Question

The area of an equilaterla triangle is increasing at a rate of 5 m^2/hr. find the rate at witch the height is chganging when the areai is:_________

Answer 1

The rate at which the height is changing is ( 5 / x ) m / hr

We know that,

Area of an equilateral triangle A = $\sqrt{3}$ $x^{2}$ / 4

h = $\sqrt{3}$ x / 2

Where,

x = Side

h = Height

Given that,

dA / dt = 5 $m^{2}$ / hr

h = $\sqrt{3}$ x / 2

Differentiate both sides with respect to t

dh / dt = ( $\sqrt{3}$  / 2 ) ( dx / dt )

dx / dt = ( 2 /  $\sqrt{3}$ ) ( dh / dt )

A = $\sqrt{3}$ $x^{2}$ / 4

Differentiate both sides with respect to t

dA / dt = ( $\sqrt{3}$ / 4 ) ( 2x ) ( dx / dt )

5 =  ( $\sqrt{3}$ / 4 ) ( 2x ) ( 2 /  $\sqrt{3}$ ) ( 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

To know more about Rate of change of height

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