Related Rates

[fgmango]

The area of an equilateral triangle is decreasing at a rate of 4 cm^2/min. Find the rate at which the length of a side is changing when the area of the triangle is 200 cm^2.

I got A=sqrt3*s^2/4

dr/dt=4cm^2/min

s=sqrt(800/sqrt(3))

dA/dt=-sqrt3/2 *s (dr/dt)

=-sqrt(3)* sqrt(800/sqrt(3))/2*(4cm^2/min)

not sure what I'm doing wrong.

 

[stapel]

You are correct that, with a side-length of "s", the area of the equilateral triangle will be:

. . . . .A_triangle = (sqrt[3]/4)s²

(Note: It is a good idea to define your variables and equations, so people know what you're referring to.)

Since the area is decreasing, the change in area with respect to time "t" must be negative.

. . . . .dA/dt = -4

(Note: Since "r" is undefined and does not appear in any of your other formulas or expressions, I have no idea what "dr/dt = 4" is meant to refer to. Sorry.)

When A = 200, then:

. . . . .200 = (sqrt[3]/4)s²

. . . . .800/sqrt[3] = s²

. . . . .4sqrt[50]/sqrt[sqrt[3]] = s

Take the area formula ("A_triangle" above) and differentiate with respect to time "t". Plug in the known values for dA/dt, A, and s. Solve for ds/dt.

Hope this helps a bit.

Eliz.

 

[fgmango]

dA/dt=-sqrt3/2*s(ds/dt)

ds/dt=-(dA/dt)/(sqrt3/2*s)

ds/dt=-200/((sqrt3/2)(4sqrt50/sqrtsqrt3))


-200* 2 sqrtsqrt3
___ ____ ________
1 sqrt3 (4sqrt50)

=100sqrtsqrt3/10sqrt15

=10sqrtsqrt3/sqrt15