Equal area property Consider the ellipse r(t) = (a cos t, b sin t), for 0 sts 27, where a and b are real numbers. Let 0 be the angle between the position vector and the x-axis. b a. Show that tan 0 = - tan t. a b. Find 0'(t). c. Recall that the area bounded by the polar curve r = on the interval [0, 0] is A(0) = "F(u))² đu. Letting f(0) f(0(1)) = |r(0(1))|, show that A'(t) = ab. d. Conclude that as an object moves around the ellipse, it sweeps out equal areas in equal times.

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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Equal area property Consider the ellipse r(t) = (a cos t, b sin t),
for 0 sts 27, where a and b are real numbers. Let 0 be the
angle between the position vector and the x-axis.
b
a. Show that tan 0 = - tan t.
a
b. Find 0'(t).
c. Recall that the area bounded by the polar curve r =
on the interval [0, 0] is A(0) = "F(u))² đu. Letting
f(0)
f(0(1)) = |r(0(1))|, show that A'(t) = ab.
d. Conclude that as an object moves around the ellipse, it sweeps
out equal areas in equal times.
Transcribed Image Text:Equal area property Consider the ellipse r(t) = (a cos t, b sin t), for 0 sts 27, where a and b are real numbers. Let 0 be the angle between the position vector and the x-axis. b a. Show that tan 0 = - tan t. a b. Find 0'(t). c. Recall that the area bounded by the polar curve r = on the interval [0, 0] is A(0) = "F(u))² đu. Letting f(0) f(0(1)) = |r(0(1))|, show that A'(t) = ab. d. Conclude that as an object moves around the ellipse, it sweeps out equal areas in equal times.
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