"The side of an equilateral triangle decreases at the rate of 2 cm/s. At what rate is the area decreasing when the area is 100 cm^2?"
Help!
let each side be x cm
given: dx/dt = -2 cm/s
find: dA/dt when A = 100
by the 30-60-90 triangle the height of the triangle will be (√3/2)x . You can also use Pythagoras to find that.
Area of triangle = (1/2)(x)(√3x/2)
= (√3/4)x^2
when A = 100
(√3/4)x^2 = 100
x^2 = 400/√3
x = 20/√(√3) = 15.1967
dA/dt = (√3/2)x dx/dt
= (√3/2)(15.1967)(-2)
= -26.32 cm^2/s
check my arithmetic
Thank you very much . This is the answer I got as well; however it is different than the answer given in the text which is 11.5 cm squared/sec. Thank you again.
Thank you!!
Thanks for this answer. I must have the same sheet, it also had 11.5 cm/s answer. I kept getting 26.32 cm/s and I couldn't figure out why...
I love you Reiny
How do you get the height of the triangle, I don't understand. can you explain it in more detail?
To find the rate at which the area of the equilateral triangle is decreasing, we need to first express the area as a function of the side length.
The area (A) of an equilateral triangle with side length (s) can be determined using the formula:
A = (√3/4) * s^2
Taking the derivative with respect to time (t) on both sides gives:
dA/dt = (√3/4) * 2s * ds/dt
Where dA/dt represents the rate at which the area is changing and ds/dt represents the rate at which the side length is changing.
Given that ds/dt = -2 cm/s (since the side length is decreasing), we can substitute this into the equation:
dA/dt = (√3/4) * 2s * (-2)
Now, we are given that the area is 100 cm^2 (A = 100). We can substitute this value into the equation and solve for dA/dt:
100 = (√3/4) * 2s * (-2)
To find s, we can isolate it by dividing both sides by (√3/4) * 2 * (-2):
s = 100 / ((√3/4) * 2 * (-2))
Simplifying further:
s = -100 / (√3)
Now, substitute this value of s back into the equation for dA/dt:
dA/dt = (√3/4) * 2 * (-100 / (√3)) * (-2)
Simplifying:
dA/dt = 400 / 4
dA/dt = 100 cm^2/s
Therefore, when the area is 100 cm^2, the area is decreasing at a rate of 100 cm^2/s.