Question

An isosceles triangle is inscribed inside a circle of radius 12 metres. Its height is increasing at a rate of 2 m/sec. At the instant when the height of the triangle is 18 m, at what rate a) its area changing? b) its perimeter changing?

The answers are 0,0. But my answers always are 24√3 and 24√3. I don't know how to get 0. Help me! Please! Tks.

Answer 1

Make a sketch.
Draw a radius to the base of the isosceles triangle.
let the base of the isosceles triangle be 2x, and let its height be h+12

area = (1/2)(2x)(h+12)
= x(h+12)

but x^2 + h^2 = 12^2
x^2 = 144 - h^2
x = (144 - h^2)^(1/2)

area = (144 - h^2)^(1/2)(h+12)

d(area)/dt = (1/2)(144-h^2)^(-1/2) (-2h)dh/dt + (144 - h^2)^(1/2) (dh/dt)
so when h = 6 and dh/dt = 2
(when h+12 = 18, my h = 6)

d(area)/dt = (1/2)(108)^(-1/2) (-16)(2) + 108^(1/2) (2)
= -16/√108 + 2√108
= (-16 + 216)/√108
= -4(4 + 54)/(6√3)
= - 116/(3√3) m^2/s

check my arithmetic, been making silly arithmetic errors lately

b)
let the side of the triangle be s
s^2 = x^2 + (h+12)^2
= 144 - h^2 + h^2 + 24h + 144
= 288 + 24h
s = √(288 + 24h) = 2(72 + 6h)^(1/2)

P = 2s + 2x
= 4(72 + 6h)^(1/2) + 2(144 - h^2)^(1/2)

take dP/dt, sub in h = 6, and dh/dt = 2