Answer :
The rate at which the height is changing is ( 5 / x ) m / hr
We know that,
Area of an equilateral triangle A = [tex]\sqrt{3}[/tex] [tex]x^{2}[/tex] / 4
h = [tex]\sqrt{3}[/tex] x / 2
Where,
x = Side
h = Height
Given that,
dA / dt = 5 [tex]m^{2}[/tex] / hr
h = [tex]\sqrt{3}[/tex] x / 2
Differentiate both sides with respect to t
dh / dt = ( [tex]\sqrt{3}[/tex] / 2 ) ( dx / dt )
dx / dt = ( 2 / [tex]\sqrt{3}[/tex] ) ( dh / dt )
A = [tex]\sqrt{3}[/tex] [tex]x^{2}[/tex] / 4
Differentiate both sides with respect to t
dA / dt = ( [tex]\sqrt{3}[/tex] / 4 ) ( 2x ) ( dx / dt )
5 = ( [tex]\sqrt{3}[/tex] / 4 ) ( 2x ) ( 2 / [tex]\sqrt{3}[/tex] ) ( dh / dt )
dh / dt = ( 5 / x ) m / hr
Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt
Therefore, the rate at which the height is changing is ( 5 / x ) m / hr
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