This question builds on an earlier \"rod problem\": A 4-centimeter rod is attach

Question

This question builds on an earlier "rod problem": 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) Let P be the point on the circle where the rod is attached. Write a formula for the slope of the tangent line to the circle at time t seconds:

(b) Write a formula for the slope of the rod at time t seconds: (Hint: You will need to recall the formula for the x-coordinate of the rod along the x-axis)

(c) Find the first time when the rod is tangent to the circle: seconds

(d) At the time in (c), what is the slope of the rod?

(e) Find the second time when the rod is tangent to the circle: seconds

(f) At the time in (e), what is the slope of the rod?

Explanation / Answer

a) for the 1st part we will have

x² + y² = 2²
implicitly:
2x dx + 2y dy = 0
dividing by dx:
2x + 2y dy/dx = 0
dividing by 2:
x + y dy/dx = 0
dy/dx = - x/y = slope
tricky question because we need this as a fuction of t.
= - (r*cos) / (r*sin) = - cot = - cot(7*t) = slope

B ) we have the points:
Point at B:
(x,y) = (2cos(7*t) + 2(4 - sin²(7*t)), 0)
and
Point on circle:
(x,y) = (2cos(7*t), 2sin(7*t))

the slope through these two points with the slope formula is:
- sin(7*t) / (4 - sin²(7*t))

c) for third part we will have this as follow

The slope found in part (b) through the point on the circle and point B was:
- sin(7*t) / (4 - sin²(7*t))
and this must be equal to the slope found in part (a):
- cot(7*t)

i.e. solve the equation:
cot(7*t) = sin(7*t) / (4 - sin²(7*t))

it is a tough equation to solve, but the solutions are:
t = 2{k + arctan[± (1 - 5)/2] } / (7)
with k = 0, 1, 2, 3, ...
and when k = 0, for the negative value we have our first value:
t = 2{ arctan[-(1 - 5)/2] } / (7)

D)slope = - cot(7*t)
slope = - cot{2(arctan[-(1 - 5)/2])}
= - 1/2

E)t = 2 {arctan[(1 - 5)/2] } / (7)

(f) At the time in (e), what is the slope of the rod?
slope = - cot(7*t)
slope = - cot{2[arctan[(1 - 5)/2] ]}
= 1/2

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