Stuck on a Related Rates Triangle Problem

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SUMMARY

The discussion focuses on solving a related rates problem involving the area of an equilateral triangle as its height increases. The height (h) is increasing at a rate of 3 cm/min, and the area (A) is expressed as A = (1/2)sh, where s is the side length. The user seeks guidance on applying the chain rule to find the rate of change of the area when the height is 5 cm. Key formulas include the relationship between height and side length, h = (1/2)√3s, which is crucial for deriving the solution.

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  • Understanding of related rates in calculus
  • Familiarity with the chain rule for differentiation
  • Knowledge of the area formula for triangles
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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

[tex]\frac{1}{2}\sqrt{3}s[/tex]

where "s" is the length of one side.

The area of this triangle is equal to

[tex]\frac{1}{2}sh[/tex]

See any substitutions?
 
[tex] \frac{dA}{dt} = \frac{dh}{dt} * \frac{dA}{dh}[/tex]
 

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