Stuck on a Related Rates Triangle Problem

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The problem involves finding the rate of change of the area of an equilateral triangle as its height increases at 3 cm/min when the height is 5 cm. The area of the triangle is given by the formula A = (1/2)sh, where "s" is the side length. The relationship between the height and the side length is expressed as h = (1/2)√3s. To solve the problem, the chain rule is applied, leading to the equation dA/dt = (dh/dt) * (dA/dh). Clarification on the correct substitutions and differentiation steps is sought to progress with the solution.
scorpa
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Hi Again,

I am doing a question on related rates that I have become stuck on.

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

I know that the area of a triangle is bh/2, but after that I am stuck :redface: I tried deriving it using the chain rule so that I could substitute h and the rate of h, but I don't think that i was doing it the right way. If anyone could direct me here I would really appreciate the help.
 
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Here are some things to consider, the height "h" of an equilateral triangle is

\frac{1}{2}\sqrt{3}s

where "s" is the length of one side.

The area of this triangle is equal to

\frac{1}{2}sh

See any substitutions?
 
<br /> \frac{dA}{dt} = \frac{dh}{dt} * \frac{dA}{dh}<br />
 
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