I need help finding the rate of change of height in this problem.

Question

The area of an equilateral triangle is increasping at 10cm²/min. Find the rate at which the height is changing when the area is 16√3.

 

 im super stuck. Please help :(

Answer 1

dh/dt = dA/dt x dh/dA           where: dA/dt = 10 cm²/min

                                                dh/dA = ?

                                                 L = length of triangle side (cm)

                                                 h = height of triangle = (L/2)*√3 cm

                                                 A = area of equilateral triangle = 16√3 cm²   (assuming cm² units in statement)

 

Need A as function of h:    

 

1)  A = (√3)/4*L²

 

2)  h = (L/2)*√3   (solve for h as function of L)

 

   L = (2/√3)*h = (2/3)*√3*h

 

Plugging 2) into 1):

 

A = (√3)/4*L² = (√3)/4*[(2/3)*√3*h]² = (√3)/4*[(4/9)*3*h²] = (√3/3)*h²

 

A = (√3/3)*h²

 

Now solve for h = f(A) since you are trying to solve for dh/dA

 

h²  =  3*A/√3 = √3*A = A*(3)^1/2

 

h = (√A)*(3)^1/4 = (A)^1/2*(3)^1/4

 

dh/dA = (3)^1/4*(1/2)*(A)^-1/2 = (3)^1/4*(1/2)/(16√3 cm²)^1/2 = (1/2)*(3)^1/4/[4*(3)^1/4 cm] = (1/8) cm^-1

 

 

dh/dt = dA/dt x dh/dA  =  (10 cm²/min

)*(1/8) cm^-1  = (5/4) cm/min   or  1.25 cm/min