Related Rates?

I will take a crack at it!!.

a) At the moment the sides are 15 inches, 1 second later the sides are 18 inches long. Therefore the ratio of: (18/15)^2 =1.2^2 =1.44 -1 x 100 = 44% increase in area 1 second after 15-inch sides.

b) The perimeter will increase by 3 inches 1 second after they are 15 inches, so that the ratio of the Perimeter =18/15 =1.2 - 1 x 100 = 20% increase 1 second after the sides are 15 inches.

c) This increase will continue to decline each and every second thereafter, as the ratios become:

21/18, 24/21, 27/24........etc.

Here is my 'guess'....

area =  $\sqrt{3}$ /4 x s^2     s = side length      the sides lenghts are function of time and = 3t   (s^2 = 9t^2)

A = sqrt(3) / 4 * 9 t^2 

dA/dt =  18 sqrt3/4  * t     at t= 5 (when sides are 15 inches)  

          = 18 sqrt(3) /4 * 5 = 38.97 in^2/sec

The perimeter.

...well there are three sides all increaseing at 3 in/ sec so perimeter is increasing at a rate 9 in/ sec 

(I think !)

I'll take a third shot at it:

(a)
A=(1/2)bh
A=(1/2)(3t)(3t)
A=(9/2)t^2

A'=(9/2)(2)t
A'=9t

The sides will be 15 inches long when A=(1/2)(15)(15)=112.5

When A=112.5 , t=5

Find A' when t=5
A'=45

45 inches per second

(b)
P=3s
P=3(3t)
P=9t

P'=9
The perimeter will always increase at a rate of 9 inches per second.

A =  (1/2)s^2sin(60)  =   [ √3/4] s^2

dA/dt  =  dA/ds * ds/dt

dA /ds =  [√3/2]s

ds/dt  =  3in/s  .....so.......

dA / dt  when s = 15  = [√3/2]*(15in)* 3in/s  = (45/2)√3 in^2 / s  ≈  38.97 in^2

If the sides are increasing at 3in/s....the perimeter increases at 9in/s