Question

An object is moving around the unit circle with parametric equations x(t)=cos(t), y(t)=sin(t), so it's location at time t is P(t)=(cos(t),sin(t)) . Assume 0 < t < π/2. At a given time t, the tangent line to the unit circle at the position P(t) will determine a right triangle in the first quadrant. (Connect the origin with the y-intercept and x-intercept of the tangent line.)

(a) The area of the right triangle is a(t)=  .

(b)

lim t → pi/2⁻a(t)=

  

(c)

lim t → 0⁺a(t)=

  

(d)

lim t → pi/4 a(t)=

  

(e) With our restriction on t, the smallest t so that a(t)=2 is  

(f) With our restriction on t, the largest t so that a(t)=2 is  

(g) The average rate of change of the area of the triangle on the time interval [π/6,π/4] is  .

(g) The average rate of change of the area of the triangle on the time interval [π/4,π/3] is  .

(h) Create a table of values to study the average rate of change of the area of the triangle on the time intervals [π/6,b], as b approaches π/6 from the right. The limiting value is  .

(i) Create a table of values to study the average rate of change of the area of the triangle on the time intervals [a,π/3], as a approaches π/3 from the left. The limiting value is  .

Answer 1

Given An object is moving around a unit circle with parametric equations: x(t) = cos t: y (t) = sin t Location at time t is P(t) = (cost, sin t) At time tthe tangent line to the unit circle at the position P(t) wll determine a right angled triangle. We have to calculate the area of the triangle We need to determine the equation of the tangent line We have x(t) cost, y(t)-sint dy' dt dx sin t and = cos t dt

Slope of tangent ах ох at cost -Sin t Equation of tangent is given by cos t Sin t ysint-sin t-xcost +cos t >ysint+xcost-1 cosect sect Coordinates of B (sect,0) Coordinates of A = (0, cosec t) Area of triangle-- bh cosec t sec t 2 sin t cost sin 2t Area of triangle a(t) in 2t

We have to determine lim ait lim a(t)= lim -sin 2t -lim -lim sin (π-2h) sin 2h >0 sin 2h We have to determine lim a(t) lim a(r) = lim -lim -o sin 2(0+h) sin 2h lim a(z) =φ t→0"

We have to determine lim a(t) linna (t)-linn sin2t Sin2 sin lim a(t) = 1 We have to find the value of smallest t sati sfying a(t) 2 sin 2t → sin 2t = sin :- The general solution is given by 2t = nT+(-1) 12 For n-0 2-1) 12 12

We are given 0 < t < If we take any other n, the value of t will be either negative or more than_ 12 Ito = 1 is the smallest value for a(t)=2 012 1× π 12 ππ ち , 212 12 12 We are given 0

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