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?
a)-3.875
b0-3.75
c)-2.25
d03.75
e)3.875
x^2 + y^2 = 1
2x dx/dt + 2y dy/dt = 0
dy/dt = (-x dx/dt)/y
= -.6(3)/.8 = - 2.25
To find the vertical velocity dy/dt at the point (.6, .8), we can use the properties of the unit circle and the relationship between dx/dt and dy/dt.
1. The unit circle represents the motion of the particle around a circle of radius 1 centered at the origin (0, 0). At any point on the unit circle, the x-coordinate represents the horizontal displacement of the particle, and the y-coordinate represents the vertical displacement.
2. Given that the point on the unit circle is (.6, .8), we can use the Pythagorean theorem to find the radius of the circle. The radius r is given by the equation r = sqrt(x^2 + y^2).
Substituting the values x = .6 and y = .8 into the equation, we get:
r = sqrt(.6^2 + .8^2)
= sqrt(0.36 + 0.64)
= sqrt(1)
= 1
3. Since the radius of the unit circle is 1, the particle is moving along the circumference of the circle. Therefore, its motion is circular, and its horizontal velocity dx/dt can be related to its vertical velocity dy/dt through the equation dy/dt = -dx/dt * (x/y).
4. Substituting the given value of dx/dt = 3 and the coordinates x = .6 and y = .8 into the equation, we get:
dy/dt = -3 * (.6/.8)
= -3 * 0.75
= -2.25
Therefore, the vertical velocity dy/dt at the point (.6, .8) is -2.25.
The answer is (c) -2.25.