A road perpendicular to a highway leads to a farmhouse located 8 mile away.An automobile travels past the farmhouse at a speed of 80 mph.
How fast is the distance between the automobile and the farmhouse increasing when the automobile is 7 miles past the intersection of the highway and the road?
The distance between the automobile and the farmhouse is increasing at a rate of ___________ mph.
Let the distance from the sideroad be x miles, let the distance between the car and the house be y miles.
You have a right-angled triangle where
x^2 + 8^2 = y^2
then 2x dx/dt = 2y dy/dt (1)
when x=7, y^2 = 49+64 = 113
so when x=7, y= √113 and dx/dt = 80
sub these values into equation (1) and solve for dy/dt
meow
How did you find 80
bark
Well, let's see! If the automobile is 7 miles past the intersection, that means it's already traveled 7 miles down the road. So, the remaining distance between the automobile and the farmhouse is 8 - 7 = 1 mile.
Since we know the automobile is traveling at a speed of 80 mph, that means it's getting 80 miles farther from the farmhouse every hour.
Therefore, the distance between the automobile and the farmhouse is increasing at a rate of 80 mph.
To solve this problem, we can use the concept of related rates. We are given the following information:
- The automobile is traveling at a constant speed of 80 mph.
- The distance between the farmhouse and the automobile is 8 miles.
- The distance between the road and the farmhouse is constant at 8 miles.
Let's assign variables to the distances to make it easier to work with:
- Let x represent the distance between the automobile and the intersection of the highway and the road.
- Let y represent the distance between the automobile and the farmhouse.
We want to find the rate at which the distance y is changing with respect to time when x = 7 miles.
To do this, we can use the Pythagorean theorem to relate x and y:
x^2 + y^2 = 8^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we are given dx/dt = 80 mph (the speed of the automobile), we can substitute this into the equation:
2x(80) + 2y(dy/dt) = 0
We can rearrange this equation to solve for dy/dt:
2y(dy/dt) = -2x(80)
dy/dt = (-2x(80))/(2y)
Now we need to find the values of x and y when the automobile is 7 miles past the intersection of the highway and the road.
When the automobile is 7 miles past the intersection, x = 7 miles.
Since the distance between the road and the farmhouse is constant at 8 miles, y = 8 - x = 8 - 7 = 1 mile.
Now we can substitute these values into the equation to find dy/dt:
dy/dt = (-2(7)(80))/(2(1))
= (-2(7)(80))/2
= (-2) * 7 * 80
= -1120 mph
Therefore, the distance between the automobile and the farmhouse is increasing at a rate of -1120 mph.
Note that the negative sign indicates that the distance is decreasing, which makes sense since the automobile is moving away from the farmhouse.