PRECALCULUS:GRAPHICAL,...-NASTA ED.
PRECALCULUS:GRAPHICAL,...-NASTA ED.
10th Edition
ISBN: 9780134672090
Author: Demana
Publisher: PEARSON
Expert Solution & Answer
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Chapter 6.3, Problem 70E
Solution

To find: the minimum distance between the Chang and Kuan.

The minimum distance between the Chang and Kuan is approximately 10.48 ft.

Given information:

The center of the first wheel is 0,20 and the second wheel is 15,15 .

For one revolution ( 2π radians) First wheel takes 12 seconds and second wheel takes 8 seconds.

At t=0 Chang’s position is at 20,20 , and Kuan’s positon is at 15,0 , since Kuan position at 6 O’clock.

Formula used:

  • The parametric equations for a point on the circle with centered h,k having radius r and making angle θ with the horizontal is x=h+rcosθ , y=k+rsinθ .
  • If x1,y1 and x2,y2 are two points. Then the distance between these points is x2x12+y2y12 .

Calculation:

The path of Chang’s position on the first wheel:

The angle covered by the first wheel per 1 sec is 2π12=π6 radians. The path of the Chang position can be computed by substituting the values h,k=0,20 , r=20 and θ=π6t in the parametric equations: x=h+rcosθ , y=k+rsinθ

  x1,y1=h+rcosθ,k+rsinθ=20cosπ6t,20+20sinπ6t

Thus, the path of the Chang’s position on the first wheel is given in parametric equations is

  x1=20cosπ6t , y1=20+20sinπ6t , t0 .

The path of Kuan’s position on the second wheel:

The angle covered by the second wheel per 1 sec is 2π8=π4 radians. The path of Kuan position can be computed by substituting the values h,k=15,15 , r=15 and θ=π4t in the parametric equations: x=h+rcosθ , y=k+rsinθ . But the initial Kuan’s position is at 15,0 , which implies that θ=π4tπ2

  x2,y2=h+rcosθ,k+rsinθ=15+15cosπ4tπ2,15+15sinπ4tπ2=15+15sinπ4t,1515cosπ4t

Thus, the path of the Kuan’s position on the second wheel is given in parametric equations is

  x2=15+15sinπ4t , y2=1515cosπ4t , t0 .

Distance between Chang and Kuan at any time:

Substitute the values x1=20cosπ6t , y1=20+20sinπ6t , x2=15+15sinπ4t , and y2=1515cosπ4t in dt=x2x12+y2y12

  dt=x2x12+y2y12=15+15sinπ4t20cosπ6t2+520sinπ6t15cosπ4t2=15+15sinπ4t20cosπ6t2+5+20sinπ6t+15cosπ4t2

Minimum distance between Chang and Kaun:

By observing the graph (see figure (1)) of the parametric equations x3=t , y3=dt , conclude that at t=12.3 the value of dt is minimum which is approximately 10.48 ft.

  PRECALCULUS:GRAPHICAL,...-NASTA ED., Chapter 6.3, Problem 70E

Figure (1)

Chapter 6 Solutions

PRECALCULUS:GRAPHICAL,...-NASTA ED.

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