Calculus/Related rates
Expert: Paul Klarreich - 3/7/2010
Question
An isosceles triangle whose base lies on the diameter of a circle is inscribed in the circle. When the area of the circle is 36pi sq.inches.. the area of the circle is increasing at 3 sq in,per sec. As the circle increases, the triangle increases accordingly so that the vertices lie along the circumference of the circle.
a) At what rate is the area of the triangle changing?
b) At what rate is the perimeter of the triangle changing?
AnswerQuestioner: steve
Country: United States
Category: Calculus
Private: No
Subject: related rate problem
Question:
An isosceles triangle whose base lies on the diameter of a circle is inscribed in the circle. When the area of the circle is 36pi sq.inches.. the area of the circle is increasing at 3 sq in,per sec. As the circle increases, the triangle increases accordingly so that the vertices lie along the circumference of the circle.
a) At what rate is the area of the triangle changing?
b) At what rate is the perimeter of the triangle changing?
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Variables:
A = area of circle.
B = area of triangle.
r = radius of circle.
P = perimeter of triangle.
Rates:
dA/dt = 3
dr/dt to be found.
DB/dt to be found.
dP/dt to be found.
Relations;
A = pi r^2
dA/dt = 2 pi r dr/dt
At time when A = 36pi, (and r = 6)
3 = 2 pi (6) dr/dt
dr/dt = 1/(4 pi)
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Now B = r^2 [ area = 1/2 base(2r) times height(r) ]
dB/dt = 2r dr/dt
dB/dt = 2(6) 1/(4 pi) etc.
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And P = 2r + r sqrt(2).
You can finish up.