The sides of an equilateral triangle are increasing at the rate of 3 inches per second. At what rate are (a) the area and (b) the perimeter increasing at the moment the sides are 15 inches long?

huzzahforbiscuits
Feb 10, 2017

#1**0 **

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.

Guest Feb 10, 2017

#2**+5 **

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 !)

ElectricPavlov
Feb 10, 2017

#3**0 **

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.

hectictar
Feb 10, 2017