A 4-centimeter rod is attached at one end A to a point on a wheel of radius 2 cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The wheel rotates counterclockwise at 3.5 rev/sec. Thus, when t=1/21 sec, the rod is situated as in the diagram at the right below. (a) How far is the right end of the rod (the point B) from the center of the wheel at time t=1/21 sec?

(b) Express the x-coordinate of the right end of the rod as a function of t:

(c) Express the speed of the right end of the rod as a function of t:

Step-by-step explanation:

The triangle that was formed has radius 2:

hypotenuse = 2

adjacent = x = 2*cosθ

opposite = y = 2*sinθ

from the first picture on the left, we see that the rod has length of 4.

We then use pythagoras theorem with respect to the triangle formed with the rod, i.e, the opposite leg, and the length of the remaining side (say, L), we have:

2²sin²θ +L² = 4²

On rearranging and making L subject of formula, we have

L = √(16 - 4sin²θ)

or

L = 2√(4 - sin²θ)

Also, the distance of point B is from the center of the circle, and it is:

2cosθ + 2√(4 - sin²θ)

We know that

3.5 revolutions per second means 3.5(2π) per second.

or

7π radians per second

So we try to write the angle formed as:

θ = 7π*t

where t is the time in seconds.

(a)

2cos(7πt) + 2√(4 - sin²(7πt))

at t = 1/21 we substitute for t, and then

2cos(π/3) + 2√(4 - sin²(π/3))

= 1 + √13 or approximately 4.6

(b)

(c)

B = 2cos(7πt) + 2√(4 - sin²(7πt))

dB/dt = - 14πsin(7πt) - { 14π*cos(7πt)sin(7πt)/√(4 - sin²(7πt)) }

yes

step-by-step explanation:

yes

the answer would be  

a.

3: 1

step-by-step explanation: