How Fast is the Area Changing in an Equilateral Triangle?

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SUMMARY

The discussion focuses on calculating the rate of change of the area of an equilateral triangle as its height increases. Given that the height (h) is increasing at a rate of 3 cm/min (dh/dt = 3 cm/min), the area (A) can be expressed as A = (1/2)bh. By using the relationship between the base (b) and height (h) derived from the properties of equilateral triangles, specifically through the Pythagorean theorem, the area can be rewritten solely in terms of height. Differentiating this area function with respect to time yields the rate of change of the area (dA/dt).

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I need help with the question: The heighth of an equilateral triangle is increasing at a rate of 3cm/min. How fast is the area changing when h is 5cm?
 
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The area of a triangle is A= 1/2*bh

You know dh/dt = 3cm/min


Try manipulating the above equations to an appropriate form where you can find the area as a function of height, and inhernetly its rate of change.
 
Write the equation, which whozum gave you, A= (1/2)bh, Use a relation between b and h that is true for all equilateral triangles (divide the triangle into two right triangles and use the Pythagorean theorem) to write A as a function of h only.
Finally, differentiate with respect to t to get a relation between dA/dt and dh/dt.
 

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