A cat walking down the street at a speed of 1 1m/s passes 2 m from a post on which there is a lamp 6 m above the ground. How fast is the distance between the cat and the lamp increasing 3 s later?
To find the rate at which the distance between the cat and the lamp is increasing, we can use the concept of related rates.
Let's denote the distance between the cat and the lamp as D. We are given that the cat is moving at a constant speed of 1 m/s, so we have dD/dt = 1 m/s (the rate at which D is changing over time is 1 m/s).
To solve the problem, we need to relate the variables D, x (horizontal distance between the cat and the lamp), and y (vertical distance between the cat and the lamp). We can use the Pythagorean theorem to do this:
D^2 = x^2 + y^2
Now, let's differentiate both sides of this equation with respect to time t:
2D*(dD/dt) = 2x*(dx/dt) + 2y*(dy/dt)
Since the height of the lamp remains constant at 6 m, we know that dy/dt = 0 (the vertical distance isn't changing). Therefore, the equation simplifies to:
2D*(dD/dt) = 2x*(dx/dt)
Plugging in the values we know, at the given instant of time:
D = 2 m (distance between the cat and the post)
x = 2 m (horizontal distance between the cat and the lamp)
dD/dt = 1 m/s (rate at which the cat is moving)
We can solve for dx/dt (the rate at which the horizontal distance is changing):
2 * 2 * 1 = 2 * dx/dt
4 = 2 * dx/dt
Dividing both sides by 2:
2 = dx/dt
So, the rate at which the horizontal distance between the cat and the lamp is changing is 2 m/s.
Therefore, 3 seconds later, the speed at which the distance between the cat and the lamp is increasing is 2 m/s.