related rates?

Question

The sides of an equilateral triangle are increasing at the rate of 27 in/sec. How fast is the triangle's area increasing when the sides of the triangle are each 18 inches long?

I got 420.88 in^2/sec

you don't have to compute it for me or anything, I just feel really uncertain of my answer, and I would like to know if I am doing it all wrong or not.

Answer 1

i got the same answer you did but I admit i have not done this type of problem in 6 years. However this webpage got the same answer I did, which gives me more confidence in it.

http://www.themathpage.com/aCalc/motion.htm

they get (a*b)*(1/2)*3^.5 where a is the length of the triangle and b is the rate of the growth rate of the sides of the triangle

Answer 2

Area of triangle = sqrt(s*(s-a)*(s-b)*(s-c)) where
s=(a+b+c)/2

In the case of an equilateral triangle :
A = sqrt(s(s-l)^3) where l=length of side and s=3l/2
so A = sqrt((3l/2)((3l/2)-l)^3)

we know dl/dt = 27
we want dA/dt
dA/dt = (dl/dt)(dA/dl)
dA/dl =(cbrt(3)l^3)/(2sqrt(l^4))
so dA/dt = 27(cbrt(3)l^3)/(2sqrt(l^4))
substitution of l=18 gives
dA/dt = 350.47 in^2/s

Answer 3

a= sqrt(3)/2*s^2
da/ds=sqrt(3)*s
da=sqrt(3)*s*ds=sqrt(3)*18*27=841.77 in^2/sec