A road perpendicular to a highway leads to a farmhouse located 5 miles away. An automobile traveling on the highway passes through this intersection at a speed of 65mph.
How fast is the distance between the automobile and the farmhouse increasing when the automobile is 1 miles past the intersection of the highway and the road?
The distance between the automobile and the farmhouse is increasing at a rate of _______ miles per hour
Let the distance on the road be x. Then the distance z to the house is
z^2 = x^2+25
z dz/dt = x dx/dt
z(1) = √26
√26 dz/dt = 1 * 65
...
To solve this problem, we can apply the concept of related rates.
Let's denote:
- x as the distance of the automobile from the intersection along the highway (in miles)
- y as the distance between the automobile and the farmhouse (in miles)
We are given that the automobile is traveling on the highway at a speed of 65 mph, which means dx/dt = 65 mph.
We need to find dy/dt, the rate at which the distance between the automobile and the farmhouse is increasing.
From the problem, we can see that there is a right triangle formed by the intersection, automobile, and the farmhouse. Let's apply the Pythagorean theorem to solve for y:
x^2 + y^2 = 5^2
x^2 + y^2 = 25
Differentiate both sides with respect to time (t):
2x(dx/dt) + 2y(dy/dt) = 0
Since we are trying to find dy/dt when the automobile is 1 mile past the intersection (x = 1), we can substitute the values into the equation:
2(1)(65) + 2y(dy/dt) = 0
Simplifying:
130 + 2y(dy/dt) = 0
Rearranging and solving for dy/dt:
2y(dy/dt) = -130
dy/dt = -130 / (2y)
dy/dt = -65 / y
The distance between the automobile and the farmhouse is increasing at a rate of -65/y miles per hour, where y is the distance between the automobile and the farmhouse.
To find the rate at which the distance between the automobile and the farmhouse is increasing, we can use the concept of related rates.
Let's denote the distance between the automobile and the farmhouse as D(t), where t represents time. We want to find dD/dt, the rate at which D is changing with respect to time.
Since we're given that the automobile is traveling on the highway at a constant speed of 65 mph, the rate of change of the distance between the automobile and the farmhouse can be found by considering the right triangle formed by the automobile, the farmhouse, and the intersection.
Let x(t) be the distance traveled by the automobile along the highway from the intersection at time t. Since the automobile is 1 mile past the intersection, we can say that x(t) = t + 1.
Using the Pythagorean theorem, we can establish a relationship between D(t), x(t), and the fixed distance between the farmhouse and the intersection:
D(t)^2 = (x(t))^2 + 5^2
D(t)^2 = (t + 1)^2 + 5^2
D(t)^2 = t^2 + 2t + 1 + 25
Differentiate both sides of the equation with respect to t:
2D(t)(dD/dt) = 2t + 2
Now we can solve for dD/dt, the rate at which the distance between the automobile and the farmhouse is increasing:
dD/dt = (2t + 2) / (2D(t))
Substituting x(t) = t + 1 and simplifying:
dD/dt = (2(t + 1) + 2) / (2D(t))
dD/dt = (2t + 4) / (2D(t))
dD/dt = (t + 2) / D(t)
To find the rate when the automobile is 1 mile past the intersection, we need to plug in t = 1 into the formula:
dD/dt = (1 + 2) / D(1)
dD/dt = 3 / D(1)
The distance D(1) can be found by substituting t = 1 into the equation for D(t):
D(1)^2 = 1^2 + 2(1) + 1 + 25
D(1)^2 = 30
D(1) ≈ √30
Now we can find the rate at which the distance between the automobile and the farmhouse is increasing when the automobile is 1 mile past the intersection:
dD/dt = 3 / D(1)
dD/dt ≈ 3 / √30
Therefore, the distance between the automobile and the farmhouse is increasing at a rate of approximately 3 / √30 miles per hour.