Question
two equal sides of an isoscles triangle with fixed based 'a' are decreasing at the rate of 9cm/second how fast is the area of triangle decreasing when the two sides are equal to 'a'
Answers
Steve
if the sides are of length s and the triangle has height h,
s^2 = (a/2)^2 + h^2
the area
z = 1/2 ah
= 1/2 a √(s^2-(a/2)^2)
dz/dt = as/(2√(s^2-(a/2)^2)) ds/dt
so, when s=a,
dz/dt = a^2/(2√(3a^2/4)) (-9)
= -9a/√3
= -3√3 a
as usual, check my math
s^2 = (a/2)^2 + h^2
the area
z = 1/2 ah
= 1/2 a √(s^2-(a/2)^2)
dz/dt = as/(2√(s^2-(a/2)^2)) ds/dt
so, when s=a,
dz/dt = a^2/(2√(3a^2/4)) (-9)
= -9a/√3
= -3√3 a
as usual, check my math