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Question

The height h of an equilateral triangle is increasing at a rate of 3cm/min. How fast is the area changing when h is 5 cm?

Answer 1

Area of equilateral triangle is

 

A = (√(3)/4)a²

 

 

height of triangle is

 

h = √(a² - (a/2)²)

h = √(a² - (a²/4))

h = √((4a² - a²) / 4)

h = √(3a²) / 2

 

 

Then solve for a:

 

2h = √(3a²)

 

2h = a√3

 

a = 2h / √3

 

a = 2h√(3) / 3

 

 

dh/dt = 3

 

h = 5

 

dA/dt = (dA/dh) * (dh / dt)

          = (dA/dh) * 3

 

 

You need to find dA/dh at h=5.  Write the area of an equilateral triangle in terms of h.  Use the derivation of the value of "a" I have given you.   Then take the derivative.  Evaluate the derivative at h=5.  Then multiply by the rate, which is 3.