^2 + y^2 = (6)^2
Differentiating both sides of the equation with respect to time (t):
2x(dx/dt) + 2y(dy/dt) = 0
Since x = 10 miles and dx/dt = -75 mph (as x is decreasing), we can substitute these values into the equation:
2(10)(-75) + 2y(dy/dt) = 0
Simplifying the equation:
-1500 + 2y(dy/dt) = 0
2y(dy/dt) = 1500
dy/dt = 1500 / (2y)
To find dy/dt when x = 10 miles, we need to find the corresponding value of y. Let's use the Pythagorean theorem again:
(10)^2 + y^2 = (6)^2
100 + y^2 = 36
y^2 = 36 - 100
y^2 = -64
Here, we encounter an issue as y^2 should not be negative. This means that for x = 10 miles, the automobile is actually beyond the farmhouse, beyond point A. Therefore, the distance between the automobile and the farmhouse is not increasing as it has already passed the farmhouse.
In conclusion, when the automobile is 10 miles past the intersection of the highway and the road, the distance between the automobile and the farmhouse is not increasing.
To solve this problem, we can use the concept of related rates. We can set up a right triangle where one leg represents the distance between the automobile and the farmhouse (which we'll call x), and the other leg represents the distance traveled by the automobile along the highway (which we'll call y).
Let's label the triangle as follows:
A
|\
| \
x | \ y
| \
|____\
B
We know that the distance between the automobile and the farmhouse (x) is fixed at 6 miles, and the automobile is traveling on the highway at a constant speed of 75 mph. We need to find how fast x is changing when y is 10 miles.
First, we need to find a relationship between x and y. By applying the Pythagorean theorem, we have:
x^2 + y^2 = h^2
Where h is the hypotenuse of the triangle, which is the distance traveled by the automobile along the highway. Since the automobile is traveling at a constant speed of 75 mph, we can express h as a function of time (t):
h = 75t
Now, we can differentiate both sides of the equation with respect to time (t):
d/dt (x^2 + y^2) = d/dt (h^2)
2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt)
Since we are interested in finding dx/dt (the rate at which the distance between the automobile and the farmhouse is changing), we can substitute the given values:
2(6)(dx/dt) + 2y(dy/dt) = 2(75t)(75)
12(dx/dt) + 2y(dy/dt) = 150t
Now, we need to find dy/dt (the rate at which the distance traveled by the automobile along the highway is changing) when y = 10. We can use the speed of the automobile, which is given as 75 mph:
dy/dt = 75
Plugging this back into the equation:
12(dx/dt) + 2(10)(75) = 150t
12(dx/dt) + 1500 = 150t
Since we want to find dx/dt when y = 10, we can substitute y = 10 and solve for t:
10^2 + y^2 = (75t)^2
100 + 10^2 = (75t)^2
100 + 100 = 5625t^2
5625t^2 = 200
t^2 = 200/5625
t^2 = 0.0356
t ≈ 0.19 (approximately)
Now, we can substitute this value of t back into the equation:
12(dx/dt) + 1500 = 150(0.19)
12(dx/dt) + 1500 = 28.5
12(dx/dt) = 28.5 - 1500
12(dx/dt) = -1471.5
dx/dt = -1471.5/12
dx/dt ≈ -122.6 mph
Therefore, the distance between the automobile and the farmhouse is decreasing at a rate of approximately 122.6 mph when the automobile is 10 miles past the intersection of the highway and the road.
if the car has gone x miles, the distance z to the farmhouse is
z^2 = x^2 + 36
so, find z(10) and then
z dz/dt = x dx/dt
and plug in your numbers.
h = highway distance from intersection
d = auto-to-farmhouse distance
by Pythagoras ... d^2 = h^2 + 6^2
differentiating implicitly
... 2 d dd/dt = 2 h dh/dt
dd/dt = h / d * dh/dt