Hi, we just learned this new section today, and I am very confused....

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?

This is a 3,4,5 right triangle.

the sin of the angle between horizontal and the resultant tangent at (.6,.8) is dy/dt/3 = -3/5
so dy/dt = -9/5

To solve this question, we need to understand the concept of velocity in relation to the unit circle.

First, let's consider the point (.6, .8) on the unit circle. This point represents the location of the particle at a particular moment. The coordinates (.6, .8) represent the x and y coordinates of the point on the circle.

To find the vertical velocity (dy/dt) at that point, we need to differentiate the y-coordinate with respect to time (t). But since we are given the horizontal velocity (dx/dt), we can use it to find dy/dt.

Now, let's calculate dy/dt using the given information.

The equation of the unit circle is x^2 + y^2 = 1. Since the particle is moving on the unit circle, the equation holds true.

Differentiating both sides of the equation with respect to time (t), we get:

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

Substituting the given values dx/dt = 3 and the point (.6, .8) into the equation, we have:

2(.6)(3) + 2(.8)(dy/dt) = 0

1.2 + 1.6(dy/dt) = 0

1.6(dy/dt) = -1.2

(dy/dt) = -1.2 / 1.6

(dy/dt) = -0.75

Therefore, the vertical velocity dy/dt at the point (.6, .8) is -0.75.