A straight driveway perpendicular to a straight highway leads to a house located one mile away. A truck travels past the house traveling 60 miles per hour. How fast is the distance between the truck and the house increasing when the truck is three miles past the intersection of the highway and road
let the truck' distance from the intersection be x miles
make a diagram and you should have a right-angles triangle with sides x,1, and let the hypotenuse be y
given dx/dt = 60 mph
find dy/dt when x = 3
y^2 = x^2 + 1
2y dy/dt = 2x dx/dt
dy/dt = (x dx/dt)/y
when x = 3, y = √10
dy/dt = 3(60)/√10 = appr. 56.92 mph
To find how fast the distance between the truck and the house is changing when the truck is three miles past the intersection, we can use related rates.
Let's denote:
- x as the distance of the truck to the intersection of the highway and the road
- y as the distance between the truck and the house
- t as time (in hours)
We are given the following information:
- The truck is traveling at a constant speed of 60 miles per hour.
- The driveway is perpendicular to the highway, and the house is located one mile away from the intersection.
- We want to find dy/dt, which represents the rate at which the distance between the truck and the house is changing with respect to time when the truck is three miles past the intersection.
First, let's draw a diagram to visualize the scenario:
House
------
/|
/ |
/ |
/ | y
/ |
/ |
/ |
/ |
/________|__
Intersection
From the diagram, we can see that the length of the driveway is √(1^2 + x^2) = √(1 + x^2). This is derived from the Pythagorean theorem.
Next, let's write an equation that relates x and y: (1 + x^2) = y^2. This equation represents the relationship between the distance of the truck from the intersection and the distance between the truck and the house.
Now, we can take the derivative of both sides of the equation with respect to time (t):
d/dt (1 + x^2) = d/dt (y^2)
The derivative of 1 + x^2 with respect to t can be expressed as 0 + 2x(dx/dt) = 2x(dx/dt), since the derivative of a constant (1) is zero.
The derivative of y^2 with respect to t can be expressed as 2y(dy/dt), using the chain rule.
Substituting these values into the equation, we get:
2x(dx/dt) = 2y(dy/dt)
Now, we can solve for dy/dt by rearranging the equation:
dy/dt = (x(dx/dt)) / y
We are given that dx/dt (the rate of change of x with respect to t) is constant and equal to 60 mph since the truck is traveling at a constant speed of 60 mph.
To find x when the truck is three miles past the intersection, we can use x = 3.
We can also substitute the value of y using the equation (1 + x^2) = y^2:
(1 + 3^2) = y^2
10 = y^2
y = √10
Now, we have all the necessary values to calculate dy/dt:
dx/dt = 60 mph
x = 3 miles
y = √10
Substituting these values into the equation:
dy/dt = (3 * 60) / √10
Evaluating this expression, we find that dy/dt ≈ 56.83 mph.
Therefore, when the truck is three miles past the intersection, the distance between the truck and the house is increasing at a rate of approximately 56.83 mph.