The fixed sides of an isosceles triangle are of length 6 cm. If the sides slide outward at a speed of 1 cm/min, at what rate is the area enclosed by the triangle changing when θ = 30°?

Question

Question: The fixed sides of an isosceles triangle are of length 6 cm. If the sides slide outward at a speed of

1 cm/min, at what rate is the area enclosed by the triangle changing when θ = 30

Answer 1

angle theta=T

Per law of cosines:

Base B is:

B^2 = 72 - 36 cosT

= 36( 2 - cos T)

so then B = 6 sqrt( 2 - cos T)

Height h = 6 cos (T/2)

area A = (1/2) B*h

(1/2) 6 sqrt(2 - cos T) (6) cos(T/2)

= 6 sqrt(2 - cos T) cos(T/2)

is the area function in terms of angle T

dA/dT = 6 sqrt(2 - cos T) (-sin T/2) (1/2) + 6 sin (2 - cos T) (sin T) (cos T/2)

= 3 sqrt(2 - cos T) (-sin T/2) + 6 sin (2 - cos T) (sin T) (cos T/2)

when T=30 = pi/6

dA/dT = 3 sqrt( 2 - rad3/2) (- sin pi/12) + 6 sin (2 - rad3/2)(1/2)(cos pi/12)

= 3 sqrt( 2 - rad3/2) (-sin pi/12) + 3 sin (2 - rad3/2)(cos pi/12)