A particle is moving around the unit circle (the circle of radius 1 centered at the origin). At the point (.6 , .8), the particle has horizontal velocity dx/dt=3. What is its vertical velocity dy/dt at that point?

I don't know what equation to use or anything for this problem. Please help :) It is much appreciated!

the equation must be

x^2 + y^2 = 1
(since .6^2 + .8^2 = 1)

2x dx/dt + 2y dy/dt = 0

divide by 2 and substitute
.6(3) + .8 dy/dt = 0
dy/dt = -1.8/.8 = - 9/4 or - 2.25

Well, it seems like the particle is having quite a "circular" adventure! Since the particle is moving around the unit circle, we know that its position on the circle can be represented by the coordinates (cosθ, sinθ), where θ is the angle made with the positive x-axis.

Now, let's take a look at the point (.6, .8) on the unit circle. We want to find the vertical velocity dy/dt at that point.

Since dx/dt = 3, this means that the rate of change of x with respect to time is 3. But we can relate the rates of change of x and y using the equation of the unit circle:

x^2 + y^2 = 1.

Differentiating both sides of the equation with respect to time, we have:

2x(dx/dt) + 2y(dy/dt) = 0.

Plugging in the values we know: x = 0.6, dx/dt = 3, we can solve for dy/dt:

2 * 0.6 * 3 + 2 * 0.8 * (dy/dt) = 0.

2.4 + 1.6(dy/dt) = 0.

1.6(dy/dt) = -2.4.

dy/dt = -2.4 / 1.6.

dy/dt = -1.5.

So, the vertical velocity dy/dt at the point (.6, .8) is -1.5. It seems like the particle is heading downwards at quite a "sigh"speed!

To solve this problem, we can use the equation relating horizontal and vertical velocities on the unit circle:

(dy/dt)/(dx/dt) = -x/y

Where x and y represent the coordinates of the particle on the unit circle.

Given that the particle is at the point (0.6, 0.8) and dx/dt = 3, we can substitute these values into the equation:

(dy/dt)/(3) = -(0.6)/(0.8)

Cross multiplying, we get:

(dy/dt) = -(0.6)/(0.8) * 3

Simplifying further, we have:

(dy/dt) = -2.25

Therefore, the vertical velocity of the particle at the point (0.6, 0.8) is -2.25.

To solve this problem, we can use the concept of differentiation and the properties of circular motion.

Let's start by understanding the problem and the given information. We are given that a particle is moving around the unit circle, which means its distance from the origin is always 1. At a specific point on the circle, designated as (0.6, 0.8), we are given the horizontal velocity of the particle, dx/dt = 3.

To find the vertical velocity of the particle, dy/dt, we need to establish a relationship between the horizontal and vertical velocities.

The relationship between dx/dt (horizontal velocity), dy/dt (vertical velocity), and the radius of the unit circle is given by the following equation:

(dx/dt)^2 + (dy/dt)^2 = r^2

In this equation, r represents the radius of the circle. Since we are dealing with the unit circle, r = 1.

Now, let's substitute the given values into the equation:

(3)^2 + (dy/dt)^2 = 1^2

Simplifying the equation gives:

9 + (dy/dt)^2 = 1

Subtracting 9 from both sides:

(dy/dt)^2 = 1 - 9

(dy/dt)^2 = -8

At this point, we run into a problem because we cannot take the square root of a negative number while working with real numbers. This is because the value of (dy/dt), the vertical velocity, does not exist at the given point on the unit circle.

Therefore, we cannot determine the vertical velocity of the particle at the point (0.6, 0.8) based on the given information.

In summary, the vertical velocity, dy/dt, cannot be determined at the point (0.6, 0.8) because the given horizontal velocity, dx/dt, does not allow the calculation of a valid vertical velocity value.