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 < ?/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.)

Answer 1

m = slope at point p = dy/dx = -cost/sint

equation of tangent at point p( x1 ,y1) = P(cost ,sint)

y -y1 =m(x-x1)

y -sint = (-cost/sint)(x-cost)

y -sint = x*cost/sint + cos^2t/sint

y intercept = sint +cos^2t/sint = 1/sint

x intercept = 1/cost

a)   a(t) =area = (y intercept)*(xintercept)/2

                       = 1/(2*sint*cost)

                       =1/(sin2t)

b) limit t-->pi/2- a(t) = limit x-->0 1/sint (pi-2x)   = - infinty

c) limit t-->0+ a(t) = limit t-->0 1/sint (2t)    = infinity

Answer 2

x^2 +y^2 =1 is equation of circle

for slope of tangent take derivative

y'=-x/y

y'=-cot t

y=(-cot)x+c

sint=(-cott)*(cost) +c

c=sint+ (cott)*(cost)

c=sint+ (cos^2t)/sint

c=(sin^2t+ cos^2t)/sint

c=cosect

y=(-cot)x+cosect

yintercept

x=0

==>cosect

xintercept

y=0

==>sect

(a) The area of the right triangle is a(t)= 0.5*xintercept*yintercept

=0.5*sect*cosect

=cosec2t

(b)

lim t -> pi/2^- a(t)=infinity

(c)

lim t -> 0⁺ a(t)=infinty

(d)

lim t -> pi/4 a(t)= cosec(2*pi/4) =1