Question

The coordinates of a particle in the metric xy-plane are differentiable functions of time t with dx/dt = −1 m/sec and dy/dt = −9 m/sec. How fast is the particle's distance from the origin changing as it passes through the point (8,6)?

Answer 1

The particle's distance from the origin is changing at a rate of √82 meters per second as it passes through the point (8,6), calculated by finding the magnitude of the velocity vector in the direction from the origin to the particle.

Step-by-step explanation:

The speed at which a particle's distance from the origin is changing is given by the rate of change of the distance with respect to time. This is essentially the magnitude of the velocity vector in the direction from the origin to the particle. Given that dx/dt = −1 m/sec and dy/dt = −9 m/sec, the particle's velocity vector is (-1, -9) at any instant in time. To find how fast the particle is moving away from the origin at the point (8,6), use the Pythagorean theorem to find the magnitude of the velocity vector.

First, calculate the distance from the origin to the particle using the particle's coordinates (8,6):

• s = √(x² + y²) = √(8² + 6²) = √(64 + 36) = √100 = 10 m

Now, calculate the rate at which this distance is changing by finding the derivative of s with respect to time:

ds/dt = √((dx/dt)² + (dy/dt)²) = √((-1)² + (-9)²) = √(1 + 81) = √82 m/sec

Therefore, the rate at which the particle's distance from the origin is changing as it passes through the point (8,6) is √82 meters per second.