#22: 2a) ANSWER:
#22: 2b) ANSWER: ± 2
/3
SOLUTION: x(t) = ( 1 + 2cost )cost = 0
implies cost = 0 or 1 + cost = 0 which implies t =
3/2, ±
2/3. Substituting
these into y(t) = ( 1 + 2cost )sint gives t = ± 2
because y is no zero at t = 3/2.
#22: 2c) ANSWER: 4cos 2
t + cost - 1
SOLUTION: dy/dt = Dt [( 1
+ 2cost ) sint ] = - 2 sint sint + (1 + 2cos t ) cost = cost + 2cos 2 t -
2sin 2 t = 4cos 2 t + cost - 1
#22: 2d) ANSWER: - sint - 4 sint
cost
SOLUTION: dx/dt = Dt [( 1
+ 2cost ) cost ] = - 2 sint cost - ( 1 + 2cost ) sint = - sint - 4 sint
cost
#22: 2e) ANSWER: - [ 4cos 2
t + cost - 1 ]/[ sint + 4 sint cost ]
SOLUTION: dy/dx = [dy/dt] / [dx/dt].
Both of these were evaluated in parts c and d. The
minus sign was factored out of dx/dt and put in front of the fraction.
#22: 2f) ANSWER: 
SOLUTION: The point of tangency is(
0,0 ) . The slope is given by dy/dx = - [ 4cos 2 t + cost - 1
]/[ sint + 4 sint cost ]. At t = 2 / 3, the slope is - 31/2. At t = - 2 / 3, the slope is
31/2.
Therefore, the two equations are y = ± 31/2 x.
#22: 2g ANSWER: (0.58 , ±
0.37) and (1.30, ± 1.76)
SOLUTION: dy/dx = 0 implies - [
4cos 2 t + cost - 1 ]/[ sint + 4 sint cost ] = 0 which in turn
implies 4cos 2 t + cost - 1 = 0. This is a quadratic equation in cost.
Therefore, cost = -0.84307
and 0.59307. Solving for t gives t = ± 2.5738 and ± 0.93593. Substituting
these
values into x(t) = ( 1 + 2cost ) cost and y(t) = ( 1 + 2cost ) sint gives the 4 points (0.58 ,
± 0.37) and (1.30, ± 1.76).
#22: 2h) ANSWER: - [ 9 + 6cost
]/[(sint + 4 sint cost) 3 ]
SOLUTION: The second derivative of y
with respect to x is the derivative with respect to x of
the dy/dx. So Dx (dy/dx) = Dt (dy/dx) / (dx/dt).
Dt (dy/dx) = Dt ( - [ 4cos 2 t + cost - 1 ]/[ sint + 4
sint cost ] ) = - [ ( sint + 4 sint cost )(-8 cost sint - sint) - (4cos 2 t + cost -
1)(cost + 4 cos 2 t - 4 sin
2 t )] /( sint + 4 sint cost ) 2 = [ 9 + 6cost ]/[(sint + 4 sint cost)
2 ] (Use technology to simplify),
Therefore, Dx (dy/dx) = { [ 9 + 6cost ]/[(sint + 4 sint cost) 2
]}/{-[ sint + 4 sint cost ]) = - [ 9 +
6cost ]/[(sint + 4 sint cost) 3 ]
#22: 2i) ANSWER: ± 16 /
31/2.
SOLUTION: Substituting t = ± 2
/ 3, into - [ 9 +
6cost ]/[(sint + 4 sint cost) 3 ] results in ± 16 / 31/2.
#22: 2j) ANSWER: ( 5 + 4 cost
)1/2 dt
SOLUTION: ds = [ (dy/dt) 2
+ (dx/dt) 2 ]1/2 dt = [ (4cos 2 t + cost - 1)
2 + (- sint - 4 sint cost) 2 ]1/2
dt = ( 5 + 4 cost )1/2 dt (Again, use technology to simplify)
#22: 2k) ANSWER: 13.365
SOLUTION: 