A. a particle moves around the circle x^2+y^2 = 1 in such a way that the x coord rate of change is dx/dt= y
B) IS THE PARTICLE MOVING CLOCKWISE OR COUNTERCLOCKWISE AROUND THE CIRCLE
C) FROM THE RT TRIANGLE ONE CAN SEE THAT SIN(THETA) = Y AND COS(THETA) = X
USE THESE RELATIONS TO SHOW THAT THE ANGLE THETA IS CHANGING AT A CONST. RATE AND THEN FIND THAT RATE
B)CLOCKWISE
In 1st quadrant (for ex.)dx/dt>0, so
x increases;
C)dx/dt=d(cos(theta))/dt=
d(cos(theta))/d(theta)*d(theta)/dt=
-sin(theta)*d(theta)/dx, so
dx/dt=-y*d(theta)/dt =>d(theta)/dt=-1
To answer part A of the question, we are given that the x-coordinate rate of change is given by dx/dt = y.
We can interpret this as the velocity in the x-direction at any given point on the circle.
To find if the particle is moving clockwise or counterclockwise around the circle, we need to evaluate the sign of dx/dt at different points on the circle.
Let's start by substituting the relation x^2 + y^2 = 1 into the equation dx/dt = y.
dx/dt = y implies dx = y * dt.
Now substitute x = cos(θ) and y = sin(θ) (as given in part C) into the above equation:
dx = sin(θ) * dt
To find out if the particle is moving clockwise or counterclockwise, we need to consider the sign of dx.
When the particle is moving counterclockwise, dx will be positive, and when moving clockwise, dx will be negative.
Using the trigonometric identity sin^2(θ) + cos^2(θ) = 1, we can rearrange the equation as:
1 = cos^2(θ) + sin^2(θ)
Rearranging further:
cos^2(θ) = 1 - sin^2(θ)
Using the relation dx = sin(θ) * dt, we can square both sides of the equation to eliminate the dt term:
dx^2 = (sin(θ) * dt)^2
= sin^2(θ) * dt^2
But dx^2 = (dx)^2, so we can rewrite the equation as:
(dx)^2 = sin^2(θ) * dt^2
Now divide both sides of the equation by dt^2:
(dx)^2 / dt^2 = sin^2(θ)
Recall that (dx/dt)^2 = (dy/dt)^2 = (dy)^2 / dt^2.
Applying this to the equation, we have:
(dy)^2 / dt^2 = sin^2(θ)
Since sin^2(θ) is always positive, this implies that (dy)^2 / dt^2 is also positive.
Therefore, (dy/dt) or y is either always positive or always negative for any angle θ. This implies that the particle is always moving either clockwise or counterclockwise around the circle, rather than switching directions.
To find the rate at which the angle θ is changing, we can differentiate the equation x = cos(θ) with respect to time t:
dx/dt = -sin(θ) * dθ/dt
Given that dx/dt = y (from part A), we can substitute y = sin(θ) and dx/dt = y into the above equation:
sin(θ) = -sin(θ) * dθ/dt
Divide both sides by -sin(θ):
1 = dθ/dt
This implies that the angle θ is changing at a constant rate of 1 unit per unit time.