Two roads intersect at right angles. At a certain moment, one bicyclist is 8 miles due NORTH of the intersection traveling TOWARDS the intersection at a rate of 16 miles/hour. A second bicyclist is 10 miles due WEST of the intersection traveling AWAY from the intersection at 12 miles/hour.
(a) How fast is distance between the two bicyclists changing at this moment?
To find how fast the distance between the two bicyclists is changing, we can use the concept of rates of change and apply the Pythagorean theorem.
Let's define the distance between the two bicyclists as D. We need to find dD/dt, the rate at which D is changing with respect to time.
From the given information, we know that one bicyclist is traveling north and the other is traveling west. This forms a right triangle, with the distance between the two bicyclists being the hypotenuse.
Using the Pythagorean theorem, we can express D in terms of the distances traveled by each bicyclist:
D^2 = (Distance North)^2 + (Distance West)^2
Substituting the given values:
D^2 = (8 miles)^2 + (-10 miles)^2
D^2 = 64 miles^2 + 100 miles^2
D^2 = 164 miles^2
Now, we can differentiate both sides of this equation with respect to time (t) using implicit differentiation:
2D * (dD/dt) = 2 * (64 miles^2 * (d/dt of miles^2) + 100 miles^2 * (d/dt of miles^2))
Differentiating each term:
2D * (dD/dt) = 2 * (64 miles^2 * 0 + 100 miles^2 * 0)
2D * (dD/dt) = 0
dD/dt = 0
Therefore, at the given moment, the distance between the two bicyclists is not changing.
I believe the correct answer is 0.635 but i got 0.625.
My final answer was -8/square root of 164. Is that correct?
make a diagram, let the distance the north biker is from the intersection be y miles, let the distance the west biker is from the intersection be x miles
Let the distance between them be D
D^2 = x^2 + y^2
2D dD/dt = 2x dx/dt + 2y dy/dt
given:
dx/dt = 12 mph
dy/dt = -16 mph (y is decreasing)
find dD/dt when x=10 and y=8
then D^2 = 100+64
D = √164
dD/dt =(2(10)(12) + 2(8)(-16))/(2√164) = .....
you finish the arithmetic