How Fast is the Area Changing in an Equilateral Triangle?

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To determine how fast the area of an equilateral triangle is changing as its height increases, the area formula A = (1/2)bh is used, with b being the base. For an equilateral triangle, the base can be expressed in terms of height using the Pythagorean theorem, leading to A as a function of height alone. By differentiating this area function with respect to time, the relationship between the rate of change of area (dA/dt) and the rate of change of height (dh/dt) can be established. Given that dh/dt is 3 cm/min when the height is 5 cm, the rate of change of the area can be calculated. This approach effectively links the geometric properties of the triangle to its dynamic changes in area.
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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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